.Twisted Extensions of Fermat’s Last TheorembyCarmen Anthony BruniA thesis submitted in partial fulfillment of the requirements forthe degree ofDoctor of PhilosophyinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)The University of British Columbia(Vancouver)April 2015c© Carmen Anthony Bruni, 2015AbstractLet x, y, z, p, n, α ∈ Z with α ≥ 1, p and n ≥ 5 primes. In 2011, Michael Bennett, FlorianLuca and Jamie Mulholland showed that the equation x3+y3 = pαzn has no pairwise coprimenonzero integer solutions provided p ≥ 5, n ≥ p2p and p /∈ S where S is the set of primes qfor which there exists an elliptic curve of conductor NE ∈ {18q, 36q, 72q} with at least onenontrivial rational 2-torsion point. In this thesis, I will present a solution that extends theresult to include a subset of the primes in S; those q ∈ S for which all curves with conductorNE ∈ {18q, 36q, 72q} with nontrivial rational 2-torsion have discriminants not of the form `2or −3m2 with `,m ∈ Z. Using a similar approach, I will classify certain integer solutions tothe equation x5 +y5 = pαzn which in part generalizes work done from Billerey and Dieulefaitin 2009. I will also discuss limitations of the methods for these equations and as they extendto further prime exponents.iiPrefaceThis dissertation is original, unpublished, independent work by the author, Carmen An-thony Bruni.iii..Table of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiiDedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 The History of the Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 A Modern Approach to Fermat’s Last Theorem . . . . . . . . . . . . . . . . 171.5 Where Do Frey-Hellegouarch Curves Come From? . . . . . . . . . . . . . . . 221.6 Ternary Fermat-Type Diophantine Equations . . . . . . . . . . . . . . . . . 261.7 Results From This Thesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Classification of Elliptic Curves With Nontrivial Rational Two Torsion 312.1 On the Q-Isomorphism Classes of Elliptic Curves With Nontrivial Rational2-Torsion and Conductor 2LqMpN . . . . . . . . . . . . . . . . . . . . . . . . 312.2 Elliptic Curves With Nontrivial Rational Two Torsion and Conductor 2aqbpc 353 Diophantine Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1313.1 S-Integer Solutions to y2 = x3 ± 2α5β . . . . . . . . . . . . . . . . . . . . . . 1313.2 Integer and Rational Solutions to y2 = x5 ± 2α5β . . . . . . . . . . . . . . . 1363.2.1 Chabauty’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.2.2 Elliptic Curve Chabauty . . . . . . . . . . . . . . . . . . . . . . . . . 1473.2.3 S-integer Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153iv3.2.4 Other Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1573.2.5 Integer Points Via Thue-Mahler Equations . . . . . . . . . . . . . . . 1573.3 Other Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1783.4 Diophantine Equations Relating to Elliptic Curves . . . . . . . . . . . . . . . 1804 Elliptic Curves With Rational Two Torsion and Conductor 18p, 36p or72p Organized by Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2135 Elliptic Curves With Rational Two Torsion and Conductor 50p, 200p or400p Organized by Primes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2356 On the Diophantine Equation x5 + y5 = pαzn . . . . . . . . . . . . . . . . . . 2547 Strengthening Results on the Diophantine Equation xq + yq = pαzn . . . . 263Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278Appendix A Final Collection of Tables . . . . . . . . . . . . . . . . . . . . . . . 288v..List of Tables2.1 Elliptic curves of conductor 2LqMpN when a4 > 0 . . . . . . . . . . . . . . . 342.2 Elliptic curves of conductor 2LqMpN when a4 < 0 . . . . . . . . . . . . . . . 353.1 Integer solutions to y2 = x3 + 2α5β with x-coordinate 2ApB, p 6= 2, 5 , B ≥ 1and A,α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1333.2 Integer solutions to y2 = x3 + 2α5β with x-coordinate 5ApB, p 6= 2, 5 , B ≥ 1and A,α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1333.3 Integer solutions to y2 = x3 + 2α5β with x-coordinate 2A5BpC , p 6= 2, 5, C ≥ 1and A,B, α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.4 Integer solutions to y2 = x3− 2α5β with x-coordinate pA, p 6= 2, 5 , A ≥ 1 andα, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1343.5 Integer solutions to y2 = x3 − 2α5β with x-coordinate 2ApB, p 6= 2, 5 , B ≥ 1and A,α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.6 Integer solutions to y2 = x3 − 2α5β with x-coordinate 5ApB, p 6= 2, 5 , B ≥ 1and A,α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1353.7 Integer solutions to y2 = x3− 2α5β with x-coordinate 2A5BpC , p 6= 2, 5, C ≥ 1and A,B, α, β ≥ 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1363.8 Rational solutions to y2 = x5 + 2α5β. . . . . . . . . . . . . . . . . . . . . . . 1373.9 Rational solutions to y2 = x5 − 2α5β. . . . . . . . . . . . . . . . . . . . . . . 1383.10 Rational solutions to y2 = x5 + 2α5β for rank 0 curves . . . . . . . . . . . . . 1423.11 Rational points on y2 = x5 + 2α5β for known rank 1 curves. . . . . . . . . . . 1463.12 Rational points on y2 = x5 − 2α5β for known rank 1 curves. . . . . . . . . . . 1473.13 Rational solutions to y2 = x5 ± 2α5β solved using Elliptic Curve Chabauty. . 1533.14 Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1613.15 Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1623.16 Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163vi3.17 Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1643.18 Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1663.19 Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1673.20 Integer solutions to x2 + 2α5β = yn with x, y ≥ 1, gcd(x, y) = 1 and n ≥ 3with α, β > 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1993.21 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + 2`5mpn = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2103.22 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2`pn = 5m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2103.23 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2`5m = pn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2113.24 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2` = 5mpn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2113.25 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equation5d2 − 2`pn = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2123.26 Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equation5d2 − δ2` = pn. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2127.1 Primes not in Pg,3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2677.2 Primes not in Pg,5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2687.3 Primes not in Pb,3 but in Pg,3. . . . . . . . . . . . . . . . . . . . . . . . . . . 2707.4 Primes not in Pb,5 but in Pg,5. . . . . . . . . . . . . . . . . . . . . . . . . . . 2717.5 Remaining primes for q = 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . 2727.6 Remaining primes for q = 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . 273A.1 Primes p of Pg,3 with p ≤ 1800 . . . . . . . . . . . . . . . . . . . . . . . . . . 288A.2 Primes p of Pb,3 with p ≤ 1800 . . . . . . . . . . . . . . . . . . . . . . . . . . 289A.3 Primes p of Pg,5 with p ≤ 320 . . . . . . . . . . . . . . . . . . . . . . . . . . 289A.4 Primes p of Pb,5 with p ≤ 320 . . . . . . . . . . . . . . . . . . . . . . . . . . 289viiAcknowledgementsMy sincerest and utmost thanks to my supervisor Professor Michael Bennett for all of hispatience and dedication to me as a Ph.D. student. I cannot express in words how much ithas meant to have someone who has believed in me as much as you have over the years. Yourguidance and help has made this project possible. I am very thankful to have a supervisorwith your vision and expertise in the field.To my supervisory committee Professors Greg Martin and Vinayak Vatsal, thank youso much for always being available for advice and for teaching. It has been great having acommittee with such a kind and generous open door policy. Your guidance has always beenappreciated over the years!I would also like to thank many professors I have had discussions with including ProfessorsNils Bruin, Samir Siksek and Michael Stoll. A thank you as well to the many people whoreply to MathOverflow posts.I would also like to thank the faculty and staff at the University of British Columbia,especially all the support staff in Math 121. Over the years I have spoken to just abouteveryone in that room and their knowledge of the hidden working of a university is so in-valuable. The University of British Columbia is very blessed to have such a great room ofpeople.A big thank you goes out to all of my friends and family back at home who have supportedme over the years. I would like to especially thank my parents for supporting me over theselast 10 years both emotionally and mentally with the highs and lows that come with agraduate degree. Without such a loving and supportive family cast I’m not sure how thiswould have been possible.To my fiance´e Jessica. I cannot put into words how much your love and support hasmeant to me over these last 5 years of my degree. Thank you so much for always beingthere to help me. Thank you for picking me up when I was down and reminding me to stayhealthy when I was too busy to remember. Thanks for helping me to explore Vancouver andviiireminding me that there is so much more to life than mathematics. Thank you for beingthere in the past and thank you in advance for being there in the future.To my friends from Waterloo, especially Vince and Vicki Chan and Faisal al-Faisal. Overthe last 10 years you have been my go to people for help and advice for all things mathrelated. Thank you for being a part of my life. This project could not have been successfulwithout such a great group of core friends.To my friends here at the University of British Columbia, thank you for everything. Icould state the names of about half the math graduate students by now over the last 5 yearshere but let me just keep it short and say that you know who you are and thank you somuch for helping me to keep my sanity intact. A big thank you especially to my office matesin Math 201 and to the MER wiki team for always being there when the stress of graduateschool was too much for me to handle alone.Finally I would like to acknowledge that this project was funded in part by both a CGS-Mscholarship and a Four Year Fellowship from the University of British Columbia for which Iam thankful for.ixDedicationPer mia nonna Giuseppina Mattina e mia zia Carmela Mattina. Il vostro amore e il durolavoro ha reso possibile questa tesi. Che tu possa riposare in pace con il comfort di Dio.xChapter 1Introduction1.1 The History of the ProblemA Diophantine equation is a polynomial defined over Z[x1, .., xn] where we seek out integer(or rational) n-tuples satisfying the polynomial. These equations are named after Diophantusof Alexandria who lived during the third century A.D. His book Arithmetica contains suchequations which he solved for using positive rational numbers. For a more thorough view ofthe history of Diophantus, see [Sch98]1.Throughout the last 2600 years of number theory, a great deal of mathematics has beencreated to solve Diophantine equations. Some of the earliest attempts at solving Diophantineequations date back to Pythagoreas. He was particularly interested in the equation x2 +y2 =z2. This equation is the Pythagorean theorem for right angled triangles, which geometricallysays that sum of the squares of the perpendicular sides of a triangle equals the square of itshypotenuse. A question one can ask is “Which right angled triangles have all integer sidelengths?” With a bit of effort, we can come up with a few simple examples, the most wellknown being (3, 4, 5) and (5, 12, 13) and then notice that integer multiples of these will alsowork such as (6, 12, 15) or (2500, 6000, 6500). In light of this, we call a solution primitiveif there is no common multiple between the three variables. Simplifying our question, wenow ask “Can we find all such primitive solutions?” This question at first might seem illposed - what happens if there are infinitely many? One cannot list all solutions if thereare infinitely many. In this case, what we can hope to find is a parameterized family ofsolutions, that is, equations for the variables that depend on parameters in such a way thatif we replace the parameters with integers, we get solutions to our original equation and thatthese solutions form a comprehensive list. It is said that Pythagoreas was the first to attempt1An English translation can be found on the author’s personal webpage at 〈http://www-irma.u-strasbg.fr/~schappa/NSch/Publications_files/Dioph.pdf〉.1such a parameterization and came up with the following [Ito¯87](x, y, z) = (2n+ 1, 2n2 + 2n, 2n2 + 2n+ 1)for a (positive) integer n. This parameterization gives an infinite family of solutions, but nota complete list. The full parameterization is possible and is given by(x, y, z) = (s2 − t2, 2st, s2 + t2)where s and t are integers. This was first given in Euclid Elements [Euc02, Book 10 Propo-sition 29].From here, it seems like a simple switch would be to turn all the squares into cubes, thatis to examine the equation x3 + y3 = z3. Thinking about this geometrically, we are askingcan we write the volume of a cube as the sum of two volumes of cubes. There are some trivialsuch examples of this problem, namely when one of the cubes has 0 length so we further saythat a solution to a Diophantine equation is trivial if xyz = 0, that is, if at least one of thecomponents is zero. Now as before, we ask “Can we classify all non-trivial primitive solutionsto this equation?” This question can be resolved in the affirmative. It is known that no suchsolutions exist, the first proof of which was given formally by Leonhard Euler in 1770. Heused the method of infinite descent, the same method that Pierre de Fermat used to solvethe case of x4 + y4 = z4 in the 1600s [Bal60, p.242]. Fermat seldom formally published anyof his proofs however stories say that whenever asked for a proof, he could always produceone [Bal60, p.242].This brings us to our title character, Pierre de Fermat. In the margin of his copy ofDiophantus’ Arithmetica2, he wrote down, after translating to modern language, that theequation xn + yn = zn had no non-trivial primitive solutions and claimed he had a remark-able proof of this fact for which the margin was too narrow to contain. Since that time,mathematicians have sought out a solution to this problem. This quest as we will see tookmany years before being completed.Years later, in 1847, Gabriel Lame´ announced a proof of Fermat’s Last Theorem usingthe idea of cyclotomic numbers [Dev90, p.278]. Let p ≥ 5 be a prime number and let ζp be aprimitive pth root of unity, that is, a strictly complex solution to xp − 1. Lame´ argues that2The term “Diophantine equation” originates from the name Diophantus.2if (x, y, z) is a solution to Fermat’s Last Theorem with exponent p a prime, thenzp = xp + yp =p−1∏i=0(x+ ζ ipy)and then one argues that each (x + ζ ipy) is a pth power times a unit. Argue similarly forxp + (−z)p and then use these two pieces of information to derive a contradiction (see [BS66,Chapter 3] for the full and complete argument). The problem with this proof is that threeyears earlier, Ernst Kummer showed that the ring Z[ζ23] fails to be a unique factorizationdomain. In one sentence, the elements of this ring do not behave like integers in the sense thatelements do not necessarily decompose into a product of prime elements uniquely, contraryto how the Fundamental Theorem of Arithmetic works for the integers. This failure preventseach factor in the product above from necessarily being a pth power as Lame´ assumes.Kummer in 1847 found a way around this. He introduced the idea of using ideals to attemptto solve this problem. Over prime ideals, unique factorization does occur as we know it andmuch of the theory of integers carries through. The concept of the class number, the size ofthe group of fractional ideals modulo the principal fractional ideals, has its roots here. Heshows that if the prime p does not divide the order of this finite group, then the Fermatequation for exponent p has only trivial solutions. These primes are called regular primesand Kummer showed that the only primes less than 100 that are not regular are 37, 59 and67.Seeing just how close Kummer was to a proof, it seemed that a solution would soonsurface. Sadly this is not the case. The next big step towards solving Fermat’s Last Theoremis due to Louis Mordell. In 1922, Mordell noticed that when dealing with a Diophantineequation, there were no known examples of equations of genus (an algebraic invariant of theequation) bigger than or equal to 2 where the number of rational points was infinite [Dev90,p.286]. He conjectured that if one considers an equation with genus strictly bigger than 1,then the equation in question must only have finitely many primitive integer solutions. Itwas not until 1983 when Gerd Faltings proved that indeed this is true [Fal83]. For his work,he was awarded the Fields Medal in 1986.Now Plu¨cker’s formula says that for n ≥ 4, the genus of the curve xn + yn = zn is atleast 2. Hence, we know that each such curve has only finitely many primitive integer points.Unfortunately, Faltings’ theorem is not effective meaning that while you know that thereare only finitely many solutions, the theorem does not give you an algorithm on how to findthem. The statement that we wish to prove is that the equation xn+yn = zn has only trivialprimitive solutions and up to this point in time, we are close, but still far away.3This brings us to the work of Andrew Wiles in 1993-1994. Our brief detour in history endsnow and I will begin to outline the proof that Wiles used to prove Fermat’s Last Theorem.This will involve techniques using elliptic curves and modular forms so we begin first bygiving a review of relevant topics.1.2 Elliptic CurvesThe information in this chapter is a summary of some of the notation in this thesis andis not meant to be an inclusive rehashing of elliptic curves. For a few excellent references,see [Kna92], [Sil09], [Was08].An elliptic curve E defined over a field K is a nonsingular curve of the formE : y2z + a1xyz + a3yz2 = x3 + a2x2z + a4xz2 + a6z3with each ai ∈ K. This is known as the Weierstrass form of an elliptic curve. In this form,we recall that there is a point at infinity given in projective coordinates by [0 : 1 : 0] ∈ E andthis is the only point that lies on the curve with z = 0. For this reason, we typically workwith the elliptic curve in the formE : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6recalling that we have an extra point given above which we call the point at infinity anddenote it by 0E or 0 if the context is clear. For such forms we have the following invariantsb2 = a21 + 4a2b4 = 2a4 + a1a3b6 = a23 + 4a6b8 = a21a6 + 4a2a6 − a1a3a4 + a2a23 − a24 =b2b6 − b244c4 = b22 − 24b4c6 = −b32 + 36b2b4 − 216b6.4We also define the discriminant of an elliptic curve as follows∆ =c34 − c261728= −b22b8 − 8b34 − 27b26 + 9b2b4b6= −a41a2a23 + a51a3a4 − a61a6 − 8a21a22a23 + a31a33 + 8a31a2a3a4 + a41a24 − 12a41a2a6 − 16a32a23+ 36a1a2a33 + 16a1a22a3a4 − 30a21a23a4 + 8a21a2a24 − 48a21a22a6 + 36a31a3a6 − 27a43+ 72a2a23a4 + 16a22a24 − 96a1a3a24 − 64a32a6 + 144a1a2a3a6 + 72a21a4a6 − 64a34 − 216a23a6+ 288a2a4a6 − 432a26.Throughout this thesis, we will be interested in rational elliptic curves with rational twotorsion. Translating so that this torsion point occurs at (x, y) = (0, 0) shifts the aboveequation to one of the formE : y2 = x3 + a2x2 + a4xand in this case, the invariants becomeb2 = 4a2b4 = 2a4b6 = 0b8 = −a24c4 = 16a22 − 48a4c6 = −64a32 + 288a2a4∆ = 16a22a24 − 64a34 = 16a24(a22 − 4a4).Definition 1.2.1. Two elliptic curves E1 and E2 are said to be isomorphic over K if thereis an admissible change of variables between them given byφ : E1 → E2(x, y) 7→ (u2x′ + r, u3y′ + su2x′ + t)where u, r, s, t ∈ K and u 6= 0. More generally, there are rational function in both directionsdefined for all points such that their composition in both directions gives the identity function.In the case where K = Q, we can perform an admissible change of variables so that thecoefficients are integers. We assume this throughout whenever we define an elliptic over Qwe have coefficients in Z. Performing the above admissible change of variables also changesthe discriminant by at most a twelfth power. There is a model of our equation such that thediscriminant is smallest possible and we shall call such a model a minimal model and such a5discriminant a minimal discriminant denoted by ∆min.One thing we will also want to do is change the field of definition from Q to a finite fieldFp where p is a prime. In doing so, we define the following types of reduction.Definition 1.2.2. We say that an elliptic curve has1. Good reduction at p if p - ∆min.2. Bad multiplicative reduction at p if p | ∆min and p - c4 (the equation of the reducedcurve has a node - a double root).3. Bad additive reduction at p if p | ∆min and p | c4 (the equation of the reduced curvehas a cusp - a triple root).On an elliptic curve, we have a group law. Any two points on an elliptic curve give riseto a third. Take the line joining the two points (or the tangent line if the two points are thesame). This line intersects the curve at a third point (or at the point at infinity).Definition 1.2.3. An isogeny over a field K between two elliptic curves defined over K is amorphism defined over K (in the sense of algebraic varieties; briefly, a rational map that isdefined everywhere) that preserves the base point (the point at infinity). Two elliptic curvesare said to be isogenous if there is a nonzero isogeny between them.Isogenies are group homomorphisms between two elliptic curves (see [Sil09, p.71 Theorem4.8]). Each isogeny induces a map called the pullbackφ∗ : K(E2)→ K(E1)F 7→ F ◦ φwhere K(Ei) is the field of rational functions F (x, y) = f(x, y)/g(x, y) such that f and ghave the same homogeneous degree, g(x, y) 6∈ I(Ei) and two functions f1/g1 and f2/g2 areidentified if f1g2 − f2g1 ∈ I(Ei) and I(Ei) is given byI(Ei) = {F ∈ K[x, y] : F (x, y) = 0 for all (x, y) ∈ Ei}.Definition 1.2.4. The degree of an isogeny isdeg φ =0 if F = 0[K(E1) : φ∗K(E2)] otherwise.A p-isogeny is a degree p isogeny.6When the field extension K(E1)/φ∗K(E2) is separable, we have that deg φ = # kerφ (areminder that this holds when charK = 0). To be more concrete, if φ : E1 → E2 is an isogenydefined over a field K and E1, E2 are elliptic curves defined over K, as outlined in [Was08,p.387], we can define rational functions f1 and f2 defined over K[x] such that(x2, y2) = φ((x1, y1)) = (f1(x1), y1f2(x1))holds for all but finitely many points (x1, y1) ∈ E1 We proceed to write f1(x) =p(x)q(x) and thenwe could have defined the degree of φ asdeg(φ) = max{deg(p(x)), deg(q(x))}.The first definition has the benefit of being more general but the downfall of being far moreabstract and often harder to work with in practice. To see the connection between definitions,let pi : E → E/{±1} ∼= P1 be the map given by pi((x, y)) = x [Sil07, Chapter 6]. Considerthe following commutative diagramEφ> EP1pi∨f1> P1pi∨Now the bottom map in the diagram is the function f1 with the exception of the point atinfinity. Now, as degrees are multiplicative, we havedeg(pi) deg(φ) = deg(pi ◦ φ) = deg(f1 ◦ pi) = deg(f1) deg(pi)and so deg(f1) = deg(φ) and deg(f1) is the degree in terms of algebraic geometry (just as wedefined it above for elliptic curves). Analyzing the bottom map, it is now easy to see thatthe degree in the abstract sense is the same as the second definition above.There is one more invariant we will need and that’s the conductor of an elliptic curve. Thisvalue measures arithmetic information about our elliptic curve in ways similar to how thediscriminant measures arithmetic information. This is an isogeny invariant. For simplicity,I will give the definition only over Q and direct the reader to [Sil94] for a more generaltreatment.Definition 1.2.5. The conductor of an elliptic curve E/Q is the integerN =∏ppfp7where the product is over all primes andfp =0 if E has good reduction at p1 if E has multiplicative reduction at p2 + δp if E has additive reduction at p.The value δp is sometimes called the wild part of the conductor.The exact definition of δp is a bit clunky and takes us astray from our real goal which is tocompute the conductor. I will however mention a formula due to Ogg and in the characteristic2 case by Saito which helps compute this value. Again since this is not entirely relevant tothe overall discussion I will only state the theorem.Theorem 1.2.6 (Ogg-Saito Formula). [Ogg67][Sai88] Let E/Q be an elliptic curve. Thenthe exponent fp of the prime p in the conductor is given byfp = vp(∆min)− cp + 1where cp is the number of components (without counting multiplicities) of the singular fibreof the Ne´ron minimal model for E at p.Again to go into details on the latter part will take us too far astray though I directthe interested reader for details on Ne´ron models or conductors to either [Mil06] or [Sil94].For now, I note that fp ≤ vp(∆min) is a consequence of the above theorem so the conductordivides the minimal discriminant. It should be noted that δp = 0 whenever p ≥ 5 as isgiven by the following upper bound due to a result of Lockhart-Rosen-Silverman [LS93] andBrumer-Kramer [BK94].Theorem 1.2.7. [Sil94, p.385] Let K/Qp be a local field with normalized valuation vK (sothat vK(p) is the ramification index of K/QP ) and let E/K be an elliptic curve. Then theexponent of the conductor of E/K is bounded byf ≤ 2 + 3vK(3) + 6vK(2)and this bound is best possible, that is, some elliptic curve E/K attains the aforementionedbound.Despite all these technicalities, the actual value of the conductor can be computed verysimply using an algorithm due to Tate [Tat75]. In fact, since then the work has been re-duced even further to simply checking congruence conditions of the coefficients in most cases8[Pap93]3 or the simplified formulas when your curve has rational two torsion [Mul06]. Fora given fixed elliptic curve, the value can be computed either in MAGMA [BCP97] or Sage[S+14].Summarizing the above, we see that for an elliptic curve over the rationals we have forp ≥ 5 and E/Q an elliptic curve with conductor N =∏p pfpfp =0 if E has good reduction at p1 if E has multiplicative reduction at p2 if E has bad reduction at p.Further, we must have that v2(N) ≤ 8, v3(N) ≤ 5. We say that the curve is semi-stable if theconductor is square free, that is, our elliptic curve has only good or multiplicative reduction.Ideally we would like to apply Tate’s algorithm to general elliptic curves where the coef-ficients depend on solutions to Diophantine equations. However computer algebra softwarecurrently cannot perform this task abstractly. One way to compute these values is throughthe following lemma.Lemma 1.2.8. [BCDY14] Suppose E and E ′ are elliptic curves given by the equationsE : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6E ′ : y2 + a′1xy + a′3y = x3 + a′2x2 + a′4x+ a′6where ai and a′i lie in a discrete valuation ring O with valuation v and uniformizer pi. Supposethatmax{v(∆E), v(∆E′)} ≤ 12kfor some positive integer k. Suppose further that v(ai − a′i) ≥ ik for i ∈ {1, 2, 3, 4, 6}. Then1. If the reduction type of E ′ is not I∗m′ for m′ > 2, then the reduction type of E is thesame as the reduction type of E ′. In this case v(NE) = v(NE′).2. If the reduction type of E ′ is I∗m′ for m′ > 2, then the reduction type of E is I∗m forsome m > 2.3. In particular, E has good reduction if and only if E ′ has good reduction.This gives us a way to compute the conductor of abstract elliptic curves by looking atall possible residue classes of the parameters modulo prime powers. Since we will need this3Errata in [Pap93]: In the column labeled Equation non minimale of table IV, the first column shouldread [4, 6,≥ 12] not [4, 6, 12].9notation for our next example and throughout the paper, we define now for any integers Mand Q, the radical of M excluding Q to beradQ(M) :=∏p -Qp |Mpwhere the product runs over primes p. Succinctly, these are the primes dividing M that donot share a prime divisor with Q.Example 1.2.9. Let ap+bp = cp for some prime p ≥ 5 and some nontrivial primitive solution(a, b, c). We assume that 2 | b and that a ≡ −1 (mod 4). Let y2 = x3 + (bp − ap)x2 − apbpxbe an elliptic curve associated to this solution. We wish to compute the conductor N of thiselliptic curve. The discriminant and c4 invariant of this elliptic curve are given by∆ = 24(abc)2p, c4 = 16(c2p − apbp).In fact, using [Kna92, p.291], we can show that the above model is guaranteed to be minimalat all primes except possibly at 2. To show this, notice that the discriminant shares primeswith rad2(abc). Thus, we are only concerned about the minimality with these primes. Foreach of these primes, notice that these primes cannot divide c4 since (a, b, c) was a primitivesolution. Thus, the equation is minimal at all primes outside of 2. It turns out that the equa-tion is not minimal at 2 and the actual minimal discriminant is given by ∆min = 2−8(abc)2pbut we will not use this fact here.For the odd primes, we see that the above equation has bad reduction at the primesdividing rad2(abc). As these primes do not divide c4, we see from [Sil09, p.45] that theseprimes have bad multiplicative reduction. Thus, we have that rad2(abc) | N . For the prime 2,we use the tables in [Mul06] since our curve has rational two torsion to see that the conductoris N = 2rad2(abc) and in fact a minimal model at 2 is given byy2 + xy = x3 +(bp − ap − 14)x2 −(apbp4)x.We note that the tables there would also give the conductor in all cases.The last major piece of discussion for elliptic curves is the trace of Frobenius. Let E/Qbe an elliptic curve and ` a prime of good reduction. The valuea`(E) = `+ 1− |E(F`)|is called the trace of Frobenius for the elliptic curve E. Traces, from a linear algebra view-point, correspond to the sum of entries on the diagonal of a matrix. In the form written10above, we see no direct relationship to linear algebra. In its simplest form, one can see thatthis is the trace of a matrix as follows. Let q = pn be a prime for this section and define theFrobenius map φq ∈ GFq := Gal(Fq/Fq) asφq((x, y)) = (xq, yq).An important fact is that this map has degree q [Sil09, p.25]. We will be most interested inthe case when q = p and so we drop the q = pn notation.Now, let ` be an odd prime for simplicity and suppose that p - `NE. Then p is a primeof good reduction and so we can consider E defined over Fp. Now, φp acts on the ` torsiongroup, denoted by E[`] = {P ∈ E(Fp) : [`]P = 0} and [`]P = P + P + ... + P a total of `times. If we pick a basis for E[`] ∼= Z/`Z × Z/`Z [Sil09, p.86], say P and Q, then we haveφp(P ) = aP + bQ and φp(Q) = cP + dQ. Then we have that(φp)` =[a bc d]and it is here that we see that Trace((φp)`) ≡ ap (mod `) and det((φp)`) ≡ p (mod `) [Was08,p.102]. In fact the above holds with p replaced by arbitrary prime powers q and ` replacedby any integer m with gcd(m, q) = 1. The above is enough for us to get what we want,however I would like to delve a bit deeper into Galois representations in order to bring tosurface some of the underlying ideas.Let E/Q be an elliptic curve. We define the representation attached to an elliptic curveas follows. First let Ta`(E) = lim←−E[`n] denote the `-adic Tate module where the projectionmaps are multiplication by `. Now, notice that any σ inside the absolute Galois groupGQ := Gal(Q/Q) acts on any E[`n] and in fact takes `n-torsion points to `n-torsion points.Hence, this σ induces an element of Aut(Ta`(E)). By picking a compatible basis for eachE[`n], that is a collection {(Pn, Qn)}n∈Z+ such that `Pn+1 = Pn and `Qn+1 = Qn, we canget an isomorphism from Ta`(E) to two copies of the `-adic integers denoted by Z2` . Thisnaturally sits inside Q2` and so combining all this gives a representation:GQ → Aut(Ta`(E)) ∼= Aut(Z2`) ↪→ Aut(Q2`) ∼= GL2(Q`).This map ρE,` : GQ → GL2(Q`) is called the representation associated to the elliptic curveE. We will also use the same symbol to denote the action from GQ to Aut(Ta`(E)). Next,we need the following definition.11Definition 1.2.10. We say ρE,` is unramified at p ifIp := {σ ∈ Dp : σ(x) ≡ x (mod p)∀x ∈ Z¯} ⊆ ker(ρ)for any maximal ideal p ∈ Z¯ over p where Dp = {σ ∈ GQ : σ(p) = p} is the decompositiongroup and Ip is called the inertia group.The decomposition group above is of particular importance to us. As the decompositiongroup sits inside GQ, it has a natural action on Aut(E[`n]) and hence on Aut(Ta`(E)). Letp - `NE be another prime where NE is the conductor of E. Then we also have an isomorphismfrom E[`n] to E¯[`n] where E¯ is the reduction of E at the good prime of reduction p as both areisomorphic to (Z/`nZ)2. This also induces an isomorphism from Aut(Ta`(E)) to Aut(Ta`(E¯)).Each σ ∈ Dp can act on each number ring Z/p via σ(α + p) = σ(α) + p since σ fixes p.As Z/p and Fp are both algebraic closures of Z/pZ ∼= Fp, they are isomorphic4. Thus, wecan map ψ : Dp → GFp and this map is a surjection [DS05, p.377]. Let φp be the Frobeniuselement of GFp which is a generator of this group. Then any preimage of φp in Dp is denotedby Frobp and is called an absolute Frobenius element over p. Notice that this element is onlywell defined up to the kernel of ψ which is equal to Ip. Thus, whenever the representationρE,` is unramified , we see that ρE,`(Frobp) is a value that is equal for any preimage choice ofφp. Similarly to the above, the group GFp also acts on E¯[`n] and hence on Aut(Ta`(E¯)) by amap we shall denote via ρ¯E¯,`. Combining all these actions gives the following commutativediagramDp > E[`n]GFpψ∨> E¯[`n]∼=∨which induces the following commutative diagramDpρE,`> Aut(Ta`(E))GFpψ∨ ρ¯E¯,`> Aut(Ta`(E¯))∼=∨From the diagrams, since the right most map is an isomorphism, we must have thatker(ψ) = Ip ⊆ ker(ρE,`) and so the representation is unramified for all primes p - `NE. As4Alternatively let p be the kernel of the reduction map from Z to Fp.12the map is commutative, we see that tr(ρE,`(Frobp)) = tr(ρ¯E,`(φp)). To evaluate the latter,we use the Weil pairing. Let (P,Q) be a basis for Aut(Ta`(E¯)). The isogeny theorem [Sil09,p.91] tells us that this ρ¯E,`(φp) =: φp,` must come from an isogeny on E¯ which we also denoteby φp. This map is given by φp(X, Y ) = (Xp, Y p) for points in E¯. Next we use the Weilpairing [Sil09, p.92-99] on E[`n] to see thate`n(P,Q)deg(φp) = e`n([deg(φp)]P,Q) by bilinearity of e`= e`n(φˆp,`φp,`P,Q) by definition of dual= e`n(φp,`P, φp,`Q) by property of Weil Pairing= e`n(aP + bQ, cP + dQ) provided φp,` =[a bc d]on the basis (P,Q)= e`n(P,Q)ad−bc since Weil pairing is bilinear and alternating= e`n(P,Q)det(φp,`) by definition from above.This means that deg(φp) ≡ det(φp,`) (mod `n) for any positive integer n. Thus in particular,we have that deg(φp) = det(φp,`) and in particular that det(φp,`) is an integer. One can alsoshow that this φp acting on E¯ has degree p and so det(φp,`) = p. Now, it is immediate thatfor any two by two matrix M =[a bc d], we havetr(M) = a+ d = 1 + (ad− bc)− ((1− a)(1− d)− bc) = 1 + det(M)− det(I −M).Thus, we havetr(ρE,`(Frobp)) = tr(ρ¯E,`(φp))= 1 + det(ρ¯E,`(φp))− det(I − ρ¯E,`(φp))= 1 + p−#E(Fp)with the last equality following from using the same argument above to see thatdeg([1]− φp) = det(I − ρ¯E,`(φp))and as the map [1]− φp is separable by [DS05, p.320], we get thatdeg([1]− φp) = degsep([1]− φp) = ker([1]− φp) = #E(Fp).Though the proof of the exact equality above is a bit scant, it is clear at the very leastthat defining the trace of Frobenius in this manner does indeed give an integer despite apriori that it is only clear that you get an element in Z` and not necessarily in Z.13A final note is that isogenous elliptic curves do in fact have equal trace of Frobeniusvalues [Kna92, p.366], a statement which can be stated as isogenous elliptic curves haveequal L-series.1.3 Modular FormsThe other key objects in the Modularity Theorem, the theorem for which these attackson Diophantine equations is based, are modular forms. These are functions on the complexupper half plane H and so are, by nature, analytic objects. In the previous section, we sawa treatment of elliptic curves that was very algebraic in nature. This joining of analysis andalgebra is one of the components that makes the Modularity Theorem such an incredible featof mathematics. We begin by describing a modular form.Definition 1.3.1. Let SL2(Z) ={M :=[a bc d]∈M2(Z) : det(M) = 1}. Then a congru-ence subgroup of SL2(Z) is a subgroup Γ ∈ SL2(Z) such that Γ(N) ⊆ Γ whereΓ(N) ={M :=[a bc d]∈ SL2(Z) : M ∼=[1 00 1](mod N)}.Definition 1.3.2. Let Γ be a congruence subgroup of SL2(Z) and let k be an integer. Aholomorphic function f : H→ C is a modular form of weight k with respect to Γ if1. f is weight-k invariant under Γ, meaning f [γ]k = f for all γ ∈ Γ wheref [γ]k(τ) = j(γ, τ)−kf(γ(τ))and j([a bc d], τ):= cτ + d.2. f [α]k is holomorphic at ∞ for all α ∈ SL2(Z) (that is, f is holomorphic at all cusps,points in Q ∪ {∞}).The first condition above gives modular forms a periodicity property as mentioned below.The second condition above actually gives us other important arithmetic information givenby Laurent series. As a congruence subgroup Γ ⊃ Γ(N) contains[1 N0 1], there must be someminimal positive integer h such that[1 h0 1]∈ Γ. For this h, we have that f(τ + h) = f(τ)by the weight-k invariance. The theory of Fourier series thus applies here and we see that14our f has a Fourier series expansionf(τ) =∞∑n=0an(f)qnh qh = e2piiτ/hThe notation here should be reminiscent of the trace of Frobenius for elliptic curves. This isno accident and we shall explore this in the next section. If in addition to the above definitionof a modular form, we have that a0(f [α]k) = 0 for all α ∈ SL2(Z), then say that our modularform f is a cusp form. This is equivalent to f being holomorphic at ∞ and the Fourierseries expansion of f satisfies |an(f)| ≤ Cnr for some constants C and r [DS05, p.200]. Ourparticular interest will be with weight 2 modular forms over the congruence subgroupΓ1(N) ={M :=[a bc d]∈ SL2(Z) : M ∼=[1 ∗0 1](mod N)}where the ∗ denotes an arbitrary element. For future reference, another often used congruencesubgroup isΓ(N) ⊆ Γ1(N) ⊆ Γ0(N) ={M :=[a bc d]∈ SL2(Z) : M ∼=[∗ ∗0 ∗](mod N)}.To define a newform, we will first need to describe the space of oldforms. An oldformcolloquially is a form that comes from a lower level than the one you are currently considering.For example, if M | N is a proper divisor (that is, M 6= N) then Sk(Γ1(M)) ⊆ Sk(Γ1(N))and so the modular forms in Sk(Γ1(N)) coming from this subset are old in the sense that ifwe were listing all modular forms by the size of the level, we already knew they existed.More precisely, there are two main ways to embed the lower subspace into the biggerone. We have described one way above. For the second way, suppose that d | (N/M) is anydivisor. Then takingf [αd]k(τ) := det(αd)k−1f(αdτ)j(αd, τ)−k = dk−1f(dτ)where αd =[d 00 1], we see that this embeds a modular form f ∈ Sk(Γ1(M)) into Sk(Γ1(Md))which is a subset of Sk(Γ1(N)). This prompts the following definition for the map id, whichcombines the above methodsid : Sk(Γ1(Nd−1))× Sk(Γ1(Nd−1))→ Sk(Γ1(N))(f, g) 7→ f + g[αd]k.15Thus, we define the space of oldforms to be the following sum of vector spacesSk(Γ1(N))old :=∑p|Nip(Sk(Γ1(Np−1))× Sk(Γ1(Np−1))).To define the space of newforms, we need to figure out a complementary space to the spaceof oldforms. This can be done by using the Petersson inner product. For any congruencesubgroup Γ of SL2(Z) and cusp forms f, g ∈ Sk(Γ), we define an inner product by〈f, g〉 :=1VΓ∫X(Γ)f(τ)g(τ)=(τ)k dµ(τ)where dµ(τ) = dx dyy2 is the hyperbolic measure and VΓ =∫X(Γ) dµ(τ). This measure isGL+2 (Q) invariant and the integrand above is continuous, bounded and Γ-invariant. Thespace of newforms is then defined to be the orthogonal complement of Sk(Γ1(N))old insideSk(Γ1(N)). SymbolicallySk(Γ1(N))new := (Sk(Γ1(N))old)⊥.By a newform, we actually mean an element of the space of newforms that is normalizedso that the first Fourier coefficient satisfies a1(f) = 1 and that it is an eigenvector for all theHecke operators Tn and 〈d〉 (such an element is also called an eigenform). To be brief, theHecke operators for f ∈ Sk(Γ1(N)) can be defined by〈d〉f := f[[a bc δ]]kwhere δ ≡ d (mod N) for d ∈ Z,[a bc δ]∈ Γ0(N) andTpf :=p−1∑j=0f1 j0 pkif p | Np−1∑j=0f1 j0 pk+ fm nN pp 00 1kif p - Nwhere mp−nN = 1. For the second Hecke operator, we can define Tn to be the multiplicativeextension of Tp under the additional rule that Tpr := TpTpr−1 − pk−1〈p〉Tpr−2 .We have introduced newforms in full generality above however we will be mainly interestedin weight two newforms for Γ0(N), that is for the space S2(Γ0(N))new. This is due to theirconnection with elliptic curves as we shall soon see. So from here on out, a newform will16refer to a normalized eigenform of S2(Γ0(N))new. The primary properties of a newform thatwe will need is that a newform is a q-expansion f := q +∑n≥2 an(f)qn, that is, a cusp formnormalized so that the first coefficient is 1, with the additional property that adjoining all thecoefficients to Q via Kf = Q(a2(f), a3(f), ...) forms a number field that is a finite and totallyreal extension of Q [DS05, p.234] and in fact, each of the ai(f) are algebraic integers. Wecall the ai(f) the Fourier coefficients of f . These coefficients satisfy |a`(f)| ≤ 2√` for prime`. This was Ramanujan’s conjecture and was proven by Deligne as a consequence of the WeilConjectures [Del74] [Del80]. Another important fact is that there are no newforms of level1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18, 22, 25, 28, 60 which can easily be checked in MAGMA(see [HK98] or [Mar05] for a formula).1.4 A Modern Approach to Fermat’s Last TheoremLet’s first start out by examining the Fermat curve x3 + y3 = z3. I want to give amodern proof as to why this curve has no non-trivial rational points via elliptic curves. Thiswill introduce some of the terminology necessary to understanding the ideas behind Wiles’approach to Fermat’s Last Theorem. Under the change of coordinates z 7→ (y+ z/3), we seethat this curve becomes x3 − y2z − yz2/3− z3/27 = 0. Homogenizing under z = 1 and thenperforming a change of coordinates to minimal Weierstrass form, we see that this curve is anelliptic curve with form y2 = x3−432. This curve has conductor 27. A check on the Cremonatables [Cre] reveals that the Mordell-Weil Group E(Q) has rank 0 and torsion subgroup oforder 3 corresponding to the point (x, y) = (12,±36) and the point at infinity. These pointscorrespond to the points (1/3, 0), (1/3,−1/3) and the point at infinity on the second curvementioned above and these lastly correspond to the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) onthe Fermat curve. Any other point on the Fermat curve would correspond to another rationalpoint on the curve y2 = x3− 432 which is impossible and we have thus shown that there areonly trivial solutions to the Fermat curve x3 + y3 = z3.The case n = 3 above gives an indication of how elliptic curves can be used in the studyof Diophantine equations. I now present a solution to Fermat’s Last Theorem based on themethod derived by Wiles. These notes are a slightly modified version of the notes producedby Samir Siksek in [Coh07b].In the background of the following will be the Modularity Theorem. This is a statementthat rational elliptic curves are in a bijection with rational modular forms. How the proof ofFermat’s Last Theorem will work is that we start with a hypothesized nontrivial primitivesolution to our Diophantine equation xn + yn = zn say (x, y, z) = (a, b, c). Next, we takethis solution and associate to it an elliptic curve called a Frey-Hellegouarch curve. Mapping17the curve under the bijection to modular forms, this form has a few special properties whichallow it to conform to the hypothesis of Ribet’s level lowering theorem. Level lowering findsan equivalent modular form, in the sense of having a congruence condition between thecoefficients modulo a prime ideal p of OKf , at a lower level that does not depend on (a, b, c).This level is equal to 2 and this is a contradiction since there are no newforms of level 2. Wewill make all of the above precise in this section. For now, we begin by stating the ModularityTheorem.Theorem 1.4.1. (The Modularity Theorem for Elliptic Curves [Wil95] [TW95] [BCDT01])Let N ≥ 1 be an integer. Then there is a one to one correspondence f 7→ Ef between rationalweight 2 newforms of level N , that is, normalized eigenforms of f ∈ S2(Γ0(N))new withrational coefficients and isogeny classes of elliptic curves E defined over Q and of conductorN . Under this correspondence, for all primes ` - N , a`(f) = a`(Ef ).As a sanity check, notice that ai(f) ∈ OKf and since they are rational, we have in factthat ai(f) ∈ Z so the terms above are indeed well defined. This is the amazing merger of tworeally different mathematical objects unifying them into one statement. One can think ofthis as a dictionary between languages, swapping between terminologies wherever thinkingof the objects in one context is easier than the other.Definition 1.4.2. Let E be an elliptic curve over Q of conductor N . Let f be a weight 2newform of level N ′ not necessarily equal to N . We will say that E arises modulo p from f(written E ∼p f) if there exists a prime ideal p ofKf above p such that a`(f) ≡ a`(E) (mod p)for all but finitely many primes `. If F is an elliptic curve of conductor N ′, then we also writeE ∼p F to mean E ∼p f where F = Ef and the f coming from the Modularity Theorem.It should be noted here that under the second interpretation, we have that ∼p forms anequivalence relation on elliptic curves. This property implies the following theorem that wewill need now and in the subsequent chapters.Theorem 1.4.3. Let E be an elliptic curve over Q of conductor N . Let f be a weight 2newform of level N ′ not necessarily equal to N . Assume that E ∼p f . There exists a primeideal p of Kf above p such that for all prime numbers `, we have1. If ` - pNN ′, then a`(f) ≡ a`(E) (mod p), that is p | NKf/Q(a`(f)− a`(E)).2. If ` || N but ` - pN ′, then a`(f) ≡ ±(`+ 1) (mod p), that is p | NKf/Q(a`(f)± (`+ 1)).where NKf/Q denotes the norm map here and throughout this article.The following is an improvement of the above theorem due to Kraus and Oesterle´ [KO92]that will help us remove the cumbersome fact that the above depends on ` - p. The annoyance18comes from the fact that p will usually be an unknown exponent and so having conditionsthat depend on this exponent is an inconvenience that we can do away with in the specialcase that the newform is rational (and so corresponds to an elliptic curve).Theorem 1.4.4. (Kraus and Oesterle´ [KO92]) Let E and F be elliptic curves defined overQ with conductors N and N ′ and assume that E ∼p F as defined above. Then for all primenumbers `, we have1. If ` - NN ′, then a`(E) ≡ a`(F ) (mod p).2. If ` || N but ` - N ′, then a`(F ) ≡ ±(`+ 1) (mod p).Next we introduce the Level Lowering Theorem of Ribet. This is a simplified modificationto suit our specific needs. It turns out that in our situation, we can reduce the general LevelLowering Theorem into the following much easier to state theorem that avoids potentialissues at p.Definition 1.4.5. Define for p a prime numberNp :=N∏q ||Np | vq(∆min)qwhere the product runs over primes q and vq(∆min) is the q-adic valuation of ∆min formallydefined asvq(n) =max{k ∈ N : qk | n} if n 6= 0∞ if n = 0.Reworded, this means vq(N) = 1 and p | vq(∆min). This Np will be the conductor of the levellowered curve. Finally we can state the following.Theorem 1.4.6. (Ribet’s Level Lowering Theorem) [Rib90] Let E be an elliptic curve definedover Q and let p ≥ 5 be a prime number. Assume that there does not exist a p-isogeny (thatis, an isogeny of degree p) defined over Q from E to some other elliptic curve and let Np beas above. Then there exists a newform f of level Np such that E ∼p f .This theorem in its original form is a statement about modular forms. The ModularityTheorem allows us to pass back and forth from elliptic curves to modular forms. See [Rib90],[Rib94] for the original phrasing. In order to apply Ribet’s Level Lowering Theorem, weneed to know when a curve has no p-isogeny. It turns out that this work was done by BarryMazur and is collected in the following theorem. It should be noted that this statement was19actually worded originally as a statement about the irreducibility of the mod p representationassociated to an elliptic curve.Theorem 1.4.7. (Mazur [Maz78]) Let E be an elliptic curve defined over Q of conductorN . Then E does not have any p-isogeny if at least one of the following conditions holds:1. p ≥ 17 and j(E) /∈ Z[12 ]2. p ≥ 11 and N is square free (that is, E is semi-stable).3. p ≥ 5, N is square free, and #E(Q)[2] = 4.The last ingredient in the proof of Fermat’s Last Theorem are Frey-Hellegouarch curves.These are elliptic curves that are associated to solutions to Diophantine equations, morespecifically, the coefficients of the elliptic curve should be related to the solution of a Dio-phantine equation. In addition, the minimal discriminant ∆min should be able to be writtenas the product of two factors. One factor should not depend on the solution of the Diophan-tine equation and the other should be a pth power (that could depend on the solution). Wewrite ∆min = C0Dp. For the primes p | D, we should also have that E has multiplicativereduction.We now have enough terminology to give a “black box” proof of Fermat’s Last Theorem.Theorem 1.4.8. Let p ≥ 5 be a prime number. The equation xp + yp + zp = 0 has nonontrivial primitive solutions.Proof. Let (a, b, c) be a nontrivial primitive solution to our equation (that is, a, b, c arepairwise coprime and abc 6= 0). Local considerations at 2 show abc is even and so withoutloss of generality, we can suppose (by changing coordinates if necessary) that b is divisibleby 2. Further, suppose a ≡ −1 (mod 4) by possibly modifying the solution to (−a,−b,−c)if necessary5. Let E be the following associated Frey-Hellegouarch curvey2 = x(x− ap)(x+ bp) = x3 + (bp − ap)x2 − apbpxComputing the minimal discriminant and the conductor using Tate’s algorithm as was donein Example 1.2.9 and computing the Np value desired in Ribet’s Level Lowering Theorem,we have∆min = 2−8(abc)2p N = 2rad2(abc) Np = 25These conditions, while seemingly not used directly, give a simplification of v2(N) in the application ofTate’s algorithm.20It is here where we need that we have a nontrivial solution for otherwise our curve abovehas discriminant zero and hence its associated defining equation has repeated roots. Beforeapplying Ribet’s theorem, we need to check that E has no p-isogenies. By construction, thiscurve has full rational 2-torsion and since N is squarefree, we can apply Mazur’s Theorem(Theorem 1.4.7) which shows that E has no p-isogenies. By Ribet’s Theorem (Theorem1.4.6), we have that there is a newform of level Np = 2 such that E ∼p f . Notice that thereare no newforms at level 2 and so we have a contradiction and thus no solution (a, b, c) canexist. This concludes the proof of Fermat’s Last Theorem, omitting the proofs of some verybig hammers that we used above. There are a few other theorems that can be importantin trying to apply the Modularity Theorem and that are used throughout this thesis. I willinclude these here. Remember that one of the key ingredients was the absence of p-isogenies.The following theorems can help to determine whether they exist.Theorem 1.4.9. Let E/Q be an elliptic curve such that its pth division polynomial is irre-ducible. Then E has no p-isogenies.Proof. Let P be a point on E. If the pth division polynomial is irreducible, then the de-nominator of the point [p]P is irreducible. Thus, there are no rational points of order p. By[Sil09, p.72], the kernel of a p-isogeny has size p. This is a contradiction since the isogenycomposed with its dual isogeny is the map [p] which as we have shown has no kernel (so if ψhas kernel of size p, then [p] = ψ˜ ◦ψ has kernel of size at least p which is a contradiction). Theorem 1.4.10. (Diamond-Kramer) [DK95] Let E be an elliptic curve defined over Q ofconductor N . If v2(N) = 3, 5, or 7, then E does not have any p-isogeny for p an odd prime.In the cases where we can reduce to a curve with complex multiplication, the followingresult might aid in further simplifying.Theorem 1.4.11. Let E and F be two elliptic curves defined over Q. Assume that F hascomplex multiplication by some imaginary quadratic field K and that E ∼p F for some primep. Then1. (Halberstadt-Kraus via Momose)[HK98] [Mom84] If p = 11 or p ≥ 17 and p splits inL, then the conductors of E and F are equal.2. (Darmon-Merel)[DM97] If p ≥ 5, p is inert, and E has a Q-rational subgroup of order2 or 3, then j(E) ∈ Z[1p ]. If in addition, p2 - N and p - N ′, where N is the conductorof E and N ′ is the conductor of F , then we have j(E) ∈ Z.Next, the following theorem whose proof can be found in [Coh07b, p.510] follows mainlyas a result of Theorem 1.4.4 and can help with bounding the exponent.21Theorem 1.4.12. Let E/Q be an elliptic curve of conductor N and let t ∈ Z be an integerdividing the rational torsion subgroup of E. Let f be a newform of level N ′. Lastly, let ` bea prime number such that `2 - N and ` - N ′. DefineS` = {a ∈ Z : −2√` ≤ a ≤ 2√` and a ≡ `+ 1 (mod t)},δf =` if f is not rational1 if f is rational,B`(f) = δfNf ((`+ 1)2 − a`(f))∏a∈S`Nf (a− a`(f))where Nf denotes the norm on Kf/Q. If E ∼p f then p | B`(f).An important question is when the above theorem can give you a bound on the exponent,that is, when B`(f) 6= 0 for at least some values of `. This occurs whenever your newform fis one of the following:1. Irrational.2. Rational, t is prime or equal to 4 and for every elliptic curve F isogenous to Ef , wehave t - #Etor(Q).3. Rational, t = 4 and for every elliptic curve F isogenous to Ef , we have F does not havefull 2 torsion.In these cases, then in fact B`(f) is nonzero for infinitely many values of `.1.5 Where Do Frey-Hellegouarch Curves Come From?There still leaves one quite important question and that is “Where do Frey-Hellegouarchcurves come from?” I will try to give an idea of where certain curves come from by motivatingthe construction of the curves used in this thesis. An excellent source of this information canbe found in [Kra99]. I will give a brief summary.First we discuss the Frey-Hellegouarch curve used in Fermat’s Last Theorem, namelyy2 = x(x− ap)(x+ bp).This curve was first noticed by Yves Hellegouarch [Hel75] while examining points of finiteorder on elliptic curves. He was one of the first people to take a solution to a Diophantineequation and associate to it an elliptic curve. In 1986, Gerhard Frey [Fre86] looked at thesame curve and noticed that it would have some strange properties should it exist (see [Fre09]22for a recent recanting of this idea). This was later formalized by Kenneth Ribet [Rib90]. Heproved a theorem that is now known as Ribet’s Level Lowering theorem. This associates toa modular form at a level containing a large pth power, a modular form at a much smallerlevel (essentially removing the pth power). His result would lead to a proof of Fermat’sLast Theorem if one could prove that there was a correspondence between elliptic curvesand modular forms. This was known as the Taniyama-Shimura-Weil Conjecture. In 1995,Andrew Wiles [Wil95] proved that this conjecture was true for semi-stable elliptic curves andthis was enough to close the proof. This conjecture is now known as the Modularity Theoremand has since been proven to be true [Wil95] [TW95] [BCDT01].Many of the other Frey-Hellegouarch curves are based on the above Frey-Hellegouarchcurve. For a Frey-Hellegouarch curve of the formY 2 = X(X − A)(X +B) = X3 + (B − A)X2 − ABX,we saw at the beginning of this section that ∆ = 24A2B2(B + A)2. In the situation above,we let A = ap and B = bp. Then since ap + bp = cp we could make the replacement and seethat the discriminant has a large pth power dividing it. Remember that to use the abovemethod we need the following:1. The Frey-Hellegouarch curve should depend on our solution to a Diophantine equation.2. The minimal discriminant ∆min should be of the form CDp where C does not dependon the solution to a Diophantine equation.3. For the primes p | D, we should also have that E has multiplicative reduction.Let’s try the technique on the Diophantine equation x3 + y3 = Czp. We consider factoringthe left hand side over Q(ζ3) (or equivalently over Q(√−3)) where ζ3 =−1−√−32 is a thirdroot of unity. This gives usx3 + y3 = (x+ y)(x2 + xy + y2) = (x+ y)(x+ ζ3y)(x+ ζ23y) = CzpNow these factors are coprime [BS66, p.157] and the class number of Q(ζ3) is 1 which allows usto write x+ ζj3y = cjzpj for j ∈ {0, 1, 2} where c0c1c2 = C are coprime factors and z0z1z2 = z.Now, if we can find factors γ0, γ1 and γ2 such that(x+ y)γ0 + (x+ ζ3y)γ1 + (x+ ζ23y)γ2 = 0then substituting the above yieldsc0zp0γ0 + c1zp1γ1 + c2zp2γ2 = 023and this should remind us of the original Frey-Hellegouarch curve we used above. Thus, wesearch for values of γj that make the above work and then see if we get a Frey-Hellegouarchcurve. The above equation can be solved provided the following system is satisfiedγ0 + γ1 + γ2 = 0γ0 + ζ3γ1 + ζ23γ2 = 0Since 1 + ζ3 + ζ23 and ζ33 = 1, we see that γ0 = 1, γ1 = ζ3 and γ2 = ζ23 gives an admissiblesolution. Inserting this information and lettingA := (x+ ζ3y)γ1 = ζ3(x+ ζ3y) B := (x+ ζ23y)γ2 = ζ23 (x+ ζ23y)we have that the above Frey-Hellegouarch curve becomesY 2 = X3 + (B − A)X2 − ABX = X3 + ((ζ23 − ζ3)(x− y))X2 −x3 + y3x+ yX.Since ζ23 − ζ3 = −2ζ3 − 1 =√−3, the above elliptic curve is not rational. However, wecan perform a quadratic twist over Q(√−3) by (−3)1/4. This is done by sending (X, Y ) 7→(X, (−3)1/4), multiplying through by (−3)3/2 and then relabeling −3Y as Y˜ and (−3)1/2Xas X˜. Performing these operations gives us the curve defined by(−3)1/2Y 2 = X3 + (−3)1/2(x− y)X2 −x3 + y3x+ yX(−3)2Y 2 = (−3)3/2X3 + (−3)2(x− y)X2 − (−3)3/2x3 + y3x+ yX(−3Y )2 = ((−3)1/2X)3 − 3(x− y)((−3)1/2X)2 + 3x3 + y3x+ y((−3)1/2X)Y˜ 2 = X˜3 − 3(x− y)X˜2 + 3(x3 + y3x+ y)X˜or, if we like, we can shift the above curve so that the 2-torsion point is at y − x viaX˜ 7→ X˜ + x− y and get the curveY˜ 2 = X˜3 + 3xyX˜ + x3 − y3It is this curve used in [BLM11] and [Mul06] and we will use this curve later in this thesis. Wecan perform the same computation with x5+y5 = Czp. Let’s factor the left hand side over thetotally real field Q(ζ5 + ζ−15 ) (or equivalently over Q(√5)). Denote by α = ζ5 + ζ−15 =−1+√52(so here we choose the primitive fifth root so that this works) and α¯ = −1−√52 . Then factoring24givesx5 + y5 = (x+ y)(x4−x3y+x2y2−xy3 + y4) = (x+ y)(x2 +αxy+ y2)(x2 + α¯xy+ y2) = Czp.Similar to the above, we solve(x+ y)2γ0 + (x2 + αxy + y2)γ1 + (x2 + α¯xy + y2)γ2 = 0where the square on the linear term above aids with the algebra. A solution to the above isgiven byγ0 = 1 γ1 = α¯ γ2 = α.SettingA := α¯(x2 + αxy + y2) B := α(x2 + α¯xy + y2)and substituting into our Frey-Hellegouarch curve givesY 2 = X3 + (B − A)X2 − ABX = X3 + (α− α¯)(x2 + y2)X2 − αα¯x5 + y5x+ yX.= X3 +√5(x2 + y2)X2 +x5 + y5x+ yX.As before, twisting at 51/4 yieldsY˜ 2 = X˜3 + 5(x2 + y2)X˜2 + 5(x5 + y5x+ y)X˜.To extend this idea further, [Fre10, Chapter 3] has used a technique similar to the casewith signature (5, 5, p) above. He exploits the fact that Q(ζ5 + ζ−15 ) is the maximal totallyreal subfield of Q(ζ5) and generalizes this idea to curves with signature (r, r, p). He thenfurther uses Hilbert modularity to get his correspondence between elliptic curves over atotally real field and Hilbert modular forms as was proven by Khare and Wintenberger[KW09a] [KW09b]. This thesis does not go into details of the methods of Hilbert modularityhowever I did want to mention that this would be a source for more interesting problems inthe future.At this time, I would like to record some more results in the same spirit as those mentionedabove here for use later.251.6 Ternary Fermat-Type Diophantine EquationsHere I describe the kinds of arguments necessary to completely solve certain types ofFermat equations. These arguments will also be used in the following chapters. Please notethat some of the notation used here involving variables is used only for the scope of thissection. To start, we will solve the Diophantine equationAxp +Byp + zp = 0 (1.1)where p ≥ 5 is prime and Ax,By and z are non-zero pairwise coprime integers and R :=AB = 2a5b. We start by assuming that (x, y, z, p) is a solution to the equation definedby 1.1. Without loss of generality, we may suppose that vq(R) < p for all primes q sinceotherwise we have via the coprime condition that one of A or B is divisible by qp and so canbe brought into the corresponding xp or yp term. Without loss of generality, we may alsoassume that Byp ≡ 0 (mod 2) and that Axp ≡ −1 (mod 4). We can associate to this curvea corresponding Frey-Hellegouarch curve given byE : Y 2 = X(X − Axp)(X +Byp)Using Tate’s algorithm (see [Coh07b, p.542], [Sil94, p.364-368], [Pap93], [Mul06]), we see thatthis curve has minimal discriminant∆min =24R2(xyz)2p if 16 - Byp2−8R2(xyz)2p if 16 | Bypand conductor N = 2αrad2(Rxyz) where α is given byα =1 if v2(R) = 0, 1 ≤ v2(R) ≤ 4 and y is even, or v2(R) ≥ 50 if v2(R) = 4 and y is odd3 if 2 ≤ v2(R) ≤ 3 and y is odd5 if v2(R) = 1 and y is odd.Next we want to apply Ribet’s Theorem first verifying that our curve has no p-isogenies.When α above equals 0 or 1, we can use Theorem 1.4.7 by noting that p ≥ 5, N is squarefree and 4 divides |Etor(Q)| (since it is factored into linear terms). When α equals 3 or 5,we can use Theorem 1.4.10 to see that E has no p-isogenies for any odd prime p. Hence, wecan apply level lowering and find a newform f ∈ Snew2 (Γ0(Np)) where Np = 2αrad2(R). AsR = 2a5b we have that rad2(R) = 1 or 5.26Now, if α = 0 or 1, we have that Np ∈ {1, 2, 5, 10} which is a contradiction since thereare no newforms at these levels. Thus, we may suppose that y is odd. If α = 3 and b = 0then Np = 8 and that is also a contradiction since there are no newforms at level 8. If α = 3but b ≥ 1 then we reduce to a curve of level Np = 40 of which there is one such isogeny class.This gives an obstruction which the method cannot overcome. These obstructions are oftenthe difficulty when solving problems with this method.The last case is when α = 5. This occurs when a = 1. If b = 0, then Np = 32 andthere is one curve at level 32. A look at the Cremona tables shows us that this curve hascomplex multiplication by Z[i]. Now we use Theorem 1.4.11. Note that the j-invariant forthe Frey-Hellegouarch curve above is given byj(E) =28(z2p − xpyp)3(xyz)p.To start, we may suppose that p 6= 5 or 13 since these cases were done by De´nes [De´n52] onthe equation xp + yp = 2zp. We proceed in cases.Case 1: p is inert in Z[i]. This occurs when p ≡ 3 (mod 4). Recall that here x, y, zare all odd. Notice that if q is a prime divisor of one of these three terms, then as x, y, zare pairwise coprime, the prime q cannot divide the numerator. So the j-invariant is not aninteger. As p is not 2 and we have already considered the case p = 5, we know that p - Npand p2 - N so this contradicts Theorem 1.4.11 unless xyz has no prime divisors. This wouldmean that |xyz| = 1.Case 2: p splits in Z[i]. This occurs when p ≡ 1 (mod 4). In this case Theorem 1.4.11immediately tells us that N = Np and so we must have that 2αrad2(Rxyz) = 2αrad2(R).This means that |xyz| = 1.In either case above, we see that |xyz| = 1. Since a = 1 and b = 0, we want to find asolution to xp + 2yp + zp = 0 with |xyz| = 1 and this can be given only by (A,B, x, y, z, p) =(1, 2,−1, 1,−1, p). This completes the proof when b = 0.If b ≥ 1, then we go to level Np = 160. By an exponent bound computation, we see thatB3(f) is nonzero and divisible only by the primes 2 and 3 for each of the three newforms oflevel 160. As all primes outside of ` = 2 or ` = 5 satisfy Theorem 1.4.12, we see that thismeans that p divides a product of twos and threes, a contradiction since p ≥ 5. This givesus the following theorem originally proven by Kraus in generality [Kra97].Theorem 1.6.1. (Kraus) The equation Axp + Byp + zp = 0 with AB = 2a5b and p ≥ 5 hasno nontrivial solutions with Ax,By and z are non-zero pairwise coprime integers when either27b = 0, a = 0, a = 1 or a ≥ 4 except for the one given by (A,B, x, y, z, p) = (1, 2,−1, 1,−1, p).Using similar techniques, Bennett and Skinner [BS04] also proved the followingTheorem 1.6.2. (Bennett and Skinner) [BS04] Let C ∈ {1, 2, 5, 10} and n ≥ 4 be an integer.Then the equationxn + yn = Cz2has no solutions in pairwise coprime integers (x, y, z) with x > y unless C = 2 and(n, x, y, z) ∈ {(5, 3,−1,±11), (4, 1,−1,±1)}.Notice that this theorem also solves the cases when C = 2a, C = 5b or C = 2a5b since ifC is not squarefree, we can write C as kδm2 with δ either 0 or 1 and k = 1, 2, 5 or 10 therebyreducing it to a case in the above theorem. Similarly, we haveTheorem 1.6.3. (Bennett and Skinner) [BS04] [Coh07b, p.507] Let p ≥ 7 be prime anda ≥ 2 or a = 0. Then the equationxp + 2ayp = z2has no solutions in pairwise coprime integers (x, y, z) with |xy| > 1 and when |xy| = 1, thenthe only solutions are (x, y, z) = (1, 1,±3).Theorem 1.6.4. (Bennett and Skinner) [BS04] Let p ≥ 7 be prime and a ≥ 6 or a = 0.Then the equationxp + 2ayp = 5z2has no solutions in pairwise coprime integers (x, y, z) with |xy| > 1.Note that in the above theorem, a = 0 was a specific case of Theorem 1.6.2. Lastly, wehaveTheorem 1.6.5. (Bennett and Skinner) [BS04] Let p ≥ 7 be prime, a ≥ 6 and b ≥ 0. Thenthe equationAxp +Byp = z2with AB = 2a5b has no nontrivial solutions with Ax,By and z are non-zero pairwise coprimeintegers.Extending the techniques above even further, we have the following results first provenby Bennett, Vatsal, and Yasdani [BVY04].28Theorem 1.6.6. (Bennett, Vatsal, Yasdani) [BVY04] Let C ∈ {1, 2, 5} and p ≥ 7 a primeand b a nonnegative integer. Then the equationxp + yp = Cbz3has no solutions in coprime integers x, y, z with |xy| > 1.Theorem 1.6.7. (Bennett, Vatsal, Yasdani) [BVY04] Let p ≥ 7 a prime and b a positiveinteger. Then the equationxp + 5byp = z3has no solutions in coprime integers x, y, z with |xy| > 1.1.7 Results From This ThesisAfter solving Fermat’s Last Theorem, what other similar types of problems can now besolved? With the armamentarium of techniques at our disposal, there have since been amultitude of generalizations of this method to approach many different types of Diophantineequations, see for example any of [BS04], [BVY04], [BLM11], [Bil07], [BD10], [DG95], [Ell04],[Kra97], [Kra98], [Sik03]. Some of these results were discussed in the previous section andwill be used later on.In this thesis, we will look at twisted extensions of Fermat’s theorem, in particular, thoseof the form xq + yq = pαzn for x, y, z, p, r, n ∈ Z with p prime, q ∈ {3, 5}, α ≥ 1 and n ≥ 5prime. Here I mention that work on the equations when α = 0 have been done by manymathematicians, including the works of [Kra98] (17 ≤ n < 104 prime), [Bru00] (n = 4, 5),[Dah08] (n = 5, 7, 11, 13) and [CS09] (infinitely many n, including a set of primes of Dirichletdensity 28219/44928) on the equation x3 + y3 = zn and an unpublished note by Darmon andKraus on the equation x5 + y5 = (2z)n for n prime. A full classification in these cases hasnot currently been discovered, though not as a result of a lack of interest.For this equation, we will discuss in Chapter 2 a classification of elliptic curves with anontrivial rational two torsion point and conductor with three distinct primes (with one ofthem being 2). From here in Chapter 3, I will present some auxiliary results on Diophantineequations to aid with our classification. Then we specify to two particular cases, curveswith conductor in the set S3,p = {18p, 36p, 72p} and those curves with conductor in S5,p ={50p, 200p, 400p}. In Chapters 4 and 5, I prove a classification on the type of primes such thatwe have an elliptic curve with conductor in Sq,p and a nontrivial rational two torsion point.We denote such primes with the notation Sq. In Chapter 6, I generalize the results from29[BLM11] first to the equation x5 + y5 = pαzn for the complement of the primes p classified inChapter 5, that is, for primes where all curves with conductor in the set S5,p do not have anelliptic curve with a nontrivial rational two torsion point. Finally, in Chapter 7, I take boththis new result and the result from the [BLM11] paper one step further and extend the resultto a subset of the primes in Sq. In this final extension I show a specific subset of Sq,p thatthe prime p must avoid in order to be in my classification. The main theorems in this thesisare given by the following. For the purposes of stating the theorems now, I will delay thedefinition of Pb,q and Pg,q until Chapter 7 but mention now that these two sets are subsetsof Sq.Theorem 1.7.1. Suppose that p ≥ 5 and that p /∈ S5. Let α ≥ 1 be an integer. Then theequationx5 + y5 = pαznhas no solutions in coprime nonzero integers x, y, z and prime n satisfying n ≥ p13pTheorem 1.7.2. Let q ∈ {3, 5} and let p ∈ Pg,q ⊆ Sq. Then p ∈ Pb,q if and only if everyelliptic curve with conductor in Sq,p with non-trivial rational two torsion has discriminantnot of the form ∆Q,q,m :=(−1q)qm2 for all integers m.Theorem 1.7.3. Let q ∈ {3, 5} and suppose that p is a prime such that p 6∈ Sq or thatp ∈ Pb,q ⊆ Sq. Then the equation xq + yq = pαzn has no nontrivial coprime integer solutions(x, y, z) where α ≥ 1 and n ≥ Cq(p) a prime. Furthermore, if the prime p avoids the lists inTables 7.1 and 7.3 (hence residing in Table 7.5) when q = 3 or in Tables 7.2 and 7.4 (henceresiding in Table 7.6) when q = 5, then we have that p 6∈ Sq or that p ∈ Pb,q ⊆ Sq.30Chapter 2Classification of Elliptic Curves WithNontrivial Rational Two Torsion2.1 On the Q-Isomorphism Classes of Elliptic CurvesWith Nontrivial Rational 2-Torsion and Conductor2LqMpNLet E be an elliptic curve over Q with rational two torsion and conductor 2LqMpN (here,we use q first because we are interested in the case when q = 5). We know via Theorem 1.2.7that the conductor of any such elliptic curve cannot be divisible by 29, 36, or `3 where ` ≥ 5is a prime. Further, by [Mul06, p. 13], we know that any elliptic curve over Q with rationaltwo torsion is such that 33 does not divide the conductor. Thus we can assume here that0 ≤ L ≤ 8 and 0 ≤M,N ≤ 2. We may also assume that E has a model given byE : y2 = x3 + ax2 + bx.For such a curve, we have that the discriminant is given by∆ = 24b2(a2 − 4b).From [Mul06, p.13-14], we see that the discriminant above and the conductor must share oddprimes and so we have thatb2(a2 − 4b) = ±2λqµpνfor some non-negative integers λ, µ, ν. Hence, we must have thatb = ±2iqjpk31for some i, j, k satisfying 0 ≤ 2i ≤ λ, 0 ≤ 2j ≤ µ and 0 ≤ 2k ≤ ν. Using this, we have thata2 = 2i+2qjpk ± 2λ−2iqµ−2jpν−2k.Now, break this down into 27 cases based on the trichotomies depending on if i+ 2 > λ− 2i,i+ 2 < λ− 2i or i+ 2 = λ− 2i (and similarly with i replaced by j or k and λ replaced by µor ν) we can create a table of Q-isomorphism classes of curves. I will demonstrate one suchcase here and for the rest I refer the interested reader to the appendix in [Mul06] thoughthe reader should be warned that there are numerous typos in the proof found there. Let’ssuppose that i+ 2 > λ− 2i, j = µ− 2j and k < ν − 2k. Factoring givesa2 = 2λ−2iqjpk(23i−λ+2 ± pν−3k).Set d := 23i−λ+2 ± pν−3k. Since a priori we do not know if q | d, we have that q(j+)/2 | awhere j ≡ (mod 2). Rewriting the above equation, we have3(a2λ/2−iq(j+)/2pk/2)2− 23i−λ+2 = ±pν−2kwith ` := 3i − λ + 2 ≥ 1 and n := ν − 2k ≥ 1. Then depending on , we have a solution toone of the following equationsd2 − 2` = ±pn or 3d2 − 2` = ±pnif = 0 in the first case or = 1 in the second case. Either way, a model for our elliptic curvecan be given byy2 = x3 + 2λ/2−iq(j+)/2pk/2dx2 + 2iqjpkx.Now, by the division algorithm, we can writeλ/2− i = 2q1 + r1j + 2= 2q2 + r2 k/2 = 2q3 + r3where r1, r2, r3 ∈ {0, 1}. Rewriting the elliptic curve givesy2 = x3 + 22q1+r1q2q2+r2p2q3+r3dx2 + 2`−2+4q1+2r1q4q2+2r2−p4q2+2r3x.If ` = 1 and r1 = 0, then the change of variables(X, Y ) =(x22(q1−1)q2(q2+(r2−1))p2q3,y23(q1−1)q3(q2+(r2−1))p3q3)32gives the Q-isomorphic curveY 2 = X3 + 22q2+(1−2)r2pr3dX2 + 23q3+2(1−2)r2p2r2XIf (`, r1) 6= (1, 0) (and so either ` > 1 or r1 > 0), we have that the change of variables(X, Y ) =(x22q1q2(q2+(r2−1))p2q3,y23q1q3(q2+(r2−1))p3q3)gives the Q-isomorphic curveY 2 = X3 + 2r1q2+(1−2)r2pr3dX2 + 2`+2r1−2q3+2(1−2)r2p2r2X.Note that when = 0, the curve above corresponds to case 4 in the tables below. When = 1, then the curve above corresponds to case 8 in the table below. To see this easily, plugin for the cases of r2 ∈ {0, 1} and ∈ {0, 1}. Modifying the above argument 26 times givesthe following theorem.Theorem 2.1.1. (Modified from [Mul06, p. 270-271, 309-310]) Suppose E is an ellipticcurve with rational two torsion and conductor 2LqMpN . Without loss of generality, E cantake the formE : y2 = x3 + a2x2 + a4x.Further assume that a4 > 0. Then, there exists an integer d and non-negative integers `,m, nsatisfying one of the equations in the second column below and the corresponding curve E isgiven by a2 and a4 given in the third and fourth columns where r1, r2, r3 ∈ {0, 1} except incases 1 through 9 (for both tables below) where if ` = 1, then r1 ∈ {1, 2} and the sign in theDiophantine equation matches that of the sign in the discriminant ∆.33Equation a2 a4 ∆1 d2 − 2`qmpn = ±1 2r1qr2pr3d 2`+2r1−2qm+2r2pn+2r3 ±22`+6r1q2m+6r2p2n+6r32 d2 − 2`qm = ±pn 2r1qr2pr3d 2`+2r1−2qm+2r2p2r3 ±22`+6r1q2m+6r2pn+6r33 d2 − 2`pn = ±qm 2r1qr2pr3d 2`+2r1−2q2r2pn+2r3 ±22`+6r1qm+6r2p2n+6r34 d2 − 2` = ±qmpn 2r1qr2pr3d 2`+2r1−2q2r2p2r3 ±22`+6r1qm+6r2pn+6r35 pd2 − 2`qm = ±1 2r1qr2pr3+1d 2`+2r1−2qm+2r2p2r3+1 ±22`+6r1q2m+6r2p6r3+36 pd2 − 2` = ±qm 2r1qr2pr3+1d 2`+2r1−2q2r2p2r3+1 ±22`+6r1qm+6r2p6r3+37 qd2 − 2`pn = ±1 2r1qr2+1pr3d 2`+2r1−2q2r2+1pn+2r3 ±22`+6r1q6r2+3p2n+6r38 qd2 − 2` = ±pn 2r1qr2+1pr3d 2`+2r1−2q2r2+1p2r3 ±22`+6r1q6r2+3pn+6r39 qpd2 − 2` = ±1 2r1qr2+1pr3+1d 2`+2r1−2q2r2+1p2r3+1 ±22`+6r1q6r2+3p6r3+310 d2 − qmpn = ±2` 2r1+1qr2pr3d 22r1qm+2r2pn+2r3 ±2`+6r1+6q2m+6r2p2n+6r311 d2 − qm = ±2`pn 2r1+1qr2pr3d 22r1qm+2r2p2r3 ±2`+6r1+6q2m+6r2pn+6r312 d2 − pn = ±2`qm 2r1+1qr2pr3d 22r1q2r2pn+2r3 ±2`+6r1+6qm+6r2p2n+6r313 d2 − 1 = ±2`qmpn 2r1+1qr2pr3d 22r1q2r2p2r3 ±2`+6r1+6qm+6r2pn+6r314 pd2 − qm = ±2` 2r1+1qr2pr3+1d 22r1qm+2r2p2r3+1 ±2`+6r1+6q2m+6r2p6r3+315 pd2 − 1 = ±2`qm 2r1+1qr2pr3+1d 22r1q2r2p2r3+1 ±2`+6r1+6qm+6r2p6r3+316 qd2 − pn = ±2` 2r1+1qr2+1pr3d 22r1q2r2+1pn+2r3 ±2`+6r1+6q6r2+3p2n+6r317 qd2 − 1 = ±2`pn 2r1+1qr2+1pr3d 22r1q2r2+1p2r3 ±2`+6r1+6q6r2+3pn+6r318 qpd2 − 1 = ±2` 2r1+1qr2+1pr3+1d 22r1q2r2+1p2r3+1 ±2`+6r1+6q6r2+3p6r3+319 2d2 − qmpn = ±1 2r1+2qr2pr3d 22r1+1qm+2r2pn+2r3 ±26r1+9q2m+6r2p2n+6r320 2d2 − qm = ±pn 2r1+2qr2pr3d 22r1+1qm+2r2p2r3 ±26r1+9q2m+6r2pn+6r321 2d2 − pn = ±qm 2r1+2qr2pr3d 22r1+1q2r2pn+2r3 ±26r1+9qm+6r2p2n+6r322 2d2 − 1 = ±qmpn 2r1+2qr2pr3d 22r1+1q2r2p2r3 ±26r1+9qm+6r2pn+6r323 2pd2 − qm = ±1 2r1+2qr2pr3+1d 22r1+1qm+2r2p2r3+1 ±26r1+9q2m+6r2p6r3+324 2pd2 − 1 = ±qm 2r1+2qr2pr3+1d 22r1+1q2r2p2r3+1 ±26r1+9qm+6r2p6r3+325 2qd2 − pn = ±1 2r1+2qr2+1pr3d 22r1+1q2r2+1pn+2r3 ±26r1+9q6r2+3p2n+6r326 2qd2 − 1 = ±pn 2r1+2qr2+1pr3d 22r1+1q2r2+1p2r3 ±26r1+9q6r2+3pn+6r327 2qpd2 − 1 = ±1 2r1+2qr2+1pr3+1d 22r1+1q2r2+1p2r3+1 ±26r1+9q6r2+3p6r3+3Table 2.1: Elliptic curves of conductor 2LqMpN when a4 > 0When a4 < 0, we have34Equation a2 a4 ∆2 d2 + 2`qm = pn 2r1qr2pr3d −2`+2r1−2qm+2r2p2r3 22`+6r1q2m+6r2pn+6r33 d2 + 2lpn = qm 2r1qr2pr3d −2`+2r1−2q2r2pn+2r3 22`+6r1qm+6r2p2n+6r34 d2 + 2l = qmpn 2r1qr2pr3d −2`+2r1−2q2r2p2r3 22`+6r1qm+6r2pn+6r36 pd2 + 2l = qm 2r1qr2pr3+1d −2`+2r1−2q2r2p2r3+1 22`+6r1qm+6r2p6r3+38 qd2 + 2l = pn 2r1qr2+1pr3d −2`+2r1−2q2r2+1p2r3 22`+6r1q6r2+3pn+6r310 d2 + qmpn = 2` 2r1+1qr2pr3d −22r1qm+2r2pn+2r3 2`+6r1+6q2m+6r2p2n+6r311 d2 + qm = 2`pn 2r1+1qr2pr3d −22r1qm+2r2p2r3 2`+6r1+6q2m+6r2pn+6r312 d2 + pn = 2`qm 2r1+1qr2pr3d −22r1q2r2pn+2r3 2`+6r1+6qm+6r2p2n+6r313 d2 + 1 = 2`qmpn 2r1+1qr2pr3d −22r1q2r2p2r3 2`+6r1+6qm+6r2pn+6r314 pd2 + qm = 2` 2r1+1qr2pr3+1d −22r1qm+2r2p2r3+1 2`+6r1+6q2m+6r2p6r3+315 pd2 + 1 = 2`qm 2r1+1qr2pr3+1d −22r1q2r2p2r3+1 2`+6r1+6qm+6r2p6r3+316 qd2 + pn = 2` 2r1+1qr2+1pr3d −22r1q2r2+1pn+2r3 2`+6r1+6q6r2+3p2n+6r317 qd2 + 1 = 2`pn 2r1+1qr2+1pr3d −22r1q2r2+1p2r3 2`+6r1+6q6r2+3pn+6r318 qpd2 + 1 = 2` 2r1+1qr2+1pr3+1d −22r1q2r2+1p2r3+1 2`+6r1+6q6r2+3p6r3+320 2d2 + qm = pn 2r1+2qr2pr3d −22r1+1qm+2r2p2r3 26r1+9q2m+6r2pn+6r321 2d2 + pn = qm 2r1+2qr2pr3d −22r1+1q2r2pn+2r3 26r1+9qm+6r2p2n+6r322 2d2 + 1 = qmpn 2r1+2qr2pr3d −22r1+1q2r2p2r3 26r1+9qm+6r2pn+6r324 2pd2 + 1 = qm 2r1+2qr2pr3+1d −22r1+1q2r2p2r3+1 26r1+9qm+6r2p6r3+326 2qd2 + 1 = pn 2r1+2qr2+1pr3d −22r1+1q2r2+1p2r3 26r1+9q6r2+3pn+6r3Table 2.2: Elliptic curves of conductor 2LqMpN when a4 < 0To avoid redundancies in the lists above, in the right hand side of the equations above,we do allow the exponent of primes on the right to be 0 and otherwise, the exponents of theDiophantine equations are at least 1. The numbering in the second list is kept to reflect thefirst list.In our case, we are interested for application purposes in the case when q = 5. Ourgoal will be to solve the above Diophantine equations in general. Then, after simplification,ideally we would be able to remove many of the conditions on the exponents to reveal a shortand succinct list of potential elliptic curves.2.2 Elliptic Curves With Nontrivial Rational Two Tor-sion and Conductor 2aqbpcIn what follows, let E be an elliptic curve of conductor 2aqbpc with nontrivial rational twotorsion where p and q are distinct odd primes. By Theorem 1.2.7, we need only to consider35values of the exponents of the conductor where0 ≤ a ≤ 8 0 ≤ b, c ≤ 2The work on curves of conductor 2a3bpc has been done already and is summarized in [Mul06,Chapter 3]. In what follows, we assume that p, q ≥ 3 are distinct primes and further that1 ≤ b, c ≤ 2. We will create a list of all such curves below satisfying this criteria sorted byconductor. From this list, we will focus on the curves of conductor 50p, 200p and 400p anduse tricks of Diophantine equations to simplify these results to get a succinct list of primes pwith an elliptic curve with nontrivial rational two torsion and conductor 50p, 200p or 400p.Before we begin, we discuss some notes on the notation.1. We are looking for curves of conductor 2aqbpc with nontrivial rational two torsion. Theidea here is to find integers `,m, n so that the curves we find fit into some appropriaterow of the tables in Section 2.1.2. We require that m ≥ 1 when b = 1 and we can allow m = 0 when b = 2 (because thenr2 = 1), which is equivalent to asking for m ≥ 2− b for b = 1, 2.3. The function ψ(t) returns the square root value congruent to 1 modulo 4 or the positivevalue if t is even.4. Some cases that were present in the tables above are not below. These omitted casesare the ones with associated Diophantine equations that cannot be solved as seen bylooking at local conditions, for example by reducing modulo a power of 2 or modulo por q.5. The tables below give minimal models at the primes that divide the conductor exceptwhen 4 does not divide the conductor. In that case, one can obtain a minimal modelas stated in [Mul06, p. 13].6. Curves denoted by E and E ′ are related by a twist by√−1.7. Curves beginning with the same letter are linked by a degree 2 isogeny as in [Coh07a,p. 532-533] or a composition of two such isogenies as in [Mul06, p.31].8. The lettering used here was inspired by [Mul06]. It is unfortunate that the letteringhere does not correspond in any way to either the lettering used there or in [Cre] butfor us, the notation is self contained and so we proceed with this in mind.Theorem 2.2.1. The elliptic curves E defined over Q of conductor qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:361. There exist integers m ≥ 2 − b and n ≥ 1 such that 26qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 · qb−1ψ(26qmpn + 1) 24qm+2(b−1)pn 212q2m+6(b−1)p2nA2 − · 2 · qb−1ψ(26qmpn + 1) q2(b−1) 212qm+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.2. There exist integers m ≥ 2 − b and n ≥ 1 such that 26qm + pn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 · qb−1ψ(26qm + pn) 24qm+2(b−1) 212q2m+6(b−1)pnB2 − · 2 · qb−1ψ(26qm + pn) q2(b−1)pn 212qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.3. There exist integers m ≥ 2−b and n ≥ 1 such that 26qm−pn is a square, p ≡ 7 (mod 8),n odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 · qb−1ψ(26qm − pn) 24qm+2(b−1) −212q2m+6(b−1)pnC2 − · 2 · qb−1ψ(26qm − pn) −q2(b−1)pn 212qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.4. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − 26qm is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 · qb−1ψ(pn − 26qm) −24qm+2(b−1) 212q2m+6(b−1)pnD2 − · 2 · qb−1ψ(pn − 26qm q2(b−1)pn −212qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.5. There exist integers m ≥ 2 − b and n ≥ 1 such that 26pn + qm is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 · qb−1ψ(26pn + qm) 24q2(b−1)pn 212qm+6(b−1)p2nE2 − · 2 · qb−1ψ(26pn + qm) qm+2(b−1) 212q2m+6(b−1)pn37where ∈ {±1} is the residue of qb−1 modulo 4.6. There exist integers m ≥ 2−b and n ≥ 1 such that 26pn−qm is a square, q ≡ 7 (mod 8),m odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 · qb−1ψ(26pn − qm) 24q2(b−1)pn −212qm+6(b−1)p2nF2 − · 2 · qb−1ψ(26pn − qm) −qm+2(b−1) 212q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.7. There exist integers m ≥ 2 − b and n ≥ 1 such that qm − 26pn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆G1 · qb−1ψ(qm − 26pn) −24q2(b−1)pn 212qm+6(b−1)p2nG2 − · 2 · qb−1ψ(qm − 26pn) qm+2(b−1) −212q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.8. There exist integers m ≥ 2 − b and n ≥ 1 such that 26 + qmpn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆H1 · qb−1ψ(26 + qmpn) 24q2(b−1) 212qm+6(b−1)pnH2 − · 2 · qb−1ψ(26 + qmpn) qm+2(b−1)pn 212q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.9. There exist integers m ≥ 2 − b and n ≥ 1 such that 26 − qmpn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆I1 · qb−1ψ(26 − qmpn) 24q2(b−1) −212qm+6(b−1)pnI2 − · 2 · qb−1ψ(26 − qmpn) −qm+2(b−1)pn 212q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.10. There exist integers m ≥ 2 − b and n ≥ 1 such that qmpn − 26 is a square, 8 | (3q +5)(q − 1), 8 | (3p + 5)(p − 1) and E is Q-isomorphic to one of the following ellipticcurves:38a2 a4 ∆J1 · qb−1ψ(qmpn − 26) −24q2(b−1) 212qm+6(b−1)pnJ2 − · 2 · qb−1ψ(qmpn − 26) qm+2(b−1)pn −212q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.In the case that b = 2, we furthermore could have one of the following conditionssatisfied:11. there exist integers n ≥ 1 and s ∈ {0, 1} such that 26pn+1q is a square, q ≡ 1 (mod 8)and E is Q-isomorphic to one of the following curves:a2 a4 ∆K1 qs+1ψ(26pn+1q ) 24q2s+1pn 212q6s+3p2nK2 −2 · qs+1ψ(26pn+1q ) q2s+1 212q6s+3pn12. there exist integers n ≥ 1 and s ∈ {0, 1} such that 26pn−1q is a square, q ≡ 7 (mod 8)and E is Q-isomorphic to one of the following curves:a2 a4 ∆L1 · qs+1ψ(26pn−1q ) 24q2s+1pn −212q6s+3p2nL2 − · 2 · qs+1ψ(26pn−1q ) −q2s+1 212q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.13. there exist integers n ≥ 1 and s ∈ {0, 1} such that 26+pnq is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆M1 · qs+1ψ(26+pnq ) 24q2s+1 212q6s+3pnM2 − · 2 · qs+1ψ(26+pnq ) q2s+1pn 212q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.14. there exist integers n ≥ 1 and s ∈ {0, 1} such that 26−pnq is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆N1 · qs+1ψ(26−pnq ) 24q2s+1 −212q6s+3pnN2 − · 2 · qs+1ψ(26−pnq ) −q2s+1pn 212q6s+3p2n39where ∈ {±1} is the residue of qs+1 modulo 4.15. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−26q is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆O1 · qs+1ψ(pn−26q ) −24q2s+1 212q6s+3pnO2 − · 2 · qs+1ψ(pn−26q ) q2s+1pn −212q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.Theorem 2.2.2. The elliptic curves E defined over Q of conductor 2qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that 2`qmpn + 1 is a square andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 · qb−1ψ(2`qmpn + 1) 2`−2qm+2(b−1)pn 22`q2m+6(b−1)p2nA2 − · 2 · qb−1ψ(2`qmpn + 1) q2(b−1) 2`+6qm+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.2. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that 2`qm + pn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 · qb−1ψ(2`qm + pn) 2`−2qm+2(b−1) 22`q2m+6(b−1)pnB2 − · 2 · qb−1ψ(2`qm + pn) q2(b−1)pn 2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.3. There exist integers ` ≥ 7, m ≥ 2 − b and n ≥ 1 such that 2`qm − pn is a square,p ≡ 7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 · qb−1ψ(2`qm − pn) 2`−2qm+2(b−1) −22`q2m+6(b−1)pnC2 − · 2 · qb−1ψ(2`qm − pn) −q2(b−1)pn 2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.404. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that pn− 2`qm is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 · qb−1ψ(pn − 2`qm) −2`−2qm+2(b−1) 22`q2m+6(b−1)pnD2 − · 2 · qb−1ψ(pn − 2`qm) q2(b−1)pn −2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.5. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that 2`pn + qm is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 · qb−1ψ(2`pn + qm) 2`−2q2(b−1)pn 22`qm+6(b−1)p2nE2 − · 2 · qb−1ψ(2`pn + qm) qm+2(b−1) 2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.6. There exist integers ` ≥ 7, m ≥ 2 − b and n ≥ 1 such that 2`pn − qm is a square,q ≡ 7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 · qb−1ψ(2`pn − qm) 2`−2q2(b−1)pn −22`qm+6(b−1)p2nF2 − · 2 · qb−1ψ(2`pn − qm) −qm+2(b−1) 2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.7. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that qm− 2`pn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆G1 · qb−1ψ(qm − 2`pn) −2`−2q2(b−1)pn 22`qm+6(b−1)p2nG2 − · 2 · qb−1ψ(qm − 2`pn) qm+2(b−1) −2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.8. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that 2` + qmpn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆H1 · qb−1ψ(2` + qmpn) 2`−2q2(b−1) 22`qm+6(b−1)pnH2 − · 2 · qb−1ψ(2` + qmpn) qm+2(b−1)pn 2`+6q2m+6(b−1)p2n41where ∈ {±1} is the residue of qb−1 modulo 4.9. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that 2`− qmpn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆I1 · qb−1ψ(2` − qmpn) 2`−2q2(b−1) −22`qm+6(b−1)pnI2 − · 2 · qb−1ψ(2` − qmpn) −qm+2(b−1)pn 2`+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.10. There exist integers ` ≥ 7, m ≥ 2− b and n ≥ 1 such that qmpn− 2` is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆J1 · qb−1ψ(qmpn − 2`) −2`−2q2(b−1) 22`qm+6(b−1)pnJ2 − · 2 · qb−1ψ(qmpn − 2`) qm+2(b−1)pn −2`+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.In the case that b = 2, we furthermore could have one of the following conditionssatisfied:11. there exist integers ` ≥ 7, n ≥ 1 and s ∈ {0, 1} such that 2`pn+1q is a square, q ≡1 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆K1 qs+1ψ(2`pn+1q ) 2`−2q2s+1pn 22`q6s+3p2nK2 −2 · qs+1ψ(2`pn+1q ) q2s+1 2`+6q6s+3pn12. there exist integers ` ≥ 7, n ≥ 1 and s ∈ {0, 1} such that 2`pn−1q is a square, q ≡7 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆L1 · qs+1ψ(2`pn−1q ) 2`−2q2s+1pn −22`q6s+3p2nL2 − · 2 · qs+1ψ(2`pn−1q ) −q2s+1 2`+6q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.13. there exist integers ` ≥ 7, n ≥ 1 and s ∈ {0, 1} such that 2`+pnq is a square and E isQ-isomorphic to one of the following curves:42a2 a4 ∆M1 · qs+1ψ(2`+pnq ) 2`−2q2s+1 22`q6s+3pnM2 − · 2 · qs+1ψ(2`+pnq ) q2s+1pn 2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.14. there exist integers ` ≥ 7, n ≥ 1 and s ∈ {0, 1} such that 2`−pnq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆N1 · qs+1ψ(2`−pnq ) 2`−2q2s+1 −22`q6s+3pnN2 − · 2 · qs+1ψ(2`−pnq ) −q2s+1pn 2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.15. there exist integers ` ≥ 7, n ≥ 1 and s ∈ {0, 1} such that pn−2`q is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆O1 · qs+1ψ(pn−2`q ) −2`−2q2s+1 22`q6s+3pnO2 − · 2 · qs+1ψ(pn−2`q ) q2s+1pn −2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.Theorem 2.2.3. The elliptic curves E defined over Q of conductor 22qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers m ≥ 2 − b and n ≥ 1 such that 4qm + pn is a square, qm ≡−1 (mod 4), n odd and p ≡ 5 (mod 8), and E is Q-isomorphic to one of the followingelliptic curves:a2 a4 ∆A1 · qb−1ψ(4qm + pn) qm+2(b−1) 24q2m+6(b−1)pnA2 − · 2 · qb−1ψ(4qm + pn) q2(b−1)pn 28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.2. There exist integers m ≥ 2 − b and n ≥ 1 such that 4qm − pn is a square, qm ≡−1 (mod 4), n odd and p ≡ 3 (mod 8), and E is Q-isomorphic to one of the followingelliptic curves:43a2 a4 ∆B1 · qb−1ψ(4qm − pn) qm+2(b−1) −24q2m+6(b−1)pnB2 − · 2 · qb−1ψ(4qm − pn) −q2(b−1)pn 28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.3. There exist integers m ≥ 2−b and n ≥ 1 such that pn−4qm is a square, qm ≡ 1 (mod 4),n odd and p ≡ 5 (mod 8), and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 · qb−1ψ(pn − 4qm) −qm+2(b−1) 24q2m+6(b−1)pnC2 − · 2 · qb−1ψ(pn − 4qm) q2(b−1)pn −28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.4. There exist integers m ≥ 2−b and n ≥ 1 such that 4pn+qm is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 qb−1ψ(4pn + qm) q2(b−1)pn 24qm+6(b−1)p2nD2 −2 · qb−1ψ(4pn + qm) qm+2(b−1) 28q2m+6(b−1)pn5. There exist integers m ≥ 2−b and n ≥ 1 such that 4pn−qm is a square, q ≡ 3 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 · qb−1ψ(4pn − qm) q2(b−1)pn −24qm+6(b−1)p2nE2 − · 2 · qb−1ψ(4pn − qm) −qm+2(b−1) 28q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.6. There exist integers m ≥ 2−b and n ≥ 1 such that qm−4pn is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 qb−1ψ(qm − 4pn) −q2(b−1)pn 24qm+6(b−1)p2nF2 −2 · qb−1ψ(qm − 4pn) qm+2(b−1) −28q2m+6(b−1)pn7. There exist integers m ≥ 2−b and n ≥ 1 such that qmpn−4 is a square, p, q ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves:44a2 a4 ∆G1 qb−1ψ(qmpn − 4) −q2(b−1) 24qm+6(b−1)pnG2 −2 · qb−1ψ(qmpn − 4) qm+2(b−1)pn −210q2m+6(b−1)p2nIn the case that b = 2, we furthermore could have one of the following conditionssatisfied:8. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn+1q is a square, pn ≡ 3 (mod 4),q ≡ 5 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆H1 qs+1ψ(4pn+1q ) q2s+1pn 24q6s+3p2nH2 −2 · qs+1ψ(4pn+1q ) q2s+1 28q6s+3pn9. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn−1q is a square, pn ≡ 1 (mod 4),q ≡ 3 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆I1 · qs+1ψ(4pn−1q ) q2s+1pn −24q6s+3p2nI2 − · 2 · qs+1ψ(4pn−1q ) −q2s+1 28q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.10. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4+pnq is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆J1 · qs+1ψ(4+pnq ) q2s+1 24q6s+3pnJ2 − · 2 · qs+1ψ(4+pnq ) q2s+1pn 28q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.11. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−4q is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆L1 qs+1ψ(pn−4q ) −q2s+1 24q6s+3pnL2 −2 · qs+1ψ(pn−4q ) q2s+1pn −28q6s+3p2n45Theorem 2.2.4. The elliptic curves E defined over Q of conductor 23qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers ` ∈ {4, 5}, m ≥ 2− b and n ≥ 1 such that 2`qmpn + 1 is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 · qb−1ψ(2`qmpn + 1) 2`−2qm+2(b−1)pn 22`q2m+6(b−1)p2nA2 − · 2 · qb−1ψ(2`qmpn + 1) q2(b−1) 2`+6qm+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.2. There exist integers m ≥ 2−b and n ≥ 1 such that 4qm+pn is a square, qm ≡ 1 (mod 4),n odd and p ≡ 5 (mod 8), and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 − · qb−1ψ(4qm + pn) qm+2(b−1) 24q2m+6(b−1)pnB2 · 2 · qb−1ψ(4qm + pn) q2(b−1)pn 28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.3. There exist integers m ≥ 2−b and n ≥ 1 such that 4qm−pn is a square, qm ≡ 1 (mod 4),n odd and p ≡ 3 (mod 8), and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 − · qb−1ψ(4qm − pn) qm+2(b−1) −24q2m+6(b−1)pnC2 · 2 · qb−1ψ(4qm − pn) −q2(b−1)pn 28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.4. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − 4qm is a square, qm ≡−1 (mod 4), n odd and p ≡ 5 (mod 8), and E is Q-isomorphic to one of the followingelliptic curves:a2 a4 ∆D1 − · qb−1ψ(pn − 4qm) −qm+2(b−1) 24q2m+6(b−1)pnD2 · 2 · qb−1ψ(pn − 4qm) q2(b−1)pn −28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.465. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2`qm + pn is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 · qb−1ψ(2`qm + pn) 2`−2qm+2(b−1) 22`q2m+6(b−1)pnE2 − · 2 · qb−1ψ(2`qm + pn) q2(b−1)pn 2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.6. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2`qm − pn is a square,p ≡ 7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 · qb−1ψ(2`qm − pn) 2`−2qm+2(b−1) −22`q2m+6(b−1)pnF2 − · 2 · qb−1ψ(2`qm − pn) −q2(b−1)pn 2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.7. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that pn − 2`qm is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆G1 · qb−1ψ(pn − 2`qm) −2`−2qm+2(b−1) 22`q2m+6(b−1)pnG2 − · 2 · qb−1ψ(pn − 2`qm) q2(b−1)pn −2`+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.8. There exist integers m ≥ 2−b and n ≥ 1 such that 4pn+qm is a square, q ≡ 5 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆H1 −qb−1ψ(4pn + qm) q2(b−1)pn 24qm+6(b−1)p2nH2 2 · qb−1ψ(4pn + qm) qm+2(b−1) 28q2m+6(b−1)pn9. There exist integers m ≥ 2−b and n ≥ 1 such that 4pn−qm is a square, q ≡ 3 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆I1 − · qb−1ψ(4pn − qm) q2(b−1)pn −24qm+6(b−1)p2nI2 · 2 · qb−1ψ(4pn − qm) −qm+2(b−1) 28q2m+6(b−1)pn47where ∈ {±1} is the residue of qb−1 modulo 4.10. There exist integers m ≥ 2−b and n ≥ 1 such that qm−4pn is a square, q ≡ 5 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆J1 −qb−1ψ(qm − 4pn) −q2(b−1)pn 24qm+6(b−1)p2nJ2 2 · qb−1ψ(qm − 4pn) qm+2(b−1) −28q2m+6(b−1)pn11. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2`pn + qm is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆K1 · qb−1ψ(2`pn + qm) 2`−2q2(b−1)pn 22`qm+6(b−1)p2nK2 − · 2 · qb−1ψ(2`pn + qm) qm+2(b−1) 2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.12. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2`pn − qm is a square,q ≡ 7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆L1 · qb−1ψ(2`pn − qm) 2`−2q2(b−1)pn −22`qm+6(b−1)p2nL2 − · 2 · qb−1ψ(2`pn − qm) −qm+2(b−1) 2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.13. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that qm − 2`pn is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆M1 · qb−1ψ(qm − 2`pn) −2`−2q2(b−1)pn 22`qm+6(b−1)p2nM2 − · 2 · qb−1ψ(qm − 2`pn) qm+2(b−1) −2`+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.14. There exist integers m ≥ 2 − b and n ≥ 1 such that 4 + qmpn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆N1 − · qb−1ψ(4 + qmpn) q2(b−1) 24qm+6(b−1)pnN2 · 2 · qb−1ψ(4 + qmpn) qm+2(b−1)pn 28q2m+6(b−1)p2n48where ∈ {±1} is the residue of qb−1 modulo 4.15. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2` + qmpn is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆O1 · qb−1ψ(2` + qmpn) 2`−2q2(b−1) 22`qm+6(b−1)pnO2 − · 2 · qb−1ψ(2` + qmpn) qm+2(b−1)pn 2`+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.16. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that 2` − qmpn is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆P1 · qb−1ψ(2` − qmpn) 2`−2q2(b−1) −22`qm+6(b−1)pnP2 − · 2 · qb−1ψ(2` − qmpn) −qm+2(b−1)pn 2`+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.17. There exist integers ` ∈ {4, 5}, m ≥ 2 − b and n ≥ 1 such that qmpn − 2` is a squareand E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆Q1 · qb−1ψ(qmpn − 2`) −2`−2q2(b−1) 22`qm+6(b−1)pnQ2 − · 2 · qb−1ψ(qmpn − 2`) qm+2(b−1)pn −2`+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.In the case that b = 2, we furthermore could have one of the following conditionssatisfied:18. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn+1q is a square, pn ≡ 1 (mod 4),q ≡ 5 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆R1 −qs+1ψ(4pn+1q ) q2s+1pn 24q6s+3p2nR2 2 · qs+1ψ(4pn+1q ) q2s+1 28q6s+3pn19. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn−1q is a square, pn ≡ 3 (mod 4),q ≡ 3 (mod 8) and E is Q-isomorphic to one of the following curves:49a2 a4 ∆S1 − · qs+1ψ(4pn−1q ) q2s+1pn −24q6s+3p2nS2 · 2 · qs+1ψ(4pn−1q ) −q2s+1 28q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.20. there exist integers ` ∈ {4, 5}, n ≥ 1 and s ∈ {0, 1} such that 2`pn+1q is a square,q ≡ 1 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆T1 qs+1ψ(2`pn+1q ) 2`−2q2s+1pn 22`q6s+3p2nT2 −2 · qs+1ψ(2`pn+1q ) q2s+1 2`+6q6s+3pn21. there exist integers ` ∈ {4, 5}, n ≥ 1 and s ∈ {0, 1} such that 2`pn−1q is a square,q ≡ 7 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆U1 · qs+1ψ(2`pn−1q ) 2`−2q2s+1pn −22`q6s+3p2nU2 − · 2 · qs+1ψ(2`pn−1q ) −q2s+1 2`+6q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.22. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4+pnq is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆V1 −qs+1ψ(4+pnq ) q2s+1 24q6s+3pnV2 2 · qs+1ψ(4+pnq ) q2s+1pn 28q6s+3p2n23. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−4q is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆W1 − · qs+1ψ(pn−4q ) −q2s+1 24q6s+3pnW2 · 2 · qs+1ψ(pn−4q ) q2s+1pn −28q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.24. there exist integers ` ∈ {4, 5}, n ≥ 1 and s ∈ {0, 1} such that 2`+pnq is a square and Eis Q-isomorphic to one of the following curves:50a2 a4 ∆X1 · qs+1ψ(2`+pnq ) 2`−2q2s+1 22`q6s+3pnX2 − · 2 · qs+1ψ(2`+pnq ) q2s+1pn 2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.25. there exist integers ` ∈ {4, 5}, n ≥ 1 and s ∈ {0, 1} such that 2`−pnq is a square and Eis Q-isomorphic to one of the following curves:a2 a4 ∆Y1 · qs+1ψ(2`−pnq ) 2`−2q2s+1 −22`q6s+3pnY2 − · 2 · qs+1ψ(2`−pnq ) −q2s+1pn 2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.26. there exist integers ` ∈ {4, 5}, n ≥ 1 and s ∈ {0, 1} such that pn−2`q is a square and Eis Q-isomorphic to one of the following curves:a2 a4 ∆Z1 · qs+1ψ(pn−2`q ) −2`−2q2s+1 22`q6s+3pnZ2 − · 2 · qs+1ψ(pn−2`q ) q2s+1pn −2`+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.Theorem 2.2.5. The elliptic curves E defined over Q of conductor 24qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that 2lqmpn + 1 is a square andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 − · qb−1ψ(2lqmpn + 1) 2l−2qm+2(b−1)pn 22lq2m+6(b−1)p2nA2 · 2 · qb−1ψ(2lqmpn + 1) q2(b−1) 2l+6qm+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.2. There exist integers m ≥ 2−b and n ≥ 1 such that 4qm+pn is a square, qm ≡ 1 (mod 4),n even or p ≡ 1 (mod 4), and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 · qb−1ψ(4qm + pn) qm+2(b−1) 24q2m+6(b−1)pnB2 − · 2 · qb−1ψ(4qm + pn) q2(b−1)pn 28qm+6(b−1)p2n51where ∈ {±1} is the residue of qm+b−1 modulo 4.3. There exist integers m ≥ 2−b and n ≥ 1 such that 4qm−pn is a square, qm ≡ 1 (mod 4),n odd and p ≡ 3 (mod 8), and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 · qb−1ψ(4qm − pn) qm+2(b−1) −24q2m+6(b−1)pnC2 − · 2 · qb−1ψ(4qm − pn) −q2(b−1)pn 28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qm+b−1 modulo 4.4. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − 4qm is a square, qm ≡−1 (mod 4), n even or p ≡ 1 (mod 4), and E is Q-isomorphic to one of the followingelliptic curves:a2 a4 ∆D1 · qb−1ψ(pn − 4qm) −qm+2(b−1) 24q2m+6(b−1)pnD2 − · 2 · qb−1ψ(pn − 4qm) q2(b−1)pn −28qm+6(b−1)p2nwhere ∈ {±1} is the residue of qm+b−1 modulo 4.5. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that 2lqm + pn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 − · qb−1ψ(2lqm + pn) 2l−2qm+2(b−1) 22lq2m+6(b−1)pnE2 · 2 · qb−1ψ(2lqm + pn) q2(b−1)pn 2l+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.6. There exist integers ` ≥ 4, m ≥ 2 − b and n ≥ 1 such that 2lqm − pn is a square,p ≡ 7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 − · qb−1ψ(2lqm − pn) 2l−2qm+2(b−1) −22lq2m+6(b−1)pnF2 · 2 · qb−1ψ(2lqm − pn) −q2(b−1)pn 2l+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.7. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that pn− 2lqm is a square and Eis Q-isomorphic to one of the following elliptic curves:52a2 a4 ∆G1 − · qb−1ψ(pn − 2lqm) −2l−2qm+2(b−1) 22lq2m+6(b−1)pnG2 · 2 · qb−1ψ(pn − 2lqm) q2(b−1)pn −2l+6qm+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.8. There exist integers m ≥ 2−b and n ≥ 1 such that 4pn+qm is a square, q ≡ 5 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆H1 qb−1ψ(4pn + qm) q2(b−1)pn 24qm+6(b−1)p2nH2 −2 · qb−1ψ(4pn + qm) qm+2(b−1) 28q2m+6(b−1)pn9. There exist integers m ≥ 2− b and n ≥ 1 such that 4pn−qm is a square, q ≡ 3 (mod 8)and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆I1 · qb−1ψ(4pn − qm) q2(b−1)pn −24qm+6(b−1)p2nI2 − · 2 · qb−1ψ(4pn − qm) −qm+2(b−1) 28q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1pn modulo 4.10. There exist integers m ≥ 2−b and n ≥ 1 such that qm−4pn is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆J1 qb−1ψ(qm − 4pn) −q2(b−1)pn 24qm+6(b−1)p2nJ2 −2 · qb−1ψ(qm − 4pn) qm+2(b−1) −28q2m+6(b−1)pn11. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that 2lpn + qm is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆K1 − · qb−1ψ(2lpn + qm) 2l−2q2(b−1)pn 22lqm+6(b−1)p2nK2 · 2 · qb−1ψ(2lpn + qm) qm+2(b−1) 2l+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.12. There exist integers ` ≥ 4, m ≥ 2 − b and n ≥ 1 such that 2lpn − qm is a square,q ≡ 7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curves:53a2 a4 ∆L1 − · qb−1ψ(2lpn − qm) 2l−2q2(b−1)pn −22lqm+6(b−1)p2nL2 · 2 · qb−1ψ(2lpn − qm) −qm+2(b−1) 2l+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.13. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that qm− 2lpn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆M1 − · qb−1ψ(qm − 2lpn) −2l−2q2(b−1)pn 22lqm+6(b−1)p2nM2 · 2 · qb−1ψ(qm − 2lpn) qm+2(b−1) −2l+6q2m+6(b−1)pnwhere ∈ {±1} is the residue of qb−1 modulo 4.14. There exist integers m ≥ 2 − b and n ≥ 1 such that 4 + qmpn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆N1 · qb−1ψ(4 + qmpn) q2(b−1) 24qm+6(b−1)pnN2 − · 2 · qb−1ψ(4 + qmpn) qm+2(b−1)pn 28q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.15. There exist integers m ≥ 2−b and n ≥ 1 such that qmpn−4 is a square, p ≡ 1 (mod 4),q ≡ 1 (mod 4) if m ≥ 1 and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆O1 − · qb−1ψ(qmpn − 4) −q2(b−1) 24qm+6(b−1)pnO2 · 2 · qb−1ψ(qmpn − 4) qm+2(b−1)pn −210q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.16. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that 2l + qmpn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆P1 − · qb−1ψ(2l + qmpn) 2l−2q2(b−1) 22lqm+6(b−1)pnP2 · 2 · qb−1ψ(2l + qmpn) qm+2(b−1)pn 2l+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.5417. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that 2l − qmpn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆Q1 − · qb−1ψ(2l − qmpn) 2l−2q2(b−1) −22lqm+6(b−1)pnQ2 · 2 · qb−1ψ(2l − qmpn) −qm+2(b−1)pn 2l+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.18. There exist integers ` ≥ 4, m ≥ 2− b and n ≥ 1 such that qlpn− 2m is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆R1 − · qb−1ψ(qlpn − 2m) −2l−2q2(b−1) 22lqm+6(b−1)pnR2 · 2 · qb−1ψ(qlpn − 2m) qm+2(b−1)pn −2l+6q2m+6(b−1)p2nwhere ∈ {±1} is the residue of qb−1 modulo 4.In the case that b = 2, we furthermore could have one of the following conditionssatisfied:19. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn+1q is a square, pn ≡ 1 (mod 4),q ≡ 5 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆S1 qs+1ψ(4pn+1q ) q2s+1pn 24q6s+3p2nS2 −2 · qs+1ψ(4pn+1q ) q2s+1 28q6s+3pn20. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4pn−1q is a square, pn ≡ 3 (mod 4),q ≡ 3 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆T1 · qs+1ψ(4pn−1q ) q2s+1pn −24q6s+3p2nT2 − · 2 · qs+1ψ(4pn−1q ) −q2s+1 28q6s+3pnwhere ∈ {±1} is the residue of qspn modulo 4.21. there exist integers ` ≥ 4, n ≥ 1 and s ∈ {0, 1} such that 2lpn+1q is a square, q ≡1 (mod 8) and E is Q-isomorphic to one of the following curves:55a2 a4 ∆U1 −qs+1ψ(2lpn+1q ) 2l−2q2s+1pn 22lq6s+3p2nU2 2 · qs+1ψ(2lpn+1q ) q2s+1 2l+6q6s+3pn22. there exist integers ` ≥ 4, n ≥ 1 and s ∈ {0, 1} such that 2lpn−1q is a square, q ≡7 (mod 8) and E is Q-isomorphic to one of the following curves:a2 a4 ∆V1 − · qs+1ψ(2lpn−1q ) 2l−2q2s+1pn −22lq6s+3p2nV2 · 2 · qs+1ψ(2lpn−1q ) −q2s+1 2l+6q6s+3pnwhere ∈ {±1} is the residue of qs+1 modulo 4.23. there exist integers n ≥ 1 and s ∈ {0, 1} such that 4+pnq is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆W1 qs+1ψ(4+pnq ) q2s+1 24q6s+3pnW2 −2 · qs+1ψ(4+pnq ) q2s+1pn 28q6s+3p2nwhere ∈ {±1} is the residue of qs modulo 4.24. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−4q is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆X1 · qs+1ψ(pn−4q ) −q2s+1 24q6s+3pnX2 − · 2 · qs+1ψ(pn−4q ) q2s+1pn −28q6s+3p2nwhere ∈ {±1} is the residue of −qs modulo 4.25. there exist integers ` ≥ 4, n ≥ 1 and s ∈ {0, 1} such that 2l+pnq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆Y1 − · qs+1ψ(2l+pnq ) 2l−2q2s+1 22lq6s+3pnY2 · 2 · qs+1ψ(2l+pnq ) q2s+1pn 2l+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.5626. there exist integers ` ≥ 4, n ≥ 1 and s ∈ {0, 1} such that 2l−pnq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆Z1 − · qs+1ψ(2l−pnq ) 2l−2q2s+1 −22lq6s+3pnZ2 · 2 · qs+1ψ(2l−pnq ) −q2s+1pn 2l+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.27. there exist integers ` ≥ 4, n ≥ 1 and s ∈ {0, 1} such that pn−2lq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆AA1 − · qs+1ψ(pn−2lq ) −2l−2q2s+1 22lq6s+3pnAA2 · 2 · qs+1ψ(pn−2lq ) q2s+1pn −2l+6q6s+3p2nwhere ∈ {±1} is the residue of qs+1 modulo 4.Theorem 2.2.6. The elliptic curves E defined over Q of conductor 25qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers m ≥ 2 − b and n ≥ 1 such that 8qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 qb−1ψ(8qmpn + 1) 2qm+2(b−1)pn 26q2m+6(b−1)p2nA2 −2 · qb−1ψ(8qmpn + 1) q2(b−1) 29qm+6(b−1)pnA1′ −qb−1ψ(8qmpn + 1) 2qm+2(b−1)pn 26q2m+6(b−1)p2nA2′ 2 · qb−1ψ(8qmpn + 1) q2(b−1) 29qm+6(b−1)pn2. There exist integers m ≥ 2 − b and n ≥ 1 such that 8qm + pn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 qb−1ψ(8qm + pn) 2qm+2(b−1) 26q2m+6(b−1)pnB2 −2 · qb−1ψ(8qm + pn) q2(b−1)pn 29qm+6(b−1)p2nB1′ −qb−1ψ(8qm + pn) 2qm+2(b−1) 26q2m+6(b−1)pnB2′ 2 · qb−1ψ(8qm + pn) q2(b−1)pn 29qm+6(b−1)p2n3. There exist integers m ≥ 2− b and n ≥ 1 such that 8qm−pn is a square, p ≡ 7 (mod 8)and E is Q-isomorphic to one of the following elliptic curves:57a2 a4 ∆C1 qb−1ψ(8qm − pn) 2qm+2(b−1) −26q2m+6(b−1)pnC2 −2 · qb−1ψ(8qm − pn) −q2(b−1)pn 29qm+6(b−1)p2nC1′ −qb−1ψ(8qm − pn) 2qm+2(b−1) −26q2m+6(b−1)pnC2′ 2 · qb−1ψ(8qm − pn) −q2(b−1)pn 29qm+6(b−1)p2n4. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − 8qm is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 qb−1ψ(pn − 8qm −2qm+2(b−1) 26q2m+6(b−1)pnD2 −2 · qb−1ψ(pn − 8qm q2(b−1)pn −29qm+6(b−1)p2nD1′ −qb−1ψ(pn − 8qm −2qm+2(b−1) 26q2m+6(b−1)pnD2′ 2 · qb−1ψ(pn − 8qm q2(b−1)pn −29qm+6(b−1)p2n5. There exist integers m ≥ 2 − b and n ≥ 1 such that 8pn + qm is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 qb−1ψ(8pn + qm) 2q2(b−1)pn 26qm+6(b−1)p2nE2 −2 · qb−1ψ(8pn + qm) qm+2(b−1) 29q2m+6(b−1)pnE1′ −qb−1ψ(8pn + qm) 2q2(b−1)pn 26qm+6(b−1)p2nE2′ 2 · qb−1ψ(8pn + qm) qm+2(b−1) 29q2m+6(b−1)pn6. There exist integers m ≥ 2−b and n ≥ 1 such that 8pn−qm is a square, q ≡ 7 (mod 8),m odd and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 qb−1ψ(8pn − qm) 2q2(b−1)pn −26qm+6(b−1)p2nF2 −2 · qb−1ψ(8pn − qm) −qm+2(b−1) 29q2m+6(b−1)pnF1′ −qb−1ψ(8pn − qm) 2q2(b−1)pn −26qm+6(b−1)p2nF2′ 2 · qb−1ψ(8pn − qm) −qm+2(b−1) 29q2m+6(b−1)pn7. There exist integers m ≥ 2 − b and n ≥ 1 such that qm − 8pn is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆G1 qb−1ψ(qm − 8pn) −2q2(b−1)pn 26qm+6(b−1)p2nG2 −2 · qb−1ψ(qm − 8pn) qm+2(b−1) −29q2m+6(b−1)pnG1′ −qb−1ψ(qm − 8pn) −2q2(b−1)pn 26qm+6(b−1)p2nG2′ 2 · qb−1ψ(qm − 8pn) qm+2(b−1) −29q2m+6(b−1)pn588. There exist integers m ≥ 2 − b and n ≥ 1 such that 8 + qmpn is a square, p ≡1, 7 (mod 8), q ≡ 1, 7 (mod 8) if m > 0, and E is Q-isomorphic to one of the followingelliptic curves:a2 a4 ∆H1 qb−1ψ(8 + qmpn) 2q2(b−1) 26qm+6(b−1)pnH2 −2 · qb−1ψ(8 + qmpn) qm+2(b−1)pn 29q2m+6(b−1)p2nH1′ −qb−1ψ(8 + qmpn) 2q2(b−1) 26qm+6(b−1)pnH2′ 2 · qb−1ψ(8 + qmpn) qm+2(b−1)pn 29q2m+6(b−1)p2n9. There exist integers m ≥ 2 − b and n ≥ 1 such that qmpn − 8 is a square, p ≡1, 3 (mod 8), q ≡ 1, 3 (mod 8) if m > 0, and E is Q-isomorphic to one of the followingelliptic curves:a2 a4 ∆I1 qb−1ψ(qmpn − 8) −2q2(b−1) 26qm+6(b−1)pnI2 −2 · qb−1ψ(qmpn − 8) qm+2(b−1)pn −29q2m+6(b−1)p2nI1′ −qb−1ψ(qmpn − 8) −2q2(b−1) 26qm+6(b−1)pnI2′ 2 · qb−1ψ(qmpn − 8) qm+2(b−1)pn −29q2m+6(b−1)p2n10. There exist integers m ≥ 2 − b and n ≥ 1 such that qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curves:a2 a4 ∆J1 2qb−1ψ(qmpn + 1) qm+2(b−1)pn 26q2m+6(b−1)p2nJ2 −4 · qb−1ψ(qmpn + 1) 4q2(b−1) 212qm+6(b−1)pnJ1′ −2qb−1ψ(qmpn + 1) qm+2(b−1)pn 26q2m+6(b−1)p2nJ2′ 4 · qb−1ψ(qmpn + 1) 4q2(b−1) 212qm+6(b−1)pn11. There exist integers m ≥ 2− b and n ≥ 1 such that qmpn−1 is a square, p ≡ 1 (mod 4)if n > 0, q ≡ 1 (mod 4) if m > 0, and E is Q-isomorphic to one of the following ellipticcurves:a2 a4 ∆K1 2qb−1ψ(qmpn − 1) q2(b−1) 26q2m+6(b−1)p2nK2 −4 · qb−1ψ(qmpn − 1) 4qm+2(b−1)pn 212qm+6(b−1)pnK1′ −2qb−1ψ(qmpn − 1) q2(b−1) 26q2m+6(b−1)p2nK2′ 4 · qb−1ψ(qmpn − 1) 4qm+2(b−1)pn 212qm+6(b−1)pn5912. There exist integers m ≥ 2 − b and n ≥ 1 such that qm + pn is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆L1 2qb−1ψ(qm + pn) q2(b−1)pn 26qm+6(b−1)p2nL2 −4 · qb−1ψ(qm + pn) 4qm+2(b−1) 212q2m+6(b−1)pnL1′ −2qb−1ψ(qm + pn) q2(b−1)pn 26qm+6(b−1)p2nL2′ 4 · qb−1ψ(qm + pn) 4qm+2(b−1) 212q2m+6(b−1)pn(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆M1 2qb−1ψ(qm + pn) qm+2(b−1) 26q2m+6(b−1)pnM2 −4 · qb−1ψ(qm + pn) 4q2(b−1)pn 212qm+6(b−1)p2nM1′ −2qb−1ψ(qm + pn) qm+2(b−1) 26q2m+6(b−1)pnM2′ 4 · qb−1ψ(qm + pn) 4q2(b−1)pn 212qm+6(b−1)p2n13. There exist integers m ≥ 2 − b and n ≥ 1 such that qm − pn is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆N1 2qb−1ψ(qm − pn) −q2(b−1)pn 26qm+6(b−1)p2nN2 −4 · qb−1ψ(qm − pn) 4qm+2(b−1) −212q2m+6(b−1)pnN1′ −2qb−1ψ(qm − pn) −q2(b−1)pn 26qm+6(b−1)p2nN2′ 4 · qb−1ψ(qm − pn) 4qm+2(b−1) −212q2m+6(b−1)pn(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆O1 2qb−1ψ(qm − pn) qm+2(b−1) −26q2m+6(b−1)pnO2 −4 · qb−1ψ(qm − pn) −4q2(b−1)pn 212qm+6(b−1)p2nO1′ −2qb−1ψ(qm − pn) qm+2(b−1) −26q2m+6(b−1)pnO2′ 4 · qb−1ψ(qm − pn) −4q2(b−1)pn 212qm+6(b−1)p2n14. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − qm is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);60a2 a4 ∆P1 2qb−1ψ(pn − qm) −q2(b−1)pn 26qm+6(b−1)p2nP2 −4 · qb−1ψ(pn − qm) 4qm+2(b−1) −212q2m+6(b−1)pnP1′ −2qb−1ψ(pn − qm) −q2(b−1)pn 26qm+6(b−1)p2nP2′ 4 · qb−1ψ(pn − qm) 4qm+2(b−1) −212q2m+6(b−1)pn(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆Q1 2qb−1ψ(pn − qm) qm+2(b−1) −26q2m+6(b−1)pnQ2 −4 · qb−1ψ(pn − qm) −4q2(b−1)pn 212qm+6(b−1)p2nQ1′ −2qb−1ψ(pn − qm) qm+2(b−1) −26q2m+6(b−1)pnQ2′ 4 · qb−1ψ(pn − qm) −4q2(b−1)pn 212qm+6(b−1)p2nIn the case that b = 2, we furthermore could have one of the following conditionssatisfied:15. there exist integers n ≥ 1 and s ∈ {0, 1} such that 8pn+1q is a square, q ≡ 1 (mod 8)and E is Q-isomorphic to one of the following curves:a2 a4 ∆R1 qs+1ψ(8pn+1q ) 2q2s+1pn 26q6s+3p2nR2 −2 · qs+1ψ(8pn+1q ) q2s+1 29q6s+3pnR1′ −qs+1ψ(8pn+1q ) 2q2s+1pn 26q6s+3p2nR2′ 2 · qs+1ψ(8pn+1q ) q2s+1 29q6s+3pn16. there exist integers n ≥ 1 and s ∈ {0, 1} such that 8pn−1q is a square, q ≡ 7 (mod 8)and E is Q-isomorphic to one of the following curves:a2 a4 ∆S1 qs+1ψ(8pn−1q ) 2q2s+1pn −26q6s+3p2nS2 −2 · qs+1ψ(8pn−1q ) −q2s+1 29q6s+3pnS1′ −qs+1ψ(8pn−1q ) 2q2s+1pn −26q6s+3p2nS2′ 2 · qs+1ψ(8pn−1q ) −q2s+1 29q6s+3pn17. there exist integers n ≥ 1 and s ∈ {0, 1} such that 8+pnq is a square and E is Q-isomorphic to one of the following curves:61a2 a4 ∆T1 qs+1ψ(8+pnq ) 2q2s+1 26q6s+3pnT2 −2 · qs+1ψ(8+pnq ) q2s+1pn 29q6s+3p2nT1′ −qs+1ψ(8+pnq ) 2q2s+1 26q6s+3pnT2′ 2 · qs+1ψ(8+pnq ) q2s+1pn 29q6s+3p2n18. there exist integers n ≥ 1 and s ∈ {0, 1} such that 8−pnq is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆U1 qs+1ψ(8−pnq ) 2q2s+1 −26q6s+3pnU2 −2 · qs+1ψ(8−pnq ) −q2s+1pn 29q6s+3p2nU1′ −qs+1ψ(8−pnq ) 2q2s+1 −26q6s+3pnU2′ 2 · qs+1ψ(8−pnq ) −q2s+1pn 29q6s+3p2n19. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−8q is a square and E is Q-isomorphic to one of the following curves:a2 a4 ∆V1 qs+1ψ(pn−8q ) −2q2s+1 26q6s+3pnV2 −2 · qs+1ψ(pn−8q ) q2s+1pn −29q6s+3p2nV1′ −qs+1ψ(pn−8q ) −2q2s+1 26q6s+3pnV2′ 2 · qs+1ψ(pn−8q ) q2s+1pn −29q6s+3p2n20. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn+1q is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆W1 qs+1ψ(pn+1q ) q2s+1pn 26q6s+3p2nW2 −2 · qs+1ψ(pn+1q ) 4q2s+1 212q6s+3pnW1′ −qs+1ψ(pn+1q ) q2s+1pn 26q6s+3p2nW2′ 2 · qs+1ψ(pn+1q ) 4q2s+1 212q6s+3pn21. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn+1q is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆X1 qs+1ψ(pn+1q ) q2s+1 26q6s+3pnX2 −2 · qs+1ψ(pn+1q ) 4q2s+1pn 212q6s+3p2nX1′ −qs+1ψ(pn+1q ) q2s+1 26q6s+3pnX2′ 2 · qs+1ψ(pn+1q ) 4q2s+1pn 212q6s+3p2n6222. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−1q is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆Y1 qs+1ψ(pn−1q ) −q2s+1 26q6s+3pnY2 −2 · qs+1ψ(pn−1q ) 4q2s+1pn −212q6s+3p2nY1′ −qs+1ψ(pn−1q ) −q2s+1 26q6s+3pnY2′ 2 · qs+1ψ(pn−1q ) 4q2s+1pn −212q6s+3p2n23. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−1q is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆Z1 qs+1ψ(pn−1q ) q2s+1pn −26q6s+3p2nZ2 −2 · qs+1ψ(pn−1q ) −4q2s+1 212q6s+3pnZ1′ −qs+1ψ(pn−1q ) q2s+1pn −26q6s+3p2nZ2′ 2 · qs+1ψ(pn−1q ) −4q2s+1 212q6s+3pnTheorem 2.2.7. The elliptic curves E defined over Q of conductor 26qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers ` ≥ 3, m ≥ 2− b and n ≥ 1 such that 2lqmpn + 1 is a square andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 2qb−1ψ(2lqmpn + 1) 2lqm+2(b−1)pn 22l+6q2m+6(b−1)p2nA2 −4 · qb−1ψ(2lqmpn + 1) 4q2(b−1) 2l+12qm+6(b−1)pnA1′ −2qb−1ψ(2lqmpn + 1) 2lqm+2(b−1)pn 22l+6q2m+6(b−1)p2nA2′ 4 · qb−1ψ(2lqmpn + 1) 4q2(b−1) 2l+12qm+6(b−1)pn2. There exist integers ` ≥ 2, m ≥ 2− b and n ≥ 1 such that 2lqm + pn is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 2qb−1ψ(2lqm + pn) 2lqm+2(b−1) 22l+6q2m+6(b−1)pnB2 −4 · qb−1ψ(2lqm + pn) 4q2(b−1)pn 2l+12qm+6(b−1)p2nB1′ −2qb−1ψ(2lqm + pn) 2lqm+2(b−1) 22l+6q2m+6(b−1)pnB2′ 4 · qb−1ψ(2lqm + pn) 4q2(b−1)pn 2l+12qm+6(b−1)p2n633. There exist integers ` ≥ 2, m ≥ 2− b and n ≥ 1 such that 2lqm− pn is a square, n odd,p ≡ 3 (mod 4) and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 2qb−1ψ(2lqm − pn) 2lqm+2(b−1) −22l+6q2m+6(b−1)pnC2 −4 · qb−1ψ(2lqm − pn) −4q2(b−1)pn 2l+12qm+6(b−1)p2nC1′ −2qb−1ψ(2lqm − pn) 2lqm+2(b−1) −22l+6q2m+6(b−1)pnC2′ 4 · qb−1ψ(2lqm − pn) −4q2(b−1)pn 2l+12qm+6(b−1)p2n4. There exist integers ` ≥ 2, m ≥ 2− b and n ≥ 1 such that pn− 2lqm is a square and Eis Q-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 2qb−1ψ(pn − 2lqm −2lqm+2(b−1) 22l+6q2m+6(b−1)pnD2 −4 · qb−1ψ(pn − 2lqm 4q2(b−1)pn −2l+12qm+6(b−1)p2nD1′ −2qb−1ψ(pn − 2lqm −2lqm+2(b−1) 22l+6q2m+6(b−1)pnD2′ 4 · qb−1ψ(pn − 2lqm 4q2(b−1)pn −2l+12qm+6(b−1)p2n5. There exist integers ` ≥ 2, m ≥ 2 − b and n ≥ 1 such that 2lpn + qm is a square,q ≡ 5 (mod 8) when l = 2 and E is Q-isomorphic to one of the following ellipticcurves:a2 a4 ∆E1 2qb−1ψ(2lpn + qm) 2lq2(b−1)pn 22l+6qm+6(b−1)p2nE2 −4 · qb−1ψ(2lpn + qm) 4qm+2(b−1) 2l+12q2m+6(b−1)pnE1′ −2qb−1ψ(2lpn + qm) 2lq2(b−1)pn 22l+6qm+6(b−1)p2nE2′ 4 · qb−1ψ(2lpn + qm) 4qm+2(b−1) 2l+12q2m+6(b−1)pn6. There exist integers ` ≥ 2, m ≥ 2 − b and n ≥ 1 such that 2lpn − qm is a square,q ≡ 7 (mod 8) when l ≥ 3 or q ≡ 3 (mod 8) if l = 2, m odd, and E is Q-isomorphic toone of the following elliptic curves:a2 a4 ∆F1 2qb−1ψ(2lpn − qm) 2lq2(b−1)pn −22l+6qm+6(b−1)p2nF2 −4 · qb−1ψ(2lpn − qm) −4qm+2(b−1) 2l+12q2m+6(b−1)pnF1′ −2qb−1ψ(2lpn − qm) 2lq2(b−1)pn −22l+6qm+6(b−1)p2nF2′ 4 · qb−1ψ(2lpn − qm) −4qm+2(b−1) 2l+12q2m+6(b−1)pn647. There exist integers ` ≥ 2, m ≥ 2 − b and n ≥ 1 such that qm − 2lpn is a square,q ≡ 5 (mod 8) when l = 2 and E is Q-isomorphic to one of the following ellipticcurves:a2 a4 ∆G1 2qb−1ψ(qm − 2lpn) −2lq2(b−1)pn 22l+6qm+6(b−1)p2nG2 −4 · qb−1ψ(qm − 2lpn) 4qm+2(b−1) −2l+12q2m+6(b−1)pnG1′ −2qb−1ψ(qm − 2lpn) −2lq2(b−1)pn 22l+6qm+6(b−1)p2nG2′ 4 · qb−1ψ(qm − 2lpn) 4qm+2(b−1) −2l+12q2m+6(b−1)pn8. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2l + qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 2qb−1ψ(2l + qmpn) 2lq2(b−1) 22l+6qm+6(b−1)pnH2 −4 · qb−1ψ(2l + qmpn) 4qm+2(b−1)pn 2l+12q2m+6(b−1)p2nH1′ −2qb−1ψ(2l + qmpn) 2lq2(b−1) 22l+6qm+6(b−1)pnH2′ 4 · qb−1ψ(2l + qmpn) 4qm+2(b−1)pn 2l+12q2m+6(b−1)p2n9. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2l − qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 2qb−1ψ(2l − qmpn) 2lq2(b−1) −22l+6qm+6(b−1)pnI2 −4 · qb−1ψ(2l − qmpn) −4qm+2(b−1)pn 2l+12q2m+6(b−1)p2nI1′ −2qb−1ψ(2l − qmpn) 2lq2(b−1) −22l+6qm+6(b−1)pnI2′ 4 · qb−1ψ(2l − qmpn) −4qm+2(b−1)pn 2l+12q2m+6(b−1)p2n10. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that qmpn − 2l is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 2qb−1ψ(qmpn − 2l) −2lq2(b−1) 22l+6qm+6(b−1)pnJ2 −4 · qb−1ψ(qmpn − 2l) 4qm+2(b−1)pn −2l+12q2m+6(b−1)p2nJ1′ −2qb−1ψ(qmpn − 2l) −2lq2(b−1) 22l+6qm+6(b−1)pnJ2′ 4 · qb−1ψ(qmpn − 2l) 4qm+2(b−1)pn −2l+12q2m+6(b−1)p2n11. There exist integers m ≥ 2 − b and n ≥ 1 such that qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curves:65a2 a4 ∆K1 2qb−1ψ(qmpn + 1) q2(b−1) 26q2m+6(b−1)p2nK2 −4 · qb−1ψ(qmpn + 1) 4qm+2(b−1)pn 212qm+6(b−1)pnK1′ −2qb−1ψ(qmpn + 1) q2(b−1) 26q2m+6(b−1)p2nK2′ 4 · qb−1ψ(qmpn + 1) 4qm+2(b−1)pn 212qm+6(b−1)pn12. There exist integers m ≥ 2−b and n ≥ 1 such that qmpn−1 is a square, p ≡ 1 (mod 4),q ≡ 1 (mod 4) if m > 0 and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆L1 2qb−1ψ(qmpn − 1) qm+2(b−1)pn 26q2m+6(b−1)p2nL2 −4 · qb−1ψ(qmpn − 1) 4q2(b−1) 212qm+6(b−1)pnL1′ −2qb−1ψ(qmpn − 1) qm+2(b−1)pn 26q2m+6(b−1)p2nL2′ 4 · qb−1ψ(qmpn − 1) 4q2(b−1) 212qm+6(b−1)pn13. There exist integers m ≥ 2 − b and n ≥ 1 such that qm + pn is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆M1 2qb−1ψ(qm + pn) qm+2(b−1) 26q2m+6(b−1)pnM2 −4 · qb−1ψ(qm + pn) 4q2(b−1)pn 212qm+6(b−1)p2nM1′ −2qb−1ψ(qm + pn) qm+2(b−1) 26q2m+6(b−1)pnM2′ 4 · qb−1ψ(qm + pn) 4q2(b−1)pn 212qm+6(b−1)p2n(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆N1 2qb−1ψ(qm + pn) q2(b−1)pn 26qm+6(b−1)p2nN2 −4 · qb−1ψ(qm + pn) 4qm+2(b−1) 212q2m+6(b−1)pnN1′ −2qb−1ψ(qm + pn) q2(b−1)pn 26qm+6(b−1)p2nN2′ 4 · qb−1ψ(qm + pn) 4qm+2(b−1) 212q2m+6(b−1)pn14. There exist integers m ≥ 2 − b and n ≥ 1 such that qm − pn is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);66a2 a4 ∆O1 2qb−1ψ(qm − pn) qm+2(b−1) −26q2m+6(b−1)pnO2 −4 · qb−1ψ(qm − pn) −4q2(b−1)pn 212qm+6(b−1)p2nO1′ −2qb−1ψ(qm − pn) qm+2(b−1) −26q2m+6(b−1)pnO2′ 4 · qb−1ψ(qm − pn) −4q2(b−1)pn 212qm+6(b−1)p2n(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆P1 2qb−1ψ(qm − pn) −q2(b−1)pn 26qm+6(b−1)p2nP2 −4 · qb−1ψ(qm − pn) 4qm+2(b−1) −212q2m+6(b−1)pnP1′ −2qb−1ψ(qm − pn) −q2(b−1)pn 26qm+6(b−1)p2nP2′ 4 · qb−1ψ(qm − pn) 4qm+2(b−1) −212q2m+6(b−1)pn15. There exist integers m ≥ 2 − b and n ≥ 1 such that pn − qm is a square and E isQ-isomorphic to one of the following elliptic curves:(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆Q1 2qb−1ψ(pn − qm) qm+2(b−1) −26q2m+6(b−1)pnQ2 −4 · qb−1ψ(pn − qm) −4q2(b−1)pn 212qm+6(b−1)p2nQ1′ −2qb−1ψ(pn − qm) qm+2(b−1) −26q2m+6(b−1)pnQ2′ 4 · qb−1ψ(pn − qm) −4q2(b−1)pn 212qm+6(b−1)p2n(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆R1 2qb−1ψ(pn − qm) −q2(b−1)pn 26qm+6(b−1)p2nR2 −4 · qb−1ψ(pn − qm) 4qm+2(b−1) −212q2m+6(b−1)pnR1′ −2qb−1ψ(pn − qm) −q2(b−1)pn 26qm+6(b−1)p2nR2′ 4 · qb−1ψ(pn − qm) 4qm+2(b−1) −212q2m+6(b−1)pnIn the case that b = 2, we furthermore could have one of the following conditionssatisfied:16. there exist integers ` ≥ 2, n ≥ 1 and s ∈ {0, 1} such that 2lpn+1q is a square, q ≡1 (mod 4) when l = 2 or q ≡ 1 (mod 8) when l ≥ 3 and E is Q-isomorphic to one ofthe following curves:67a2 a4 ∆S1 2qs+1ψ(2lpn+1q ) 2lq2s+1pn 22l+6q6s+3p2nS2 −4 · qs+1ψ(2lpn+1q ) 4q2s+1 2l+12q6s+3pnS1′ −2qs+1ψ(2lpn+1q ) 2lq2s+1pn 22l+6q6s+3p2nS2′ 4 · qs+1ψ(2lpn+1q ) 4q2s+1 2l+12q6s+3pn17. there exist integers ` ≥ 2, n ≥ 1 and s ∈ {0, 1} such that 2lpn−1q is a square, q ≡3 (mod 4) when l = 2 or q ≡ 7 (mod 8) when l ≥ 3 and E is Q-isomorphic to one ofthe following curves:a2 a4 ∆T1 2qs+1ψ(2lpn−1q ) 2lq2s+1pn −22l+6q6s+3p2nT2 −4 · qs+1ψ(2lpn−1q ) −4q2s+1 2l+12q6s+3pnT1′ −2qs+1ψ(2lpn−1q ) 2lq2s+1pn −22l+6q6s+3p2nT2′ 4 · qs+1ψ(2lpn−1q ) −4q2s+1 2l+12q6s+3pn18. there exist integers ` ≥ 2, n ≥ 1 and s ∈ {0, 1} such that 2l+pnq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆U1 2qs+1ψ(2l+pnq ) 2lq2s+1 22l+6q6s+3pnU2 −4 · qs+1ψ(2l+pnq ) 4q2s+1pn 2l+12q6s+3p2nU1′ −2qs+1ψ(2l+pnq ) 2lq2s+1 22l+6q6s+3pnU2′ 4 · qs+1ψ(2l+pnq ) 4q2s+1pn 2l+12q6s+3p2n19. there exist integers ` ≥ 2, n ≥ 1 and s ∈ {0, 1} such that 2l−pnq is a square and E isQ-isomorphic to one of the following curves:a2 a4 ∆V1 2qs+1ψ(2l−pnq ) 2lq2s+1 −22l+6q6s+3pnV2 −4 · qs+1ψ(2l−pnq ) −4q2s+1pn 2l+12q6s+3p2nV1′ −2qs+1ψ(2l−pnq ) 2lq2s+1 −22l+6q6s+3pnV2′ 4 · qs+1ψ(2l−pnq ) −4q2s+1pn 2l+12q6s+3p2n20. there exist integers ` ≥ 2, n ≥ 1 and s ∈ {0, 1} such that pn−2lq is a square and E isQ-isomorphic to one of the following curves:68a2 a4 ∆W1 2qs+1ψ(pn−2lq ) −2lq2s+1 22l+6q6s+3pnW2 −4 · qs+1ψ(pn−2lq ) 4q2s+1pn −2l+12q6s+3p2nW1′ −2qs+1ψ(pn−2lq ) −2lq2s+1 22l+6q6s+3pnW2′ 4 · qs+1ψ(pn−2lq ) 4q2s+1pn −2l+12q6s+3p2n21. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn+1q is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆X1 qs+1ψ(pn+1q ) q2s+1pn 26q6s+3p2nX2 −2 · qs+1ψ(pn+1q ) 4q2s+1 212q6s+3pnX1′ −qs+1ψ(pn+1q ) q2s+1pn 26q6s+3p2nX2′ 2 · qs+1ψ(pn+1q ) 4q2s+1 212q6s+3pn22. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn+1q is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆Y1 qs+1ψ(pn+1q ) q2s+1 26q6s+3pnY2 −2 · qs+1ψ(pn+1q ) 4q2s+1pn 212q6s+3p2nY1′ −qs+1ψ(pn+1q ) q2s+1 26q6s+3pnY2′ 2 · qs+1ψ(pn+1q ) 4q2s+1pn 212q6s+3p2n23. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−1q is a square, q ≡ 3 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆Z1 qs+1ψ(pn−1q ) −q2s+1 26q6s+3pnZ2 −2 · qs+1ψ(pn−1q ) 4q2s+1pn −212q6s+3p2nZ1′ −qs+1ψ(pn−1q ) −q2s+1 26q6s+3pnZ2′ 2 · qs+1ψ(pn−1q ) 4q2s+1pn −212q6s+3p2n24. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−1q is a square, q ≡ 1 (mod 4) andE is Q-isomorphic to one of the following curves:a2 a4 ∆AA1 qs+1ψ(pn−1q ) q2s+1pn −26q6s+3p2nAA2 −2 · qs+1ψ(pn−1q ) −4q2s+1 212q6s+3pnAA1′ −qs+1ψ(pn−1q ) q2s+1pn −26q6s+3p2nAA2′ 2 · qs+1ψ(pn−1q ) −4q2s+1 212q6s+3pn69Theorem 2.2.8. The elliptic curves E defined over Q of conductor 27qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers m ≥ 2 − b and n ≥ 1 such that 2qmpn − 1 is a square, p, q ≡1 (mod 4), t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆A1 2t+1qb−1ψ(2qmpn − 1) 21+2tqm+2(b−1)pn −28+6tq2m+6(b−1)p2nA2 −22−t · qb−1ψ(2qmpn − 1) −22−2tq2(b−1) 213−6tqm+6(b−1)pnA1′ −2t+1qb−1ψ(2qmpn − 1) 21+2tqm+2(b−1)pn −28+6tq2m+6(b−1)p2nA2′ 22−t · qb−1ψ(2qmpn − 1) −22−2tq2(b−1) 213−6tqm+6(b−1)pn2. There exist integers m ≥ 2−b and n ≥ 1 such that 2qm+pn is a square, p ≡ 3 (mod 4),n odd, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆B1 2t+1qb−1ψ(2qm + pn) 21+2tqm+2(b−1) 28+6tq2m+6(b−1)pnB2 −22−t · qb−1ψ(2qm + pn) 22−2tq2(b−1)pn 213−6tqm+6(b−1)p2nB1′ −2t+1qb−1ψ(2qm + pn) 21+2tqm+2(b−1) 28+6tq2m+6(b−1)pnB2′ 22−t · qb−1ψ(2qm + pn) 22−2tq2(b−1)pn 213−6tqm+6(b−1)p2n3. There exist integers m ≥ 2− b and n ≥ 1 such that 2qm − pn is a square, n is even orp ≡ 1 (mod 4), t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆C1 2t+1qb−1ψ(2qm − pn) 21+2tqm+2(b−1) −28+6tq2m+6(b−1)pnC2 −22−t · qb−1ψ(2qm − pn) −22−2tq2(b−1)pn 213−6tqm+6(b−1)p2nC1′ −2t+1qb−1ψ(2qm − pn) 21+2tqm+2(b−1) −28+6tq2m+6(b−1)pnC2′ 22−t · qb−1ψ(2qm − pn) −22−2tq2(b−1)pn 213−6tqm+6(b−1)p2n4. There exist integers m ≥ 2−b and n ≥ 1 such that pn−2qm is a square, p ≡ 3 (mod 4),n odd, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆D1 2t+1qb−1ψ(pn − 2qm) −21+2tqm+2(b−1) 28+6tq2m+6(b−1)pnD2 −22−t · qb−1ψ(pn − 2qm) 22−2tq2(b−1)pn −213−6tqm+6(b−1)p2nD1′ −2t+1qb−1ψ(pn − 2qm) −21+2tqm+2(b−1) 28+6tq2m+6(b−1)pnD2′ 22−t · qb−1ψ(pn − 2qm) 22−2tq2(b−1)pn −213−6tqm+6(b−1)p2n705. There exist integers m ≥ 2− b and n ≥ 1 such that 2pn + qm is a square, t ∈ {0, 1} andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆E1 2t+1qb−1ψ(2pn + qm) 21+2tq2(b−1)pn 28+6tqm+6(b−1)p2nE2 −22−t · qb−1ψ(2pn + qm) 22−2tqm+2(b−1) 213−6tq2m+6(b−1)pnE1′ −2t+1qb−1ψ(2pn + qm) 21+2tq2(b−1)pn 28+6tqm+6(b−1)p2nE2′ 22−t · qb−1ψ(2pn + qm) 22−2tqm+2(b−1) 213−6tq2m+6(b−1)pn6. There exist integers m ≥ 2− b and n ≥ 1 such that 2pn− qm is a square, t ∈ {0, 1} andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆F1 2t+1qb−1ψ(2pn − qm) 21+2tq2(b−1)pn −28+6tqm+6(b−1)p2nF2 −22−t · qb−1ψ(2pn − qm) −22−2tqm+2(b−1) 213−6tq2m+6(b−1)pnF1′ −2t+1qb−1ψ(2pn − qm) 21+2tq2(b−1)pn −28+6tqm+6(b−1)p2nF2′ 22−t · qb−1ψ(2pn − qm) −22−2tqm+2(b−1) 213−6tq2m+6(b−1)pn7. There exist integers m ≥ 2− b and n ≥ 1 such that qm− 2pn is a square, t ∈ {0, 1} andE is Q-isomorphic to one of the following elliptic curves:a2 a4 ∆G1 2t+1qb−1ψ(qm − 2pn) −21+2tq2(b−1)pn 28+6tqm+6(b−1)p2nG2 −22−t · qb−1ψ(qm − 2pn) 22−2tqm+2(b−1) −213−6tq2m+6(b−1)pnG1′ −2t+1qb−1ψ(qm − 2pn) −21+2tq2(b−1)pn 28+6tqm+6(b−1)p2nG2′ 22−t · qb−1ψ(qm − 2pn) 22−2tqm+2(b−1) −213−6tq2m+6(b−1)pn8. There exist integers m ≥ 2 − b and n ≥ 1 such that 2 + qmpn is a square, p ≡ 1 or7 (mod 8), q ≡ 1 or 7 (mod 8) if m > 0, t ∈ {0, 1} and E is Q-isomorphic to one ofthe following elliptic curves:a2 a4 ∆H1 2t+1qb−1ψ(2 + qmpn) 21+2tq2(b−1) 28+6tqm+6(b−1)p2nH2 −22−t · qb−1ψ(2 + qmpn) 22−2tqm+2(b−1)pn 213−6tq2m+6(b−1)pnH1′ −2t+1qb−1ψ(2 + qmpn) 21+2tq2(b−1) 28+6tqm+6(b−1)p2nH2′ 22−t · qb−1ψ(2 + qmpn) 22−2tqm+2(b−1)pn 213−6tq2m+6(b−1)pn9. There exist integers m ≥ 2 − b and n ≥ 1 such that qmpn − 2 is a square, p ≡ 1 or3 (mod 8), q ≡ 1 or 3 (mod 8) if m > 0, t ∈ {0, 1} and E is Q-isomorphic to one ofthe following elliptic curves:71a2 a4 ∆I1 2t+1qb−1ψ(qmpn − 2) −21+2tq2(b−1) 28+6tqm+6(b−1)p2nI2 −22−t · qb−1ψ(qmpn − 2) 22−2tqm+2(b−1)pn −213−6tq2m+6(b−1)pnI1′ −2t+1qb−1ψ(qmpn − 2) −21+2tq2(b−1) 28+6tqm+6(b−1)p2nI2′ 22−t · qb−1ψ(qmpn − 2) 22−2tqm+2(b−1)pn −213−6tq2m+6(b−1)pnIn the case that b = 2, we furthermore could have one of the following conditionssatisfied:10. there exist integers n ≥ 1 and s ∈ {0, 1} such that 2pn+1q is a square, q ≡ 3 (mod 4),t ∈ {0, 1} and E is Q-isomorphic to one of the following curves:a2 a4 ∆J1 2t+1qs+1ψ(2pn+1q ) 21+2tq2s+1pn 28+6tq6s+3p2nJ2 −22−t · qs+1ψ(2pn+1q ) 22−2tq2s+1 213−6tq6s+3pnJ1′ −2t+1qs+1ψ(2pn+1q ) 21+2tq2s+1pn 28+6tq6s+3p2nJ2′ 22−t · qs+1ψ(2pn+1q ) 22−2tq2s+1 213−6tq6s+3pn11. there exist integers n ≥ 1 and s ∈ {0, 1} such that 2pn−1q is a square, q ≡ 1 (mod 4),t ∈ {0, 1} and E is Q-isomorphic to one of the following curves:a2 a4 ∆K1 2t+1qs+1ψ(2pn−1q ) 21+2tq2s+1pn −28+6tq6s+3p2nK2 −22−t · qs+1ψ(2pn−1q ) −22−2tq2s+1 213−6tq6s+3pnK1′ −2t+1qs+1ψ(2pn−1q ) 21+2tq2s+1pn −28+6tq6s+3p2nK2′ 22−t · qs+1ψ(2pn−1q ) −22−2tq2s+1 213−6tq6s+3pn12. there exist integers n ≥ 1 and s ∈ {0, 1} such that 2+pnq is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following curves:a2 a4 ∆L1 2t+1qb−1ψ(2+pnq ) 21+2tq2(b−1) 28+6tq6s+3pnL2 −22−t · qb−1ψ(2+pnq ) 22−2tq2(b−1)pn 213−6tq6s+3p2nL1′ −2t+1qb−1ψ(2+pnq ) 21+2tq2(b−1) 28+6tq6s+3pnL2′ 22−t · qb−1ψ(2+pnq ) 22−2tq2(b−1)pn 213−6tq6s+3p2n13. there exist integers n ≥ 1 and s ∈ {0, 1} such that pn−2q is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following curves:72a2 a4 ∆M1 2t+1qb−1ψ(pn−2q ) −21+2tqm+2(b−1) 28+6tq6s+3pnM2 −22−t · qb−1ψ(pn−2q ) 22−2tq2(b−1)pn −213−6tq6s+3p2nM1′ −2t+1qb−1ψ(pn−2q ) −21+2tqm+2(b−1) 28+6tq6s+3pnM2′ 22−t · qb−1ψ(pn−2q ) 22−2tq2(b−1)pn −213−6tq6s+3p2nTheorem 2.2.9. The elliptic curves E defined over Q of conductor 28qbp and having at leastone rational point of order 2 are the ones such that one of the following conditions is satisfied:1. There exist integers m ≥ 0 and n ≥ 0 such that qmpn+12 is a square, p, q ≡ 1, 7 (mod 8),t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 2t+2qb−1ψ( qmpn+12 ) 21+2tqm+2(b−1)pn −29+6tq2m+6(b−1)p2nA2 −23−tqb−1ψ( qmpn+12 ) −23−2tq2(b−1) 215−6tqm+6(b−1)pnA1′ −2t+2qb−1ψ( qmpn+12 ) 21+2tqm+2(b−1)pn −29+6tq2m+6(b−1)p2nA2′ 23−tqb−1ψ( qmpn+12 ) −23−2tq2(b−1) 215−6tqm+6(b−1)pn2. There exist integers m ≥ 0 and n ≥ 0 such that qmpn−12 is a square, p, q ≡ 1, 3 (mod 8),t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 2t+2qb−1ψ( qmpn−12 ) 21+2tqm+2(b−1)pn −29+6tq2m+6(b−1)p2nB2 −23−tqb−1ψ( qmpn−12 ) −23−2tq2(b−1) 215−6tqm+6(b−1)pnB1′ −2t+2qb−1ψ( qmpn−12 ) 21+2tqm+2(b−1)pn −29+6tq2m+6(b−1)p2nB2′ 23−tqb−1ψ( qmpn−12 ) −23−2tq2(b−1) 215−6tqm+6(b−1)pn3. There exist integers m ≥ 0 and n ≥ 0 such that qm+pn2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 2t+2qb−1ψ( qm+pn2 ) 21+2tqm+2(b−1) 29+6tq2m+6(b−1)pnC2 −23−tqb−1ψ( qm+pn2 ) 23−2tq2(b−1)pn 215−6tqm+6(b−1)p2nC1′ −2t+2qb−1ψ( qm+pn2 ) 21+2tqm+2(b−1) 29+6tq2m+6(b−1)pnC2′ 23−tqb−1ψ( qm+pn2 ) 23−2tq2(b−1)pn 215−6tqm+6(b−1)p2n4. There exist integers m ≥ 0 and n ≥ 0 such that qm−pn2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curves73a2 a4 ∆D1 2t+2qb−1ψ( qm−pn2 ) 21+2tqm+2(b−1) −29+6tq2m+6(b−1)pnD2 −23−tqb−1ψ( qm−pn2 ) −23−2tq2(b−1)pn 215−6tqm+6(b−1)p2nD1′ −2t+2qb−1ψ( qm−pn2 ) 21+2tqm+2(b−1) −29+6tq2m+6(b−1)pnD2′ 23−tqb−1ψ( qm−pn2 ) −23−2tq2(b−1)pn 215−6tqm+6(b−1)p2n5. There exist integers m ≥ 0 and n ≥ 0 such that pn−qm2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 2t+2qb−1ψ(pn−qm2 ) 21+2tq2(b−1)pn −29+6tqm+6(b−1)p2nE2 −23−tqb−1ψ(pn−qm2 ) −23−2tqm+2(b−1) 215−6tq2m+6(b−1)pnE1′ −2t+2qb−1ψ(pn−qm2 ) 21+2tq2(b−1)pn −29+6tqm+6(b−1)p2nE2′ 23−tqb−1ψ(pn−qm2 ) −23−2tqm+2(b−1) 215−6tq2m+6(b−1)pnIn the case that b = 2, we furthermore could have one of the following conditionssatisfied:6. There exist integers m ≥ 0 and n ≥ 0 such that pn+12q is a square, pn ≡ 1 (mod 4),s, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn 29+6tq6s+3p2nF2 −23−tqs+1pψ(pn+1q ) 23−2tq2s+1 215−6tq6s+3pnF1′ −2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn 29+6tq6s+3p2nF2′ 23−tqs+1pψ(pn+1q ) 23−2tq2s+1 215−6tq6s+3pn7. There exist integers m ≥ 0 and n ≥ 0 such that pn−12q is a square, pn ≡ 3 (mod 4),s, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn −29+6tq6s+3p2nG2 −23−tqs+1pψ(pn+1q ) −23−2tq2s+1 215−6tq6s+3pnG1′ −2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn −29+6tq6s+3p2nG2′ 23−tqs+1pψ(pn+1q ) −23−2tq2s+1 215−6tq6s+3pnTheorem 2.2.10. The elliptic curves E defined over Q of conductor q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:741. There exist integers m ≥ 0 and n ≥ 0 such that 26qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 · qpψ(26qmpn + 1) 24qm+2pn+2 212q2m+6p2n+6A2 − · 2 · qpψ(26qmpn + 1) q2p2 212qm+6pn+6where ∈ {±1} is the residue of qp modulo 4.2. There exist integers m ≥ 0 and n ≥ 0 such that 26qm + pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 · qpψ(26qm + pn) 24qm+2p2 212q2m+6pn+6B2 − · 2 · qpψ(26qm + pn) q2pn+2 212qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.3. There exist integers m ≥ 0 and n ≥ 0 such that 26qm − pn is a square, p ≡ 7 (mod 8),n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 · qpψ(2`qm − pn) 24qm+2p2 −212q2m+6pn+6C2 − · 2 · qpψ(26qm − pn) −q2pn+2 212qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.4. There exist integers m ≥ 0 and n ≥ 0 such that pn − 26qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 · qpψ(pn − 26qm) −24qm+2p2 212q2m+6pn+6D2 − · 2 · qpψ(pn − 26qm) q2pn+2 −212qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.5. There exist integers m ≥ 0 and n ≥ 0 such that 26pn + qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 · qpψ(26pn + qm) 24q2pn+2 212qm+6p2n+6E2 − · 2 · qpψ(26pn + qm) qm+2p2 212q2m+6pn+675where ∈ {±1} is the residue of qp modulo 4.6. There exist integers m ≥ 0 and n ≥ 0 such that 26pn − qm is a square, q ≡ 7 (mod 8),m odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 · qpψ(26pn − qm) 24q2pn+2 −212qm+6p2n+6F2 − · 2 · qpψ(26pn − qm) −qm+2p2 212q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.7. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that qm − 26pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 · qpψ(qm − 26pn) −24q2pn+2 212qm+6p2n+6G2 − · 2 · qpψ(qm − 26pn) qm+2p2 −212q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.8. There exist integers m ≥ 0 and n ≥ 0 such that 26 + qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 · qpψ(26 + qmpn) 24q2p2 212qm+6pn+6H2 − · 2 · qpψ(26 + qmpn) qm+2pn+2 212q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.9. There exist integers m ≥ 0 and n ≥ 0 such that 26 − qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 · qpψ(26 − qmpn) 24q2p2 −212qm+6pn+6I2 − · 2 · qpψ(26 − qmpn) −qm+2pn+2 212q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.10. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 26 is a square and E isQ-isomorphic to one of the following elliptic curves76a2 a4 ∆J1 · qpψ(qmpn − 26) −24q2p2 212qm+6pn+6J2 − · 2 · qpψ(qmpn − 26) qm+2pn+2 −212q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.11. There exist integers m ≥ 0 and r ∈ {0, 1} such that 26qm+1p is a square, p ≡ 1 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆K1 · qpr+1ψ(26qm+1p ) 24qm+2p2r+1 212q2m+6p6r+3K2 − · 2 · qpr+1ψ(26qm+1p ) q2p2r+1 212qm+6p6r+3where ∈ {±1} is the residue of q modulo 4.12. There exist integers m ≥ 0 and r ∈ {0, 1} such that 26qm−1p is a square, p ≡ 7 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 · qpr+1ψ(26qm−1p ) 24qm+2p2r+1 −212q2m+6p6r+3L2 − · 2 · qpr+1ψ(26qm−1p ) −q2p2r+1 212qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.13. There exist integers m ≥ 0 and r ∈ {0, 1} such that 26+qmp is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 · qpr+1ψ(26+qmp ) 24q2p2r+1 212qm+6p6r+3M2 − · 2 · qpr+1ψ(26+qmp ) qm+2p2r+1 212q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.14. There exist integers m ≥ 0 and r ∈ {0, 1} such that 26−qmp is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 · qpr+1ψ(26−qmp ) 24q2p2r+1 −212qm+6p6r+3N2 − · 2 · qpr+1ψ(26−qmp ) −qm+2p2r+1 212q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.7715. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−26p is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆O1 · qpr+1ψ( qm−26p ) −24q2p2r+1 212qm+6p6r+3O2 − · 2 · qpr+1ψ( qm−26p ) qm+2p2r+1 −212q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.16. There exist integers n ≥ 0 and s ∈ {0, 1} such that 26pn+1q is a square, q ≡ 1 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆P1 · qs+1pψ(26pn+1q ) 24q2s+1pn+2 212q6s+3p2n+6P2 − · 2 · qs+1pψ(26pn+1q ) q2s+1p2 212q6s+3pn+6where ∈ {±1} is the residue of p modulo 4.17. There exist integers n ≥ 0 and s ∈ {0, 1} such that 26pn−1q is a square, q ≡ 7 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Q1 · qs+1pψ(26pn−1q ) 24q2s+1pn+2 −212q6s+3p2n+6Q2 − · 2 · qs+1pψ(26pn−1q ) −q2s+1p2 212q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.18. There exist integers n ≥ 0 and s ∈ {0, 1} such that 26+pnq is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆R1 · qs+1pψ(26+pnq ) 24q2s+1p2 212q6s+3pn+6R2 − · 2 · qs+1pψ(26+pnq ) q2s+1pn+2 212q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.19. There exist integers n ≥ 0 and s ∈ {0, 1} such that 26−pnq is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆S1 · qs+1pψ(26−pnq ) 24q2s+1p2 −212q6s+3pn+6S2 − · 2 · qs+1pψ(26−pnq ) −q2s+1pn+2 212q6s+3p2n+678where ∈ {±1} is the residue of qs+1p modulo 4.20. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−26q is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆T1 · qs+1pψ(pn−26q ) −24q2s+1p2 212q6s+3pn+6T2 − · 2 · qs+1pψ(pn−26q ) q2s+1pn+2 −212q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.21. There exist integers r, s ∈ {0, 1} such that 26+1qp is a square and E is Q-isomorphic toone of the following elliptic curvea2 a4 ∆U1 · qs+1pr+1ψ(26+1qp ) 24q2s+1p2r+1 212q6s+3p6r+3U2 − · 2 · qs+1pr+1ψ(26+1qp ) q2s+1p2r+1 212q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.22. There exist integers r, s ∈ {0, 1} such that 26−1qp is a square and E is Q-isomorphic toone of the following elliptic curvea2 a4 ∆V1 · qs+1pr+1ψ(26−1qp ) 24q2s+1p2r+1 −212q6s+3p6r+3V2 − · 2 · qs+1pr+1ψ(26−1qp ) −q2s+1p2r+1 212q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.Theorem 2.2.11. The elliptic curves E defined over Q of conductor 2q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2`qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 · qpψ(2`qmpn + 1) 2`−2qm+2pn+2 22`q2m+6p2n+6A2 − · 2 · qpψ(2`qmpn + 1) q2p2 2`+6qm+6pn+6where ∈ {±1} is the residue of qp modulo 4.792. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2`qm + pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 · qpψ(2`qm + pn) 2`−2qm+2p2 22`q2m+6pn+6B2 − · 2 · qpψ(2`qm + pn) q2pn+2 2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.3. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2`qm − pn is a square, p ≡7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 · qpψ(2`qm − pn) 2`−2qm+2p2 −22`q2m+6pn+6C2 − · 2 · qpψ(2`qm − pn) −q2pn+2 2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.4. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that pn − 2`qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 · qpψ(pn − 2`qm) −2`−2qm+2p2 22`q2m+6pn+6D2 − · 2 · qpψ(pn − 2`qm) q2pn+2 −2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.5. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2`pn + qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 · qpψ(2`pn + qm) 2`−2q2pn+2 22`qm+6p2n+6E2 − · 2 · qpψ(2`pn + qm) qm+2p2 2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.6. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2`pn − qm is a square, q ≡7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 · qpψ(2`pn − qm) 2`−2q2pn+2 −22`qm+6p2n+6F2 − · 2 · qpψ(2`pn − qm) −qm+2p2 2`+6q2m+6pn+680where ∈ {±1} is the residue of qp modulo 4.7. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that qm − 2`pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 · qpψ(qm − 2`pn) −2`−2q2pn+2 22`qm+6p2n+6G2 − · 2 · qpψ(qm − 2`pn) qm+2p2 −2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.8. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2` + qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 · qpψ(2` + qmpn) 2`−2q2p2 22`qm+6pn+6H2 − · 2 · qpψ(2` + qmpn) qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.9. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 2` − qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 · qpψ(2` − qmpn) 2`−2q2p2 −22`qm+6pn+6I2 − · 2 · qpψ(2` − qmpn) −qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.10. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that qmpn − 2` is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 · qpψ(qmpn − 2`) −2`−2q2p2 22`qm+6pn+6J2 − · 2 · qpψ(qmpn − 2`) qm+2pn+2 −2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.11. There exist integers ` ≥ 7, m ≥ 0 and r ∈ {0, 1} such that 2`qm+1p is a square, p ≡1 (mod 8) and E is Q-isomorphic to one of the following elliptic curves81a2 a4 ∆K1 · qpr+1ψ(2`qm+1p ) 2`−2qm+2p2r+1 22`q2m+6p6r+3K2 − · 2 · qpr+1ψ(2`qm+1p ) q2p2r+1 2`+6qm+6p6r+3where ∈ {±1} is the residue of q modulo 4.12. There exist integers ` ≥ 7, m ≥ 0 and r ∈ {0, 1} such that 2`qm−1p is a square, p ≡7 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 · qpr+1ψ(2`qm−1p ) 2`−2qm+2p2r+1 −22`q2m+6p6r+3L2 − · 2 · qpr+1ψ(2`qm−1p ) −q2p2r+1 2`+6qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.13. There exist integers ` ≥ 7, m ≥ 0 and r ∈ {0, 1} such that 2`+qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 · qpr+1ψ(2`+qmp ) 2`−2q2p2r+1 22`qm+6p6r+3M2 − · 2 · qpr+1ψ(2`+qmp ) qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.14. There exist integers ` ≥ 7, m ≥ 0 and r ∈ {0, 1} such that 2`−qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 · qpr+1ψ(2`−qmp ) 2`−2q2p2r+1 −22`qm+6p6r+3N2 − · 2 · qpr+1ψ(2`−qmp ) −qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.15. There exist integers ` ≥ 7, m ≥ 0 and r ∈ {0, 1} such that qm−2`p is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆O1 · qpr+1ψ( qm−2`p ) −2`−2q2p2r+1 22`qm+6p6r+3O2 − · 2 · qpr+1ψ( qm−2`p ) qm+2p2r+1 −2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.8216. There exist integers ` ≥ 7, n ≥ 0 and s ∈ {0, 1} such that 2`pn+1q is a square, q ≡1 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆P1 · qs+1pψ(2`pn+1q ) 2`−2q2s+1pn+2 22`q6s+3p2n+6P2 − · 2 · qs+1pψ(2`pn+1q ) q2s+1p2 2`+6q6s+3pn+6where ∈ {±1} is the residue of p modulo 4.17. There exist integers ` ≥ 7, n ≥ 0 and s ∈ {0, 1} such that 2`pn−1q is a square, q ≡7 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Q1 · qs+1pψ(2`pn−1q ) 2`−2q2s+1pn+2 −22`q6s+3p2n+6Q2 − · 2 · qs+1pψ(2`pn−1q ) −q2s+1p2 2`+6q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.18. There exist integers ` ≥ 7, n ≥ 0 and s ∈ {0, 1} such that 2`+pnq is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆R1 · qs+1pψ(2`+pnq ) 2`−2q2s+1p2 22`q6s+3pn+6R2 − · 2 · qs+1pψ(2`+pnq ) q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.19. There exist integers ` ≥ 7, n ≥ 0 and s ∈ {0, 1} such that 2`−pnq is a square and E isQ-isomorphic to one of the following elliptic curvea2 a4 ∆S1 · qs+1pψ(2`−pnq ) 2`−2q2s+1p2 −22`q6s+3pn+6S2 − · 2 · qs+1pψ(2`−pnq ) −q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.20. There exist integers ` ≥ 7, n ≥ 0 and s ∈ {0, 1} such that pn−2`q is a square and E isQ-isomorphic to one of the following elliptic curvea2 a4 ∆T1 · qs+1pψ(pn−2`q ) −2`−2q2s+1p2 22`q6s+3pn+6T2 − · 2 · qs+1pψ(pn−2`q ) q2s+1pn+2 −2`+6q6s+3p2n+683where ∈ {±1} is the residue of qs+1p modulo 4.21. There exist integers ` ≥ 7 and r, s ∈ {0, 1} such that 2`+1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆U1 · qs+1pr+1ψ(2`+1qp ) 2`−2q2s+1p2r+1 22`q6s+3p6r+3U2 − · 2 · qs+1pr+1ψ(2`+1qp ) q2s+1p2r+1 2`+6q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.22. There exist integers ` ≥ 7 and r, s ∈ {0, 1} such that 2`−1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆V1 · qs+1pr+1ψ(2`−1qp ) 2`−2q2s+1p2r+1 −22`q6s+3p6r+3V2 − · 2 · qs+1pr+1ψ(2`−1qp ) −q2s+1p2r+1 2`+6q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.Theorem 2.2.12. The elliptic curves E defined over Q of conductor 22q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers m ≥ 0 and n ≥ 0 such that 4qm+pn is a square, qm ≡ −1 (mod 4),p ≡ 5 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 · qpψ(4qm + pn) qm+2p2 24q2m+6pn+6A2 − · 2 · qpψ(4qm + pn) q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.2. There exist integers m ≥ 0 and n ≥ 0 such that 4qm−pn is a square, qm ≡ −1 (mod 4),p ≡ 3 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 · qpψ(4qm − pn) qm+2p2 −24q2m+6pn+6B2 − · 2 · qpψ(4qm − pn) −q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.843. There exist integers m ≥ 0 and n ≥ 0 such that pn − 4qm is a square, qm ≡ 1 (mod 4),p ≡ 5 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 · qpψ(pn − 4qm) −qm+2p2 24q2m+6pn+6C2 − · 2 · qpψ(pn − 4qm) q2pn+2 −28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.4. There exist integers m ≥ 0 and n ≥ 0 such that 4pn + qm is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 · qpψ(4pn + qm) q2pn+2 24qm+6p2n+6D2 − · 2 · qpψ(4pn + qm) qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.5. There exist integers m ≥ 0 and n ≥ 0 such that 4pn − qm is a square, q ≡ 3 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 · qpψ(4pn − qm) q2pn+2 −24qm+6p2n+6E2 − · 2 · qpψ(4pn − qm) −qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.6. There exist integers m ≥ 0 and n ≥ 0 such that qm − 4pn is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 · qpψ(qm − 4pn) −q2pn+2 24qm+6p2n+6F2 − · 2 · qpψ(qm − 4pn) qm+2p2 −28q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.7. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 4 is a square, p, q ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 qpψ(qmpn − 4) −q2p2 24qm+6pn+6G2 −2 · qpψ(qmpn − 4) qm+2pn+2 −28q2m+6p2n+6858. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm+1p is a square, p ≡ 5 (mod 8),qm ≡ 3 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 · qpr+1ψ(4qm+1p ) qm+2p2r+1 24q2m+6p6r+3H2 − · 2 · qpr+1ψ(4qm+1p ) q2p2r+1 28qm+6p6r+3where ∈ {±1} is the residue of q modulo 4.9. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm−1p is a square, p ≡ 3 (mod 8),qm ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 · qpr+1ψ(4qm−1p ) qm+2p2r+1 −24q2m+6p6r+3I2 − · 2 · qpr+1ψ(4qm−1p ) −q2p2r+1 28qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.10. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4+qmp is a square, p ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 · qpr+1ψ(4+qmp ) q2p2r+1 24qm+6p6r+3J2 − · 2 · qpr+1ψ(4+qmp ) qm+2p2r+1 28q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.11. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−4p is a square, p ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆K1 · qpr+1ψ( qm−4p ) −q2p2r+1 24qm+6p6r+3K2 − · 2 · qpr+1ψ( qm−4p ) qm+2p2r+1 −28q2m+6p6r+3where ∈ {±1} is the residue of q modulo 4.12. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn+1q is a square, q ≡ 5 (mod 8),pn ≡ 3 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 · qs+1pψ(4pn+1q ) q2s+1pn+2 24q6s+3p2n+6L2 − · 2 · qs+1pψ(4pn+1q ) q2s+1p2 28q6s+3pn+686where ∈ {±1} is the residue of p modulo 4.13. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn−1q is a square, q ≡ 3 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 · qs+1pψ(4pn−1q ) q2s+1pn+2 −24q6s+3p2n+6M2 − · 2 · qs+1pψ(4pn−1q ) −q2s+1p2 28q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.14. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4+pnq is a square, q ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 · qs+1pψ(4+pnq ) q2s+1p2 24q6s+3pn+6N2 − · 2 · qs+1pψ(4+pnq ) q2s+1pn+2 28q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.15. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−4q is a square, q ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆O1 · qs+1pψ(pn−4q ) −q2s+1p2 24q6s+3pn+6O2 − · 2 · qs+1pψ(pn−4q ) q2s+1pn+2 −28q6s+3p2n+6where ∈ {±1} is the residue of p modulo 4.Theorem 2.2.13. The elliptic curves E defined over Q of conductor 23q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2`qmpn + 1 is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 · qpψ(2`qmpn + 1) 2`−2qm+2pn+2 22`q2m+6p2n+6A2 − · 2 · qpψ(2`qmpn + 1) q2p2 2`+6qm+6pn+6where ∈ {±1} is the residue of qp modulo 4.872. There exist integers m ≥ 0 and n ≥ 0 such that 4qm + pn is a square, qm ≡ 1 (mod 4),p ≡ 5 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 − · qpψ(4qm + pn) qm+2p2 24q2m+6pn+6B2 · 2 · qpψ(4qm + pn) q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.3. There exist integers m ≥ 0 and n ≥ 0 such that 4qm − pn is a square, qm ≡ 1 (mod 4),p ≡ 3 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 − · qpψ(4qm − pn) qm+2p2 −24q2m+6pn+6C2 · 2 · qpψ(4qm − pn) −q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.4. There exist integers m ≥ 0 and n ≥ 0 such that pn−4qm is a square, qm ≡ −1 (mod 4),p ≡ 5 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 − · qpψ(pn − 4qm) −qm+2p2 24q2m+6pn+6D2 · 2 · qpψ(pn − 4qm) q2pn+2 −28qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.5. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2`qm + pn is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 · qpψ(2`qm + pn) 2`−2qm+2p2 22`q2m+6pn+6E2 − · 2 · qpψ(2`qm + pn) q2pn+2 2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.6. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2`qm − pn is a square,p ≡ 7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 · qpψ(2`qm − pn) 2`−2qm+2p2 −22`q2m+6pn+6F2 − · 2 · qpψ(2`qm − pn) −q2pn+2 2`+6qm+6p2n+688where ∈ {±1} is the residue of qp modulo 4.7. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that pn − 2`qm is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 · qpψ(pn − 2`qm) −2`−2qm+2p2 22`q2m+6pn+6G2 − · 2 · qpψ(pn − 2`qm) q2pn+2 −2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 41.8. There exist integers m ≥ 0 and n ≥ 0 such that 4pn + qm is a square, q ≡ 5 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 − · qpψ(4pn + qm) q2pn+2 24qm+6p2n+6H2 · 2 · qpψ(4pn + qm) qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of p modulo 4.9. There exist integers m ≥ 0 and n ≥ 0 such that 4pn − qm is a square, q ≡ 3 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 − · qpψ(4pn − qm) q2pn+2 −24qm+6p2n+6I2 · 2 · qpψ(4pn − qm) −qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.10. There exist integers m ≥ 0 and n ≥ 0 such that qm − 4pn is a square, q ≡ 5 (mod 8),pn ≡ −1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 − · qpψ(qm − 4pn) −q2pn+2 24qm+6p2n+6J2 · 2 · qpψ(qm − 4pn) qm+2p2 −28q2m+6pn+6where ∈ {±1} is the residue of p modulo 4.11. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2`pn + qm is a square andE is Q-isomorphic to one of the following elliptic curves1Note the typo in [Mul06, p.113]89a2 a4 ∆K1 · qpψ(2`pn + qm) 2`−2q2pn+2 22`qm+6p2n+6K2 − · 2 · qpψ(2`pn + qm) qm+2p2 2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.12. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2`pn − qm is a square,q ≡ 7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 · qpψ(2`pn − qm) 2`−2q2pn+2 −22`qm+6p2n+6L2 − · 2 · qpψ(2`pn − qm) −qm+2p2 2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.13. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that qm − 2`pn is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 · qpψ(qm − 2`pn) −2`−2q2pn+2 22`qm+6p2n+6M2 − · 2 · qpψ(qm − 2`pn) qm+2p2 −2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.14. There exist integers m ≥ 0 and n ≥ 0 such that 4 + qmpn is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 − · qpψ(4 + qmpn) q2p2 24qm+6pn+6N2 · 2 · qpψ(4 + qmpn) qm+2pn+2 28q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.15. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2` + qmpn is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆O1 · qpψ(2` + qmpn) 2`−2q2p2 22`qm+6pn+6O2 − · 2 · qpψ(2` + qmpn) qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.9016. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that 2` − qmpn is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆P1 · qpψ(2` − qmpn) 2`−2q2p2 −22`qm+6pn+6P2 − · 2 · qpψ(2` − qmpn) −qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.17. There exist integers ` ∈ {4, 5}, m ≥ 0 and n ≥ 0 such that qmpn − 2` is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Q1 · qpψ(qmpn − 2`) −2`−2q2p2 22`qm+6pn+6Q2 − · 2 · qpψ(qmpn − 2`) qm+2pn+2 −2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.18. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm+1p is a square, p ≡ 5 (mod 8),qm ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆R1 −qpr+1ψ(4qm+1p ) qm+2p2r+1 24q2m+6p6r+3R2 2 · qpr+1ψ(4qm+1p ) q2p2r+1 28qm+6p6r+319. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm−1p is a square, p ≡ 3 (mod 8),qm ≡ 3 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆S1 − · qpr+1ψ(4qm−1p ) qm+2p2r+1 −24q2m+6p6r+3S2 · 2 · qpr+1ψ(4qm−1p ) −q2p2r+1 28qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.20. There exist integers ` ∈ {4, 5}, m ≥ 0 and r ∈ {0, 1} such that 2`qm+1p is a square,p ≡ 1 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆T1 · qpr+1ψ(2`qm+1p ) 2`−2qm+2p2r+1 22`q2m+6p6r+3T2 − · 2 · qpr+1ψ(2`qm+1p ) q2p2r+1 2`+6qm+6p6r+391where ∈ {±1} is the residue of q modulo 4.21. There exist integers ` ∈ {4, 5}, m ≥ 0 and r ∈ {0, 1} such that 2`qm−1p is a square,p ≡ 7 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆U1 · qpr+1ψ(2`qm−1p ) 2`−2qm+2p2r+1 −22`q2m+6p6r+3U2 − · 2 · qpr+1ψ(2`qm−1p ) −q2p2r+1 2`+6qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.22. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4+qmp is a square, p ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆V1 − · qpr+1ψ(4+qmp ) q2p2r+1 24qm+6p6r+3V2 · 2 · qpr+1ψ(4+qmp ) qm+2p2r+1 28q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.23. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−4p is a square, p ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆W1 − · qpr+1ψ( qm−4p ) −q2p2r+1 24qm+6p6r+3W2 · 2 · qpr+1ψ( qm−4p ) qm+2p2r+1 −28q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.24. There exist integers ` ∈ {4, 5}, m ≥ 0 and r ∈ {0, 1} such that 2`+qmp is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆X1 · qpr+1ψ(2`+qmp ) 2`−2q2p2r+1 22`qm+6p6r+3X2 − · 2 · qpr+1ψ(2`+qmp ) qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.25. There exist integers ` ∈ {4, 5}, m ≥ 0 and r ∈ {0, 1} such that 2`−qmp is a square andE is Q-isomorphic to one of the following elliptic curves92a2 a4 ∆Y1 · qpr+1ψ(2`−qmp ) 2`−2q2p2r+1 −22`qm+6p6r+3Y2 − · 2 · qpr+1ψ(2`−qmp ) −qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.26. There exist integers ` ∈ {4, 5}, m ≥ 0 and r ∈ {0, 1} such that qm−2`p is a square andE is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Z1 · qpr+1ψ( qm−2`p ) −2`−2q2p2r+1 22`qm+6p6r+3Z2 − · 2 · qpr+1ψ( qm−2`p ) qm+2p2r+1 −2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.27. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn+1q is a square, q ≡ 5 (mod 8),pn ≡ 1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AA1 − · qs+1pψ(4pn+1q ) q2s+1pn+2 24q6s+3p2n+6AA2 · 2 · qs+1pψ(4pn+1q ) q2s+1p2 28q6s+3pn+6where ∈ {±1} is the residue of p modulo 4.28. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn−1q is a square, q ≡ 3 (mod 8),pn ≡ 3 (mod 4) and E is Q-isomorphic to one of the following elliptic curves 2a2 a4 ∆AB1 − · qs+1pψ(4pn−1q ) q2s+1pn+2 −24q6s+3p2n+6AB2 · 2 · qs+1pψ(4pn−1q ) −q2s+1p2 28q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.29. There exist integers ` ∈ {4, 5}, n ≥ 0 and s ∈ {0, 1} such that 2`pn+1q is a square,q ≡ 1 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AC1 · qs+1pψ(2`pn+1q ) 2`−2q2s+1pn+2 22`q6s+3p2n+6AC2 − · 2 · qs+1pψ(2`pn+1q ) q2s+1p2 2`+6q6s+3pn+62Typo in [Mul06, p. 116] the two a2 values there are reversed. Also there are two Z curves.93where ∈ {±1} is the residue of p modulo 4.30. There exist integers ` ∈ {4, 5}, n ≥ 0 and s ∈ {0, 1} such that 2`pn−1q is a square,q ≡ 7 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AD1 · qs+1pψ(2`pn−1q ) 2`−2q2s+1pn+2 −22`q6s+3p2n+6AD2 − · 2 · qs+1pψ(2`pn−1q ) −q2s+1p2 2`+6q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.31. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4+pnq is a square, q ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AE1 − · qs+1pψ(4+pnq ) q2s+1p2 24q6s+3pn+6AE2 · 2 · qs+1pψ(4+pnq ) q2s+1pn+2 28q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.32. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−4q is a square, q ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AF1 − · qs+1pψ(pn−4q ) −q2s+1p2 24q6s+3pn+6AF2 · 2 · qs+1pψ(pn−4q ) q2s+1pn+2 −28q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.33. There exist integers ` ∈ {4, 5}, n ≥ 0 and s ∈ {0, 1} such that 2`+pnq is a square and Eis Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AG1 · qs+1pψ(2`+pnq ) 2`−2q2s+1p2 22`q6s+3pn+6AG2 − · 2 · qs+1pψ(2`+pnq ) q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.34. There exist integers ` ∈ {4, 5}, n ≥ 0 and s ∈ {0, 1} such that 2`−pnq is a square and Eis Q-isomorphic to one of the following elliptic curve94a2 a4 ∆AH1 · qs+1pψ(2`−pnq ) 2`−2q2s+1p2 −22`q6s+3pn+6AH2 − · 2 · qs+1pψ(2`−pnq ) −q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.35. There exist integers ` ∈ {4, 5}, n ≥ 0 and s ∈ {0, 1} such that pn−2`q is a square and Eis Q-isomorphic to one of the following elliptic curvea2 a4 ∆AI1 · qs+1pψ(pn−2`q ) −2`−2q2s+1p2 22`q6s+3pn+6AI2 − · 2 · qs+1pψ(pn−2`q ) q2s+1pn+2 −2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.36. We have ` = 5, p, q ∈ {3, 11} distinct, r, s ∈ {0, 1} and E is Q-isomorphic to one ofthe following elliptic curvea2 a4 ∆AJ1 · qs+1pr+1ψ(25+1qp ) 23q2s+1p2r+1 210q6s+3p6r+3AJ2 − · 2 · qs+1pr+1ψ(25+1qp ) q2s+1p2r+1 211q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.37. We have ` = 4, p, q ∈ {3, 5} distinct, r, s ∈ {0, 1} and E is Q-isomorphic to one of thefollowing elliptic curvea2 a4 ∆AK1 · qs+1pr+1ψ(24−1qp ) 4q2s+1p2r+1 −28q6s+3p6r+3AK2 − · 2 · qs+1pr+1ψ(24−1qp ) −q2s+1p2r+1 210q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.Theorem 2.2.14. The elliptic curves E defined over Q of conductor 24q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2`qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curves95a2 a4 ∆A1 − · qpψ(2`qmpn + 1) 2`−2qm+2pn+2 22`q2m+6p2n+6A2 · 2 · qpψ(2`qmpn + 1) q2p2 2`+6qm+6pn+6where ∈ {±1} is the residue of qp modulo 4.2. There exist integers m ≥ 0 and n ≥ 0 such that 4qm + pn is a square, qm ≡ 1 (mod 4),n even or p ≡ 1 (mod 4), and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 · qpψ(4qm + pn) qm+2p2 24q2m+6pn+6B2 − · 2 · qpψ(4qm + pn) q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qm+1p modulo 4.3. There exist integers m ≥ 0 and n ≥ 0 such that 4qm − pn is a square, qm ≡ 1 (mod 4),n odd and p ≡ 3 (mod 8), and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 · qpψ(4qm − pn) qm+2p2 −24q2m+6pn+6C2 − · 2 · qpψ(4qm − pn) −q2pn+2 28qm+6p2n+6where ∈ {±1} is the residue of qm+1p modulo 4.4. There exist integers m ≥ 0 and n ≥ 0 such that pn−4qm is a square, qm ≡ −1 (mod 4),n even or p ≡ 1 (mod 4), and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 · qpψ(pn − 4qm) −qm+2p2 24q2m+6pn+6D2 − · 2 · qpψ(pn − 4qm) q2pn+2 −28qm+6p2n+6where ∈ {±1} is the residue of qm+1p modulo 4.5. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2`qm + pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 − · qpψ(2`qm + pn) 2`−2qm+2p2 22`q2m+6pn+6E2 · 2 · qpψ(2`qm + pn) q2pn+2 2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.966. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2`qm − pn is a square, p ≡7 (mod 8), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 − · qpψ(2`qm − pn) 2`−2qm+2p2 −22`q2m+6pn+6F2 · 2 · qpψ(2`qm − pn) −q2pn+2 2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.7. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that pn − 2`qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 − · qpψ(pn − 2`qm) −2`−2qm+2p2 22`q2m+6pn+6G2 · 2 · qpψ(pn − 2`qm) q2pn+2 −2`+6qm+6p2n+6where ∈ {±1} is the residue of qp modulo 4.8. There exist integers m ≥ 0 and n ≥ 0 such that 4pn + qm is a square, q ≡ 5 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 · qpψ(4pn + qm) q2pn+2 24qm+6p2n+6H2 − · 2 · qpψ(4pn + qm) qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of pn+1 modulo 4.9. There exist integers m ≥ 0 and n ≥ 0 such that 4pn − qm is a square, q ≡ 3 (mod 8),and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 · qpψ(4pn − qm) q2pn+2 −24qm+6p2n+6I2 − · 2 · qpψ(4pn − qm) −qm+2p2 28q2m+6pn+6where ∈ {±1} is the residue of qpn+1 modulo 4.10. There exist integers m ≥ 0 and n ≥ 0 such that qm − 4pn is a square, q ≡ 5 (mod 8),and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 · qpψ(qm − 4pn) −q2pn+2 24qm+6p2n+6J2 − · 2 · qpψ(qm − 4pn) qm+2p2 −28q2m+6pn+697where ∈ {±1} is the residue of −pn+1 modulo 4.11. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2`pn + qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆K1 − · qpψ(2`pn + qm) 2`−2q2pn+2 22`qm+6p2n+6K2 · 2 · qpψ(2`pn + qm) qm+2p2 2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.12. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2`pn − qm is a square, q ≡7 (mod 8), m odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 − · qpψ(2`pn − qm) 2`−2q2pn+2 −22`qm+6p2n+6L2 · 2 · qpψ(2`pn − qm) −qm+2p2 2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.13. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that qm − 2`pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 − · qpψ(qm − 2`pn) −2`−2q2pn+2 22`qm+6p2n+6M2 · 2 · qpψ(qm − 2`pn) qm+2p2 −2`+6q2m+6pn+6where ∈ {±1} is the residue of qp modulo 4.14. There exist integers m ≥ 0 and n ≥ 0 such that 4 + qmpn is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 · qpψ(4 + qmpn) q2p2 24qm+6pn+6N2 − · 2 · qpψ(4 + qmpn) qm+2pn+2 28q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.15. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 4 is a square, p ≡ 1 (mod 4) ifn ≥ 1, q ≡ 1 (mod 4) if m ≥ 1, and E is Q-isomorphic to one of the following ellipticcurves98a2 a4 ∆O1 −qpψ(qmpn − 4) −q2p2 24qm+6pn+6O2 · 2 · qpψ(qmpn − 4) qm+2pn+2 −28q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.16. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2` + qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆P1 − · qpψ(2` + qmpn) 2`−2q2p2 22`qm+6pn+6P2 · 2 · qpψ(2` + qmpn) qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.17. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that 2` − qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆Q1 − · qpψ(2` − qmpn) 2`−2q2p2 −22`qm+6pn+6Q2 · 2 · qpψ(2` − qmpn) −qm+2pn+2 2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.18. There exist integers ` ≥ 4, m ≥ 0 and n ≥ 0 such that qmpn − 2` is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆R1 − · qpψ(qmpn − 2`) −2`−2q2p2 22`qm+6pn+6R2 · 2 · qpψ(qmpn − 2`) qm+2pn+2 −2`+6q2m+6p2n+6where ∈ {±1} is the residue of qp modulo 4.19. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm+1p is a square, p ≡ 5 (mod 8)and E is Q-isomorphic to one of the following elliptic curves 3a2 a4 ∆S1 · qpr+1ψ(4qm+1p ) qm+2p2r+1 24q2m+6p6r+3S2 − · 2 · qpr+1ψ(4qm+1p ) q2p2r+1 28qm+6p6r+33Typo in [Mul06, p. 120], the 2m should be removed.99where ∈ {±1} is the residue of qm+1pr modulo 4.20. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4qm−1p is a square, p ≡ 3 (mod 8)and E is Q-isomorphic to one of the following elliptic curves 4a2 a4 ∆T1 · qpr+1ψ(4qm−1p ) qm+2p2r+1 −24q2m+6p6r+3T2 − · 2 · qpr+1ψ(4qm−1p ) −q2p2r+1 28qm+6p6r+3where ∈ {±1} is the residue of qm+1pr modulo 4.21. There exist integers ` ≥ 4, m ≥ 0 and r ∈ {0, 1} such that 2`qm+1p is a square, p ≡1 (mod 8) and E is Q-isomorphic to one of the following elliptic curves 5a2 a4 ∆U1 − · qpr+1ψ(2`qm+1p ) 2`−2qm+2p2r+1 22`q2m+6p6r+3U2 · 2 · qpr+1ψ(2`qm+1p ) q2p2r+1 2`+6qm+6p6r+3where ∈ {±1} is the residue of q modulo 4.22. There exist integers ` ≥ 4, m ≥ 0 and r ∈ {0, 1} such that 2`qm−1p is a square, p ≡7 (mod 8) and E is Q-isomorphic to one of the following elliptic curves 6a2 a4 ∆V1 − · qpr+1ψ(2`qm−1p ) 2`−2qm+2p2r+1 −22`q2m+6p6r+3V2 · 2 · qpr+1ψ(2`qm−1p ) −q2p2r+1 2`+6qm+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.23. There exist integers m ≥ 0 and r ∈ {0, 1} such that 4+qmp is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆W1 · qpr+1ψ(4+qmp ) q2p2r+1 24qm+6p6r+3W2 − · 2 · qpr+1ψ(4+qmp ) qm+2p2r+1 28q2m+6p6r+3where ∈ {±1} is the residue of qpr modulo 4.4Typo in [Mul06, p. 121], the 2m should be removed.5Typo in [Mul06, p. 121], the 2m should be 2m−2.6Typo in [Mul06, p. 121], the 2m should be 2m−2.10024. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−4p is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆X1 · qpr+1ψ( qm−4p ) −q2p2r+1 24qm+6p6r+3X2 − · 2 · qpr+1ψ( qm−4p ) qm+2p2r+1 −28q2m+6p6r+3where ∈ {±1} is the residue of −qpr modulo 4.25. There exist integers ` ≥ 4, m ≥ 0 and r ∈ {0, 1} such that 2`+qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆Y1 − · qpr+1ψ(2`+qmp ) 2`−2q2p2r+1 22`qm+6p6r+3Y2 · 2 · qpr+1ψ(2`+qmp ) qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.26. There exist integers ` ≥ 4, m ≥ 0 and r ∈ {0, 1} such that 2`−qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆Z1 − · qpr+1ψ(2`−qmp ) 2`−2q2p2r+1 −22`qm+6p6r+3Z2 · 2 · qpr+1ψ(2`−qmp ) −qm+2p2r+1 2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.27. There exist integers ` ≥ 4, m ≥ 0 and r ∈ {0, 1} such that qm−2`p is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆AA1 − · qpr+1ψ( qm−2`p ) −2`−2q2p2r+1 22`qm+6p6r+3AA2 · 2 · qpr+1ψ( qm−2`p ) qm+2p2r+1 −2`+6q2m+6p6r+3where ∈ {±1} is the residue of qpr+1 modulo 4.28. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn+1q is a square, q ≡ 5 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AB1 · qs+1pψ(4pn+1q ) q2s+1pn+2 24q6s+3p2n+6AB2 − · 2 · qs+1pψ(4pn+1q ) q2s+1p2 28q6s+3pn+6101where ∈ {±1} is the residue of qspn+1 modulo 4.29. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4pn−1q is a square, q ≡ 3 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AC1 · qs+1pψ(4pn−1q ) q2s+1pn+2 −24q6s+3p2n+6AC2 − · 2 · qs+1pψ(4pn−1q ) −q2s+1p2 28q6s+3pn+6where ∈ {±1} is the residue of qspn+1 modulo 4.30. There exist integers ` ≥ 4, n ≥ 0 and s ∈ {0, 1} such that 2`pn+1q is a square, q ≡1 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AD1 − · qs+1pψ(2`pn+1q ) 2`−2q2s+1pn+2 22`q6s+3p2n+6AD2 · 2 · qs+1pψ(2`pn+1q ) q2s+1p2 2`+6q6s+3pn+6where ∈ {±1} is the residue of p modulo 4.31. There exist integers ` ≥ 4, n ≥ 0 and s ∈ {0, 1} such that 2`pn−1q is a square, q ≡7 (mod 8) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AE1 − · qs+1pψ(2`pn−1q ) 2`−2q2s+1pn+2 −22`q6s+3p2n+6AE2 · 2 · qs+1pψ(2`pn−1q ) −q2s+1p2 2`+6q6s+3pn+6where ∈ {±1} is the residue of qs+1p modulo 4.32. There exist integers n ≥ 0 and s ∈ {0, 1} such that 4+pnq is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆AF1 · qs+1pψ(4+pnq ) q2s+1p2 24q6s+3pn+6AF2 − · 2 · qs+1pψ(4+pnq ) q2s+1pn+2 28q6s+3p2n+6where ∈ {±1} is the residue of qsp modulo 4.33. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−4q is a square and E is Q-isomorphic to one of the following elliptic curve102a2 a4 ∆AG1 · qs+1pψ(pn−4q ) −q2s+1p2 24q6s+3pn+6AG2 − · 2 · qs+1pψ(pn−4q ) q2s+1pn+2 −28q6s+3p2n+6where ∈ {±1} is the residue of −qsp modulo 4.34. There exist integers ` ≥ 4, n ≥ 0 and s ∈ {0, 1} such that 2`+pnq is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆AH1 − · qs+1pψ(2`+pnq ) 2`−2q2s+1p2 22`q6s+3pn+6AH2 · 2 · qs+1pψ(2`+pnq ) q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.35. There exist integers ` ≥ 4, n ≥ 0 and s ∈ {0, 1} such that 2`−pnq is a square and E isQ-isomorphic to one of the following elliptic curvea2 a4 ∆AI1 − · qs+1pψ(2`−pnq ) 2`−2q2s+1p2 −22`q6s+3pn+6AI2 · 2 · qs+1pψ(2`−pnq ) −q2s+1pn+2 2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.36. There exist integers ` ≥ 4, n ≥ 0 and s ∈ {0, 1} such that pn−2`q is a square and E isQ-isomorphic to one of the following elliptic curvea2 a4 ∆AJ1 − · qs+1pψ(pn−2`q ) −2`−2q2s+1p2 22`q6s+3pn+6AJ2 · 2 · qs+1pψ(pn−2`q ) q2s+1pn+2 −2`+6q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.37. There exist integers ` ≥ 4 and r, s ∈ {0, 1} such that 2`+1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AK1 − · qs+1pr+1ψ(2`+1qp ) 2`−2q2s+1p2r+1 22`q6s+3p6r+3AK2 · 2 · qs+1pr+1ψ(2`+1qp ) q2s+1p2r+1 2`+6q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.10338. There exist integers ` ≥ 4 and r, s ∈ {0, 1} such that 2`−1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AL1 − · qs+1pr+1ψ(2`−1qp ) 2`−2q2s+1p2r+1 −22`q6s+3p6r+3AL2 · 2 · qs+1pr+1ψ(2`−1qp ) −q2s+1p2r+1 2`+6q6s+3p6r+3where ∈ {±1} is the residue of qs+1pr+1 modulo 4.Theorem 2.2.15. The elliptic curves E defined over Q of conductor 25q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers ` ≥ 7, m ≥ 0 and n ≥ 0 such that 8qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 qpψ(8qmpn + 1) 2qm+2pn+2 26q2m+6p2n+6A2 −2 · qpψ(8qmpn + 1) q2p2 29qm+6pn+6A1′ −qpψ(8qmpn + 1) 2qm+2pn+2 26q2m+6p2n+6A2′ 2 · qpψ(8qmpn + 1) q2p2 29qm+6pn+6where ∈ {±1} is the residue of qp modulo 4.2. There exist integers m ≥ 0 and n ≥ 0 such that 8qm + pn is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 qpψ(8qm + pn) 2qm+2p2 26q2m+6pn+6B2 −2 · qpψ(8qm + pn) q2pn+2 29qm+6p2n+6B1′ −qpψ(8qm + pn) 2qm+2p2 26q2m+6pn+6B2′ 2 · qpψ(8qm + pn) q2pn+2 29qm+6p2n+63. There exist integers m ≥ 0 and n ≥ 0 such that 8qm − pn is a square, p ≡ 7 (mod 8),n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 qpψ(8qm − pn) 2qm+2p2 −26q2m+6pn+6C2 −2 · qpψ(8qm − pn) −q2pn+2 29qm+6p2n+6C1′ −qpψ(8qm − pn) 2qm+2p2 −26q2m+6pn+6C2′ 2 · qpψ(8qm − pn) −q2pn+2 29qm+6p2n+61044. There exist integers m ≥ 0 and n ≥ 0 such that pn − 8qm is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 qpψ(pn − 8qm) −2qm+2p2 26q2m+6pn+6D2 −2 · qpψ(pn − 8qm) q2pn+2 −29qm+6p2n+6D1′ −qpψ(pn − 8qm) −2qm+2p2 26q2m+6pn+6D2′ 2 · qpψ(pn − 8qm) q2pn+2 −29qm+6p2n+65. There exist integers m ≥ 0 and n ≥ 0 such that 8pn + qm is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 qpψ(8pn + qm) 2q2pn+2 26qm+6p2n+6E2 −2 · qpψ(8pn + qm) qm+2p2 29q2m+6pn+6E1′ −qpψ(8pn + qm) 2q2pn+2 26qm+6p2n+6E2′ 2 · qpψ(8pn + qm) qm+2p2 29q2m+6pn+66. There exist integers m ≥ 0 and n ≥ 0 such that 8pn − qm is a square, q ≡ 7 (mod 8),m odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆F1 qpψ(8pn − qm) 2q2pn+2 −26qm+6p2n+6F2 −2 · qpψ(8pn − qm) −qm+2p2 29q2m+6pn+6F1′ −qpψ(8pn − qm) 2q2pn+2 −26qm+6p2n+6F2′ 2 · qpψ(8pn − qm) −qm+2p2 29q2m+6pn+67. There exist integers m ≥ 0 and n ≥ 0 such that qm − 8pn is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 qpψ(qm − 8pn) −2q2pn+2 26qm+6p2n+6G2 −2 · qpψ(qm − 8pn) qm+2p2 −29q2m+6pn+6G1′ −qpψ(qm − 8pn) −2q2pn+2 26qm+6p2n+6G2′ 2 · qpψ(qm − 8pn) qm+2p2 −29q2m+6pn+68. There exist integers m ≥ 0 and n ≥ 0 such that 8 + qmpn is a square, p ≡ 1, 7 (mod 8)if n > 0, q ≡ 1, 7 (mod 8) if m > 0, and E is Q-isomorphic to one of the followingelliptic curves105a2 a4 ∆H1 qpψ(8 + qmpn) 2q2p2 26qm+6pn+6H2 −2 · qpψ(8 + qmpn) qm+2pn+2 29q2m+6p2n+6H1′ −qpψ(8 + qmpn) 2q2p2 26qm+6pn+6H2′ 2 · qpψ(8 + qmpn) qm+2pn+2 29q2m+6p2n+69. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 8 is a square, p ≡ 1, 3 (mod 8)if n > 0, q ≡ 1, 3 (mod 8) if m > 0, and E is Q-isomorphic to one of the followingelliptic curvesa2 a4 ∆I1 qpψ(qmpn − 8) −2q2p2 26qm+6pn+6I2 −2 · qpψ(qmpn − 8) qm+2pn+2 −29q2m+6p2n+6I1′ −qpψ(qmpn − 8) −2q2p2 26qm+6pn+6I2′ 2 · qpψ(qmpn − 8) qm+2pn+2 −29q2m+6p2n+610. There exist integers m ≥ 0 and n ≥ 0 such that qmpn + 1 is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 2qpψ(qmpn + 1) qm+2pn+2 26q2m+6p2n+6J2 −4 · qpψ(qmpn + 1) 4q2p2 212qm+6pn+6J1′ −2qpψ(qmpn + 1) qm+2pn+2 26q2m+6p2n+6J2′ 4 · qpψ(qmpn + 1) 4q2p2 212qm+6pn+611. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 1 is a square, p ≡ 1 (mod 4) ifn > 0, q ≡ 1 (mod 4) if m > 0, and E is Q-isomorphic to one of the following ellipticcurvesa2 a4 ∆K1 2qpψ(qmpn − 1) q2p2 26q2m+6p2n+6K2 −4 · qpψ(qmpn − 1) 4qm+2pn+2 212qm+6pn+6K1′ −2qpψ(qmpn − 1) q2p2 26q2m+6p2n+6K2′ 4 · qpψ(qmpn − 1) 4qm+2pn+2 212qm+6pn+612. There exist integers m ≥ 0 and n ≥ 0 such that qm + pn is a square and E is Q-isomorphic to one of the following elliptic curves(a) m is even or q ≡ 1 (mod 4);106a2 a4 ∆L1 2qpψ(qm + pn) q2pn+2 26qm+6p2n+6L2 −4 · qpψ(qm + pn) 4qm+2p2 212q2m+6pn+6L1′ −2qpψ(qm + pn) q2pn+2 26qm+6p2n+6L2′ 4 · qpψ(qm + pn) 4qm+2p2 212q2m+6pn+6(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆M1 2qpψ(qm + pn) qm+2p2 26q2m+6pn+6M2 −4 · qpψ(qm + pn) 4q2pn+2 212qm+6p2n+6M1′ −2qpψ(qm + pn) qm+2p2 26q2m+6pn+6M2′ 4 · qpψ(qm + pn) 4q2pn+2 212qm+6p2n+613. There exist integers m ≥ 0 and n ≥ 0 such that qm − pn is a square and E is Q-isomorphic to one of the following elliptic curves(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆N1 2qpψ(qm − pn) −q2pn+2 26qm+6p2n+6N2 −4 · qpψ(qm − pn) 4qm+2 −212q2m+6pn+2N1′ −2qpψ(qm − pn) −q2pn+2 26qm+6p2n+6N2′ 4 · qpψ(qm − pn) 4qm+2 −212q2m+6pn+2(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆O1 2qpψ(qm − pn) qm+2p2 −26q2m+6pn+6O2 −4 · qpψ(qm − pn) −4q2pn+2 212qm+6p2n+6O1′ −2qpψ(qm − pn) qm+2p2 −26q2m+6pn+6O2′ 4 · qpψ(qm − pn) −4q2pn+2 212qm+6p2n+614. There exist integers m ≥ 0 and n ≥ 0 such that pn − qm is a square and E is Q-isomorphic to one of the following elliptic curves(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆P1 2qpψ(pn − qm) −q2pn+2 26qm+6p2n+6P2 −4 · qpψ(pn − qm) 4qm+2p2 −212q2m+6pn+6P1′ −2qpψ(pn − qm) −q2pn+2 26qm+6p2n+6P2′ 4 · qpψ(pn − qm) 4qm+2p2 −212q2m+6pn+6107(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆Q1 2qpψ(pn − qm) qm+2p2 −26q2m+6pn+6Q2 −4 · qpψ(pn − qm) −4q2pn+2 212qm+6p2n+6Q1′ −2qpψ(pn − qm) qm+2p2 −26q2m+6pn+6Q2′ 4 · qpψ(pn − qm) −4q2pn+2 212qm+6p2n+615. There exist integers m ≥ 0 and r ∈ {0, 1} such that 8qm+1p is a square, p ≡ 1 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆R1 qpr+1ψ(8qm+1p ) 2qm+2p2r+1 26q2m+6p6r+3R2 −2 · qpr+1ψ(8qm+1p ) q2p2r+1 29qm+6p6r+3R1′ −qpr+1ψ(8qm+1p ) 2qm+2p2r+1 26q2m+6p6r+3R2′ 2 · qpr+1ψ(8qm+1p ) q2p2r+1 29qm+6p6r+316. There exist integers m ≥ 0 and r ∈ {0, 1} such that 8qm−1p is a square, p ≡ 7 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆S1 qpr+1ψ(8qm−1p ) 2qm+2p2r+1 −26q2m+6p6r+3S2 −2 · qpr+1ψ(8qm−1p ) −q2p2r+1 29qm+6p6r+3S1′ −qpr+1ψ(8qm−1p ) 2qm+2p2r+1 −26q2m+6p6r+3S2′ 2 · qpr+1ψ(8qm−1p ) −q2p2r+1 29qm+6p6r+317. There exist integers m ≥ 0 and r ∈ {0, 1} such that 8+qmp is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆T1 qpr+1ψ(8+qmp ) 2q2p2r+1 26qm+6p6r+3T2 −2 · qpr+1ψ(8+qmp ) qm+2p2r+1 29q2m+6p6r+3T1′ −qpr+1ψ(8+qmp ) 2q2p2r+1 26qm+6p6r+3T2′ 2 · qpr+1ψ(8+qmp ) qm+2p2r+1 29q2m+6p6r+318. There exist integers m ≥ 0 and r ∈ {0, 1} such that 8−qmp is a square and E is Q-isomorphic to one of the following elliptic curves108a2 a4 ∆U1 qpr+1ψ(8−qmp ) 2q2p2r+1 −26qm+6p6r+3U2 −2 · qpr+1ψ(8−qmp ) −qm+2p2r+1 29q2m+6p6r+3U1′ −qpr+1ψ(8−qmp ) 2q2p2r+1 −26qm+6p6r+3U2′ 2 · qpr+1ψ(8−qmp ) −qm+2p2r+1 29q2m+6p6r+319. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−8p is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆V1 qpr+1ψ( qm−8p ) −2q2p2r+1 26qm+6p6r+3V2 −2 · qpr+1ψ( qm−8p ) qm+2p2r+1 −29q2m+6p6r+3V1′ −qpr+1ψ( qm−8p ) −2q2p2r+1 26qm+6p6r+3V2′ 2 · qpr+1ψ( qm−8p ) qm+2p2r+1 −29q2m+6p6r+320. There exist integers n ≥ 0 and s ∈ {0, 1} such that 8pn+1q is a square, q ≡ 1 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆W1 qs+1pψ(8pn+1q ) 2q2s+1pn+2 26q6s+3p2n+6W2 −2 · qs+1pψ(8pn+1q ) q2s+1p2 29q6s+3pn+6W1′ −qs+1pψ(8pn+1q ) 2q2s+1pn+2 26q6s+3p2n+6W2′ 2 · qs+1pψ(8pn+1q ) q2s+1p2 29q6s+3pn+621. There exist integers n ≥ 0 and s ∈ {0, 1} such that 8pn−1q is a square, q ≡ 7 (mod 8)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆X1 qs+1pψ(8pn−1q ) 2q2s+1pn+2 −26q6s+3p2n+6X2 −2 · qs+1pψ(8pn−1q ) −q2s+1p2 29q6s+3pn+6X1′ −qs+1pψ(8pn−1q ) 2q2s+1pn+2 −26q6s+3p2n+6X2′ 2 · qs+1pψ(8pn−1q ) −q2s+1p2 29q6s+3pn+622. There exist integers n ≥ 0 and s ∈ {0, 1} such that 8+pnq is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Y1 qs+1pψ(8+pnq ) 2q2s+1p2 26q6s+3pn+6Y2 −2 · qs+1pψ(8+pnq ) q2s+1pn+2 29q6s+3p2n+6Y1′ −qs+1pψ(8+pnq ) 2q2s+1p2 26q6s+3pn+6Y2′ 2 · qs+1pψ(8+pnq ) q2s+1pn+2 29q6s+3p2n+6109where ∈ {±1} is the residue of qs+1p modulo 4.23. There exist integers n ≥ 0 and s ∈ {0, 1} such that 8−pnq is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆Z1 qs+1pψ(8−pnq ) 2q2s+1p2 −26q6s+3pn+6Z2 −2 · qs+1pψ(8−pnq ) −q2s+1pn+2 29q6s+3p2n+6Z1′ −qs+1pψ(8−pnq ) 2q2s+1p2 −26q6s+3pn+6Z2′ 2 · qs+1pψ(8−pnq ) −q2s+1pn+2 29q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.24. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−8q is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AA1 qs+1pψ(pn−8q ) −2q2s+1p2 26q6s+3pn+6AA2 −2 · qs+1pψ(pn−8q ) q2s+1pn+2 −29q6s+3p2n+6AA1′ −qs+1pψ(pn−8q ) −2q2s+1p2 26q6s+3pn+6AA2′ 2 · qs+1pψ(pn−8q ) q2s+1pn+2 −29q6s+3p2n+6where ∈ {±1} is the residue of qs+1p modulo 4.25. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm+1p is a square, qm ≡ −1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) p ≡ 1 (mod 4);a2 a4 ∆AB1 2qpr+1ψ( qm+1p ) qm+2p2r+1 26q2m+6p6r+3AB2 −4 · qpr+1ψ( qm+1p ) 4q2p2r+1 212qm+6p6r+3AB1′ −2qpr+1ψ( qm+1p ) qm+2p2r+1 26q2m+6p6r+3AB2′ 4 · qpr+1ψ( qm+1p ) 4q2p2r+1 212qm+6p6r+3(b) p ≡ −1 (mod 4);a2 a4 ∆AC1 2qpr+1ψ( qm+1p ) q2p2r+1 26qm+6p6r+3AC2 −4 · qpr+1ψ( qm+1p ) 4qm+2p2r+1 212q2m+6p6r+3AC1′ −2qpr+1ψ( qm+1p ) q2p2r+1 26qm+6p6r+3AC2′ 4 · qpr+1ψ( qm+1p ) 4qm+2p2r+1 212q2m+6p6r+311026. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−1p is a square, qm ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) p ≡ 1 (mod 4);a2 a4 ∆AD1 2qpr+1ψ( qm−1p ) −q2p2r+1 −26qm+6p6r+3AD2 −4 · qpr+1ψ( qm−1p ) 4qm+2p2r+1 212q2m+6p6r+3AD1′ −2qpr+1ψ( qm−1p ) −q2p2r+1 −26qm+6p6r+3AD2′ 4 · qpr+1ψ( qm−1p ) 4qm+2p2r+1 212q2m+6p6r+3(b) p ≡ −1 (mod 4);a2 a4 ∆AE1 2qpr+1ψ( qm−1p ) qm+2p2r+1 −26q2m+6p6r+3AE2 −4 · qpr+1ψ( qm−1p ) −4q2p2r+1 212qm+6p6r+3AE1′ −2qpr+1ψ( qm−1p ) qm+2p2r+1 −26q2m+6p6r+3AE2′ 4 · qpr+1ψ( qm−1p ) −4q2p2r+1 212qm+6p6r+327. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn+1q is a square, pn ≡ −1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) q ≡ 1 (mod 4);a2 a4 ∆AF1 2qs+1pψ(pn+1q ) q2s+1pn+2 26q6s+3p2n+6AF2 −4 · qs+1pψ(pn+1q ) 4q2s+1p2 212q6s+3pn+6AF1′ −2qs+1pψ(pn+1q ) q2s+1pn+2 26q6s+3p2n+6AF2′ 4 · qs+1pψ(pn+1q ) 4q2s+1p2 212q6s+3pn+6(b) q ≡ −1 (mod 4);a2 a4 ∆AG1 2qs+1pψ(pn+1q ) q2s+1p2 26q6s+3pn+6AG2 −4 · qs+1pψ(pn+1q ) 4q2s+1pn+2 212q6s+3p2n+6AG1′ −2qs+1pψ(pn+1q ) q2s+1p2 26q6s+3pn+6AG2′ 4 · qs+1pψ(pn+1q ) 4q2s+1pn+2 212q6s+3p2n+628. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−1q is a square, pn ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) q ≡ 1 (mod 4);111a2 a4 ∆AH1 2qs+1pψ(pn−1q ) −q2s+1p2 −26q6s+3pn+6AH2 −4 · qs+1pψ(pn−1q ) 4q2s+1pn+2 212q6s+3p2n+6AH1′ −2qs+1pψ(pn−1q ) −q2s+1p2 −26q6s+3pn+6AH2′ 4 · qs+1pψ(pn−1q ) 4q2s+1pn+2 212q6s+3p2n+6(b) q ≡ −1 (mod 4);a2 a4 ∆AI1 2qs+1pψ(pn−1q ) q2s+1pn+2 −26q6s+3p2n+6AI2 −4 · qs+1pψ(pn−1q ) −4q2s+1p2 212q6s+3pn+6AI1′ −2qs+1pψ(pn−1q ) q2s+1pn+2 −26q6s+3p2n+6AI2′ 4 · qs+1pψ(pn−1q ) −4q2s+1p2 212q6s+3pn+629. There exist integers r, s ∈ {0, 1} and E is Q-isomorphic to one of the following ellipticcurve(a) qp ≡ 1 (mod 4);a2 a4 ∆AJ1 0 −q2s+1p2r+1 26q6s+3p6r+3AJ2 0 4q2s+1p2r+1 −212q6s+3p6r+3(b) qp ≡ −1 (mod 4);a2 a4 ∆AK1 0 q2s+1p2r+1 −26q6s+3p6r+3AK2 0 −4q2s+1p2r+1 212q6s+3p6r+3Theorem 2.2.16. The elliptic curves E defined over Q of conductor 26q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers ` ≥ 3, m ≥ 0 and n ≥ 0 such that 2`qmpn + 1 is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 2qpψ(2`qmpn + 1) 2`qm+2pn+2 22`+6q2m+6p2n+6A2 −4qpψ(2`qmpn + 1) 4q2p2 2`+12qm+6pn+6A1′ −2qpψ(2`qmpn + 1) 2`qm+2pn+2 22`+6q2m+6p2n+6A2′ 4qpψ(2`qmpn + 1) 4q2p2 2`+12qm+6pn+61122. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2`qm + pn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 2qpψ(2`qm + pn) 2`qm+2p2 22`+6q2m+6pn+6B2 −4qpψ(2`qm + pn) 4q2pn+2 2`+12qm+6p2n+6B1′ −2qpψ(2`qm + pn) 2`qm+2p2 22`+6q2m+6pn+6B2′ 4qpψ(2`qm + pn) 4q2pn+2 2`+12qm+6p2n+63. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2`qm − pn is a square, p ≡3 (mod 4), n odd and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 2qpψ(2`qm − pn) 2`qm+2p2 −22`+6q2m+6pn+6C2 −4qpψ(2`qm − pn) −4q2pn+2 2`+12qm+6p2n+6C1′ −2qpψ(2`qm − pn) 2`qm+2p2 −22`+6q2m+6pn+6C2′ 4qpψ(2`qm − pn) −4q2pn+2 2`+12qm+6p2n+64. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that pn − 2`qm is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 2qpψ(pn − 2`qm) −2`qm+2p2 22`+6q2m+6pn+6D2 −4qpψ(pn − 2`qm) 4q2pn+2 −2`+12qm+6p2n+6D1′ −2qpψ(pn − 2`qm) −2`qm+2p2 22`+6q2m+6pn+6D2′ 4qpψ(pn − 2`qm) 4q2pn+2 −2`+12qm+6p2n+65. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2`pn + qm is a square, q ≡5 (mod 8) if ` = 2 and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 2qpψ(2`pn + qm) 2`q2pn+2 22`+6qm+6p2n+6E2 −4 · qpψ(2`pn + qm) 4qm+2p2 2`+12q2m+6pn+6E1′ −2qpψ(2`pn + qm) 2`q2pn+2 22`+6qm+6p2n+6E2′ 4 · qpψ(2`pn + qm) 4qm+2p2 2`+12q2m+6pn+66. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2`pn − qm is a square, q ≡7 (mod 8) when ` ≥ 3 or q ≡ 3 (mod 8) if ` = 2, m odd and E is Q-isomorphic to oneof the following elliptic curves113a2 a4 ∆F1 2qpψ(2`pn − qm) 2`q2pn+2 −22`+6qm+6p2n+6F2 −4 · qpψ(2`pn − qm) −4qm+2p2 2`+12q2m+6pn+6F1′ −2qpψ(2`pn − qm) 2`q2pn+2 −22`+6qm+6p2n+6F2′ 4 · qpψ(2`pn − qm) −4qm+2p2 2`+12q2m+6pn+67. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that qm − 2`pn is a square, q ≡5 (mod 8) if ` = 2 and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 2qpψ(qm − 2`pn) −2`q2pn+2 22`+6qm+6p2n+6G2 −4 · qpψ(qm − 2`pn) 4qm+2p2 −2`+12q2m+6pn+6G1′ −2qpψ(qm − 2`pn) −2`q2pn+2 22`+6qm+6p2n+6G2′ 4 · qpψ(qm − 2`pn) 4qm+2p2 −2`+12q2m+6pn+68. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2` + qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 2qpψ(2` + qmpn) 2`q2p2 22`+6qm+6pn+6H2 −4 · qpψ(2` + qmpn) 4qm+2pn+2 2`+12q2m+6p2n+6H1′ −2qpψ(2` + qmpn) 2`q2p2 22`+6qm+6pn+6H2′ 4 · qpψ(2` + qmpn) 4qm+2pn+2 2`+12q2m+6p2n+69. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that 2` − qmpn is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 2qpψ(2` − qmpn) 2`q2p2 −22`+6qm+6pn+6I2 −4 · qpψ(2` − qmpn) −4qm+2pn+2 2`+12q2m+6p2n+6I1′ −2qpψ(2` − qmpn) 2`q2p2 −22`+6qm+6pn+6I2′ 4 · qpψ(2` − qmpn) −4qm+2pn+2 2`+12q2m+6p2n+610. There exist integers ` ≥ 2, m ≥ 0 and n ≥ 0 such that qmpn − 2` is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆J1 2qpψ(qmpn − 2`) −2`q2p2 22`+6qm+6pn+6J2 −4 · qpψ(qmpn − 2`) 4qm+2pn+2 −2`+12q2m+6p2n+6J1′ −2qpψ(qmpn − 2`) −2`q2p2 22`+6qm+6pn+6J2′ 4 · qpψ(qmpn − 2`) 4qm+2pn+2 −2`+12q2m+6p2n+611411. There exist integers m ≥ 0 and n ≥ 0 such that qmpn + 1 is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆K1 2qpψ(qmpn + 1) q2p2 26q2m+6p2n+6K2 −4 · qpψ(qmpn + 1) 4qm+2pn+2 212qm+6pn+6K1′ −2qpψ(qmpn + 1) q2p2 26q2m+6p2n+6K2′ 4 · qpψ(qmpn + 1) 4qm+2pn+2 212qm+6pn+612. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 1 is a square, p ≡ 1 (mod 4) ifn > 0, q ≡ 1 (mod 4) if m > 0, and E is Q-isomorphic to one of the following ellipticcurvesa2 a4 ∆L1 2qpψ(qmpn − 1) qm+2pn+2 26q2m+6p2n+6L2 −4 · qpψ(qmpn − 1) 4q2p2 212qm+6pn+6L1′ −2qpψ(qmpn − 1) qm+2pn+2 26q2m+6p2n+6L2′ 4 · qpψ(qmpn − 1) 4q2p2 212qm+6pn+613. There exist integers m ≥ 0 and n ≥ 0 such that qm + pn is a square and E is Q-isomorphic to one of the following elliptic curves(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆M1 2qpψ(qm + pn) qm+2p2 26q2m+6pn+6M2 −4 · qpψ(qm + pn) 4q2pn+2 212qm+6p2n+6M1′ −2qpψ(qm + pn) qm+2p2 26q2m+6pn+6M2′ 4 · qpψ(qm + pn) 4q2pn+2 212qm+6p2n+6(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆N1 2qpψ(qm + pn) q2pn+2 26qm+6p2n+6N2 −4 · qpψ(qm + pn) 4qm+2p2 212q2m+6pn+6N1′ −2qpψ(qm + pn) q2pn+2 26qm+6p2n+6N2′ 4 · qpψ(qm + pn) 4qm+2p2 212q2m+6pn+614. There exist integers m ≥ 0 and n ≥ 0 such that qm − pn is a square and E is Q-isomorphic to one of the following elliptic curves115(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆O1 2qpψ(qm − pn) qm+2p2 −26q2m+6pn+6O2 −4 · qpψ(qm − pn) −4q2pn+2 212qm+6p2n+6O1′ −2qpψ(qm − pn) qm+2p2 −26q2m+6pn+6O2′ 4 · qpψ(qm − pn) −4q2pn+2 212qm+6p2n+6(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆P1 2qpψ(qm − pn) −q2pn+2 26qm+6p2n+6P2 −4 · qpψ(qm − pn) 4qm+2p2 −212q2m+6pn+6P1′ −2qpψ(qm − pn) −q2pn+2 26qm+6p2n+6P2′ 4 · qpψ(qm − pn) 4qm+2p2 −212q2m+6pn+615. There exist integers m ≥ 0 and n ≥ 0 such that pn − qm is a square and E is Q-isomorphic to one of the following elliptic curves(a) m is even or q ≡ 1 (mod 4);a2 a4 ∆Q1 2qpψ(pn − qm) qm+2p2 −26q2m+6pn+6Q2 −4 · qpψ(pn − qm) −4q2pn+2 212qm+6p2n+6Q1′ −2qpψ(pn − qm) qm+2p2 −26q2m+6pn+6Q2′ 4 · qpψ(pn − qm) −4q2pn+2 212qm+6p2n+6(b) m is odd and q ≡ 3 (mod 4);a2 a4 ∆R1 2qpψ(pn − qm) −q2pn+2 26qm+6p2n+6R2 −4 · qpψ(pn − qm) 4qm+2 −212q2m+6pn+6R1′ −2qpψ(pn − qm) −q2pn+2 26qm+6p2n+6R2′ 4 · qpψ(pn − qm) 4qm+2 −212q2m+6pn+616. There exist integers ` ≥ 2, m ≥ 0 and r ∈ {0, 1} such that 2`qm+1p is a square, p ≡1 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆S1 2qpr+1ψ(2`qm+1p ) 2`qm+2p2r+1 22`+6q2m+6p6r+3S2 −4 · qpr+1ψ(2`qm+1p ) 4q2p2r+1 2`+12qm+6p6r+3S1′ −2qpr+1ψ(2`qm+1p ) 2`qm+2p2r+1 22`+6q2m+6p6r+3S2′ 4 · qpr+1ψ(2`qm+1p ) 4q2p2r+1 2`+12qm+6p6r+311617. There exist integers ` ≥ 2, m ≥ 0 and r ∈ {0, 1} such that 2`qm−1p is a square, p ≡3 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆T1 2qpr+1ψ(2`qm−1p ) 2`qm+2p2r+1 −22`+6q2m+6p6r+3T2 −4 · qpr+1ψ(2`qm−1p ) −4q2p2r+1 2`+12qm+6p6r+3T1′ −2qpr+1ψ(2`qm−1p ) 2`qm+2p2r+1 −22`+6q2m+6p6r+3T2′ 4 · qpr+1ψ(2`qm−1p ) −4q2p2r+1 2`+12qm+6p6r+318. There exist integers ` ≥ 2, m ≥ 0 and r ∈ {0, 1} such that 2`+qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆U1 2qpr+1ψ(2`+qmp ) 2`q2p2r+1 22`+6qm+6p6r+3U2 −4 · qpr+1ψ(2`+qmp ) 4qm+2p2r+1 2`+12q2m+6p6r+3U1′ −2qpr+1ψ(2`+qmp ) 2`q2p2r+1 22`+6qm+6p6r+3U2′ 4 · qpr+1ψ(2`+qmp ) 4qm+2p2r+1 2`+12q2m+6p6r+319. There exist integers ` ≥ 2, m ≥ 0 and r ∈ {0, 1} such that 2`−qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆V1 2qpr+1ψ(2`−qmp ) 2`q2p2r+1 −22`+6qm+6p6r+3V2 −4 · qpr+1ψ(2`−qmp ) −4qm+2p2r+1 2`+12q2m+6p6r+3V1′ −2qpr+1ψ(2`−qmp ) 2`q2p2r+1 −22`+6qm+6p6r+3V2′ 4 · qpr+1ψ(2`−qmp ) −4qm+2p2r+1 2`+12q2m+6p6r+320. There exist integers ` ≥ 2, m ≥ 0 and r ∈ {0, 1} such that qm−2`p is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆W1 2qpr+1ψ( qm−2`p ) −2`q2p2r+1 22`+6qm+6p6r+3W2 −4 · qpr+1ψ( qm−2`p ) 4qm+2p2r+1 −2`+12q2m+6p6r+3W1′ −2qpr+1ψ( qm−2`p ) −2`q2p2r+1 22`+6qm+6p6r+3W2′ 4 · qpr+1ψ( qm−2`p ) 4qm+2p2r+1 −2`+12q2m+6p6r+321. There exist integers ` ≥ 2, n ≥ 0 and s ∈ {0, 1} such that 2`pn+1q is a square, q ≡1 (mod 4) and E is Q-isomorphic to one of the following elliptic curves117a2 a4 ∆X1 2qs+1pψ(2`pn+1q ) 2`q2s+1pn+2 22`+6q6s+3p2n+6X2 −4 · qs+1pψ(2`pn+1q ) 4q2s+1p2 2`+12q6s+3pn+6X1′ −2qs+1pψ(2`pn+1q ) 2`q2s+1pn+2 22`+6q6s+3p2n+6X2′ 4 · qs+1pψ(2`pn+1q ) 4q2s+1p2 2`+12q6s+3pn+622. There exist integers ` ≥ 2, n ≥ 0 and s ∈ {0, 1} such that 2`pn−1q is a square, q ≡3 (mod 4) and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆Y1 2qs+1pψ(2`pn−1q ) 2`q2s+1pn+2 −22`+6q6s+3p2n+6Y2 −4 · qs+1pψ(2`pn−1q ) −4q2s+1p2 2`+12q6s+3pn+6Y1′ −2qs+1pψ(2`pn−1q ) 2`q2s+1pn+2 −22`+6q6s+3p2n+6Y2′ 4 · qs+1pψ(2`pn−1q ) −4q2s+1p2 2`+12q6s+3pn+623. There exist integers ` ≥ 2, n ≥ 0 and s ∈ {0, 1} such that 2`+pnq is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆Z1 2qs+1pψ(2`+pnq ) 2`q2s+1p2 22`+6q6s+3pn+6Z2 −4 · qs+1pψ(2`+pnq ) 4q2s+1pn+2 2`+12q6s+3p2n+6Z1′ −2qs+1pψ(2`+pnq ) 2`q2s+1p2 22`+6q6s+3pn+6Z2′ 4 · qs+1pψ(2`+pnq ) 4q2s+1pn+2 2`+12q6s+3p2n+624. There exist integers ` ≥ 2, n ≥ 0 and s ∈ {0, 1} such that 2`−pnq is a square and E isQ-isomorphic to one of the following elliptic curvea2 a4 ∆AA1 2qs+1pψ(2`−pnq ) 2`q2s+1p2 −22`+6q6s+3pn+6AA2 −4 · qs+1pψ(2`−pnq ) −4q2s+1pn+2 2`+12q6s+3p2n+6AA1′ −2qs+1pψ(2`−pnq ) 2`q2s+1p2 −22`+6q6s+3pn+6AA2′ 4 · qs+1pψ(2`−pnq ) −4q2s+1pn+2 2`+12q6s+3p2n+625. There exist integers ` ≥ 2, n ≥ 0 and s ∈ {0, 1} such that pn−2`q is a square and E isQ-isomorphic to one of the following elliptic curve118a2 a4 ∆AB1 2qs+1pψ(pn−2`q ) −2`q2s+1p2 22`+6q6s+3pn+6AB2 −4 · qs+1pψ(pn−2`q ) 4q2s+1pn+2 −2`+12q6s+3p2n+6AB1′ −2qs+1pψ(pn−2`q ) −2`q2s+1p2 22`+6q6s+3pn+6AB2′ 4 · qs+1pψ(pn−2`q ) 4q2s+1pn+2 −2`+12q6s+3p2n+626. There exist integers ` ≥ 2 and r, s ∈ {0, 1} such that 2`+1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AC1 2qs+1pr+1ψ(2`+1qp ) 2`q2s+1p2r+1 22`+6q6s+3p6r+3AC2 −4 · qs+1pr+1ψ(2`+1qp ) 4q2s+1p2r+1 2`+12q6s+3p6r+3AC1′ −2qs+1pr+1ψ(2`+1qp ) 2`q2s+1p2r+1 22`+6q6s+3p6r+3AC2′ 4 · qs+1pr+1ψ(2`+1qp ) 4q2s+1p2r+1 2`+12q6s+3p6r+327. There exist integers ` ≥ 2 and r, s ∈ {0, 1} such that 2`−1qp is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆AD1 2qs+1pr+1ψ(2`−1qp ) 2`q2s+1p2r+1 −22`+6q6s+3p6r+3AD2 −4 · qs+1pr+1ψ(2`−1qp ) −4q2s+1p2r+1 2`+12q6s+3p6r+3AD1′ −2qs+1pr+1ψ(2`−1qp ) 2`q2s+1p2r+1 −22`+6q6s+3p6r+3AD2′ 4 · qs+1pr+1ψ(2`−1qp ) −4q2s+1p2r+1 2`+12q6s+3p6r+328. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm+1p is a square, qm ≡ −1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) p ≡ 1 (mod 4);a2 a4 ∆AE1 2qpr+1ψ( qm+1p ) q2p2r+1 26qm+6p6r+3AE2 −4 · qpr+1ψ( qm+1p ) 4qm+2p2r+1 212q2m+6p6r+3AE1′ −2qpr+1ψ( qm+1p ) q2p2r+1 26qm+6p6r+3AE2′ 4 · qpr+1ψ( qm+1p ) 4qm+2p2r+1 212q2m+6p6r+3(b) p ≡ −1 (mod 4);a2 a4 ∆AF1 2qpr+1ψ( qm+1p ) qm+2p2r+1 26q2m+6p6r+3AF2 −4 · qpr+1ψ( qm+1p ) 4q2p2r+1 212qm+6p6r+3AF1′ −2qpr+1ψ( qm+1p ) qm+2p2r+1 26q2m+6p6r+3AF2′ 4 · qpr+1ψ( qm+1p ) 4q2p2r+1 212qm+6p6r+311929. There exist integers m ≥ 0 and r ∈ {0, 1} such that qm−1p is a square, qm ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) p ≡ 1 (mod 4);a2 a4 ∆AG1 2qpr+1ψ( qm−1p ) qm+2p2r+1 −26q2m+6p6r+3AG2 −4 · qpr+1ψ( qm−1p ) −4q2p2r+1 212qm+6p6r+3AG1′ −2qpr+1ψ( qm−1p ) qm+2p2r+1 −26q2m+6p6r+3AG2′ 4 · qpr+1ψ( qm−1p ) −4q2p2r+1 212qm+6p6r+3(b) p ≡ −1 (mod 4);a2 a4 ∆AH1 2qpr+1ψ( qm−1p ) −q2p2r+1 −26qm+6p6r+3AH2 −4 · qpr+1ψ( qm−1p ) 4qm+2p2r+1 212q2m+6p6r+3AH1′ −2qpr+1ψ( qm−1p ) −q2p2r+1 −26qm+6p6r+3AH2′ 4 · qpr+1ψ( qm−1p ) 4qm+2p2r+1 212q2m+6p6r+330. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn+1q is a square, pn ≡ −1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) q ≡ 1 (mod 4);a2 a4 ∆AI1 2qs+1pψ(pn+1q ) q2s+1p2 26q6s+3pn+6AI2 −4 · qs+1pψ(pn+1q ) 4q2s+1pn+2 212q6s+3p2n+6AI1′ −2qs+1pψ(pn+1q ) q2s+1p2 26q6s+3pn+6AI2′ 4 · qs+1pψ(pn+1q ) 4q2s+1pn+2 212q6s+3p2n+6(b) q ≡ −1 (mod 4);a2 a4 ∆AJ1 2qs+1pψ(pn+1q ) q2s+1pn+2 26q6s+3p2n+6AJ2 −4 · qs+1pψ(pn+1q ) 4q2s+1p2 212q6s+3pn+6AJ1′ −2qs+1pψ(pn+1q ) q2s+1pn+2 26q6s+3p2n+6AJ2′ 4 · qs+1pψ(pn+1q ) 4q2s+1p2 212q6s+3pn+631. There exist integers n ≥ 0 and s ∈ {0, 1} such that pn−1q is a square, pn ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curves(a) q ≡ 1 (mod 4);120a2 a4 ∆AK1 2qs+1pψ(pn−1q ) q2s+1pn+2 −26q6s+3p2n+6AK2 −4 · qs+1pψ(pn−1q ) −4q2s+1p2 212q6s+3pn+6AK1′ −2qs+1pψ(pn−1q ) q2s+1pn+2 −26q6s+3p2n+6AK2′ 4 · qs+1pψ(pn−1q ) −4q2s+1p2 212q6s+3pn+6(b) q ≡ −1 (mod 4);a2 a4 ∆AL1 2qs+1pψ(pn−1q ) −q2s+1p2 −26q6s+3pn+6AL2 −4 · qs+1pψ(pn−1q ) 4q2s+1pn+2 212q6s+3p2n+6AL1′ −2qs+1pψ(pn−1q ) −q2s+1p2 −26q6s+3pn+6AL2′ 4 · qs+1pψ(pn−1q ) 4q2s+1pn+2 212q6s+3p2n+632. There exist integers r, s ∈ {0, 1} and E is Q-isomorphic to one of the following ellipticcurve(a) qp ≡ 1 (mod 4);a2 a4 ∆AM1 0 q2s+1p2r+1 −26q6s+3p6r+3AM2 0 −4q2s+1p2r+1 212q6s+3p6r+3(b) qp ≡ −1 (mod 4);a2 a4 ∆AN1 0 −q2s+1p2r+1 26q6s+3p6r+3AN2 0 4q2s+1p2r+1 −212q6s+3p6r+3Theorem 2.2.17. The elliptic curves E defined over Q of conductor 27q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers m ≥ 0 and n ≥ 0 such that 2qmpn−1 is a square, p, q ≡ 1 (mod 4),t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 2t+1qpψ(2qmpn − 1) 21+2tqm+2pn+2 −28+6tq2m+6p2n+6A2 −22−t · qpψ(2qmpn − 1) −22−2tq2p2 213−6tqm+6pn+6A1′ −2t+1qpψ(2qmpn − 1) 21+2tqm+2pn+2 −28+6tq2m+6p2n+6A2′ 22−t · qpψ(2qmpn − 1) −22−2tq2p2 213−6tqm+6pn+61212. There exist integers m ≥ 0 and n ≥ 0 such that 2`qm + pn is a square, p ≡ 3 (mod 4),n odd, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 2t+1qpψ(2qm + pn) 21+2tqm+2p2 28+6tq2m+6pn+6B2 −22−tqpψ(2qm + pn) 22−2tq2pn+2 213−6tqm+6p2n+6B1′ −2t+1qpψ(2qm + pn) 21+2tqm+2p2 28+6tq2m+6pn+6B2′ 22−tqpψ(2qm + pn) 22−2tq2pn+2 213−6tqm+6p2n+63. There exist integers m ≥ 0 and n ≥ 0 such that 2`qm− pn is a square, either n even orp ≡ 1 (mod 4), t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 2t+1qpψ(2qm − pn) 21+2tqm+2p2 −28+6tq2m+6pn+6C2 −22−tqpψ(2qm − pn) −22−2tq2pn+2 213−6tqm+6p2n+6C1′ −2t+1qpψ(2qm − pn) 21+2tqm+2p2 −28+6tq2m+6pn+6C2′ 22−tqpψ(2qm − pn) −22−2tq2pn+2 213−6tqm+6p2n+64. There exist integers m ≥ 0 and n ≥ 0 such that pn − 2`qm is a square, p ≡ 3 (mod 4),n odd, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 2t+1qpψ(pn − 2qm) −21+2tqm+2p2 28+6tq2m+6pn+6D2 −22−tqpψ(pn − 2qm) 22−2tq2pn+2 −213−6tqm+6p2n+6D1′ −2t+1qpψ(pn − 2qm) −21+2tqm+2p2 28+6tq2m+6pn+6D2′ 22−tqpψ(pn − 2qm) 22−2tq2pn+2 −213−6tqm+6p2n+65. There exist integers m ≥ 0 and n ≥ 0 such that 2pn + qm is a square, t ∈ {0, 1} and Eis Q-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 2t+1qpψ(2pn + qm) 21+2tq2pn+2 28+6tqm+6p2n+6E2 −22−t · qpψ(2pn + qm) 22−2tqm+2p2 213−6tq2m+6pn+6E1′ −2t+1qpψ(2pn + qm) 21+2tq2pn+2 28+6tqm+6p2n+6E2′ 22−t · qpψ(2pn + qm) 22−2tqm+2p2 213−6tq2m+6pn+66. There exist integers m ≥ 0 and n ≥ 0 such that 2pn − qm is a square, t ∈ {0, 1} and Eis Q-isomorphic to one of the following elliptic curves122a2 a4 ∆F1 2t+1qpψ(2pn − qm) 21+2tq2pn+2 −28+6tqm+6p2n+6F2 −22−t · qpψ(2pn − qm) −22−2tqm+2p2 213−6tq2m+6pn+6F1′ −2t+1qpψ(2pn − qm) 21+2tq2pn+2 −28+6tqm+6p2n+6F2′ 22−t · qpψ(2pn − qm) −22−2tqm+2p2 213−6tq2m+6pn+67. There exist integers m ≥ 0 and n ≥ 0 such that qm − 2pn is a square, t ∈ {0, 1} and Eis Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 2t+1qpψ(qm − 2pn) −21+2tq2pn+2 28+6tqm+6p2n+6G2 −22−t · qpψ(qm − 2pn) 22−2tqm+2p2 −213−6tq2m+6pn+6G1′ −2t+1qpψ(qm − 2pn) −21+2tq2pn+2 28+6tqm+6p2n+6G2′ 22−t · qpψ(qm − 2pn) 22−2tqm+2p2 −213−6tq2m+6pn+68. There exist integers m ≥ 0 and n ≥ 0 such that 2+qmpn is a square, p ≡ 1 or 7 (mod 8)if n > 0, q ≡ 1 or 7 (mod 8) if m > 0, t ∈ {0, 1} and E is Q-isomorphic to one of thefollowing elliptic curvesa2 a4 ∆H1 2t+1qpψ(2` + qmpn) 21+2tq2p2 28+6tqm+6pn+6H2 −22−t · qpψ(2` + qmpn) 22−2tqm+2pn+2 213−6tq2m+6p2n+6H1′ −2t+1qpψ(2` + qmpn) 21+2tq2p2 28+6tqm+6pn+6H2′ 22−t · qpψ(2` + qmpn) 22−2tqm+2pn+2 213−6tq2m+6p2n+69. There exist integers m ≥ 0 and n ≥ 0 such that qmpn − 2 is a square, p ≡ 1 or3 (mod 8), q ≡ 1 or 3 (mod 8) if m > 0, t ∈ {0, 1} and E is Q-isomorphic to one ofthe following elliptic curvesa2 a4 ∆I1 2t+1qpψ(qmpn − 2`) −21+2tq2p2 28+6tqm+6pn+6I2 −22−t · qpψ(qmpn − 2`) 22−2tqm+2pn+2 −213−6tq2m+6p2n+6I1′ −2t+1qpψ(qmpn − 2`) −21+2tq2p2 28+6tqm+6pn+6I2′ 22−t · qpψ(qmpn − 2`) 22−2tqm+2pn+2 −213−6tq2m+6p2n+610. There exist integers m ≥ 0 and r, t ∈ {0, 1} such that 2qm+1p is a square, p ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curves123a2 a4 ∆J1 2t+1qpr+1ψ(2qm+1p ) 21+2tqm+2p2r+1 28+6tq2m+6p6r+3J2 −22−2t · qpr+1ψ(2qm+1p ) 22−2tq2p2r+1 213−6tqm+6p6r+3J1′ −2t+1qpr+1ψ(2qm+1p ) 21+2tqm+2p2r+1 28+6tq2m+6p6r+3J2′ 22−2t · qpr+1ψ(2qm+1p ) 22−2tq2p2r+1 213−6tqm+6p6r+311. There exist integers m ≥ 0 and r, t ∈ {0, 1} such that 2qm−1p is a square, p ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆K1 2t+1qpr+1ψ(2qm−1p ) 21+2tqm+2p2r+1 −28+6tq2m+6p6r+3K2 −22−2t · qpr+1ψ(2qm−1p ) −22−2tq2p2r+1 213−6tqm+6p6r+3K1′ −2t+1qpr+1ψ(2qm−1p ) 21+2tqm+2p2r+1 −28+6tq2m+6p6r+3K2′ 22−2t · qpr+1ψ(2qm−1p ) −22−2tq2p2r+1 213−6tqm+6p6r+312. There exist integers m ≥ 0 and r, t ∈ {0, 1} such that 2+qmp is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆L1 2t+1qpr+1ψ(2+qmp ) 21+2tq2p2r+1 28+6tqm+6p6r+3L2 −22−2t · qpr+1ψ(2+qmp ) 22−2tqm+2p2r+1 213−6tq2m+6p6r+3L1′ −2t+1qpr+1ψ(2+qmp ) 21+2tq2p2r+1 28+6tqm+6p6r+3L2′ 22−2t · qpr+1ψ(2+qmp ) 22−2tqm+2p2r+1 213−6tq2m+6p6r+313. There exist integers m ≥ 0 and r, t ∈ {0, 1} such that qm−2p is a square and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆M1 2t+1qpr+1ψ( qm−2p ) −21+2tq2p2r+1 28+6tqm+6p6r+3M2 −22−2t · qpr+1ψ( qm−2p ) 22−2tqm+2p2r+1 −213−6tq2m+6p6r+3M1′ −2t+1qpr+1ψ( qm−2p ) −21+2tq2p2r+1 28+6tqm+6p6r+3M2′ 22−2t · qpr+1ψ( qm−2p ) 22−2tqm+2p2r+1 −213−6tq2m+6p6r+314. There exist integers n ≥ 0 and s, t ∈ {0, 1} such that 2pn+1q is a square, q ≡ 3 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆N1 2t+1qs+1pψ(2pn+1q ) 21+2tq2s+1pn+2 28+6tq6s+3p2n+6N2 −22−2t · qs+1pψ(2pn+1q ) 22−2tq2s+1p2 213−6tq6s+3pn+6N1′ −2t+1qs+1pψ(2pn+1q ) 21+2tq2s+1pn+2 28+6tq6s+3p2n+6N2′ 22−2t · qs+1pψ(2pn+1q ) 22−2tq2s+1p2 213−6tq6s+3pn+612415. There exist integers n ≥ 0 and s, t ∈ {0, 1} such that 2pn−1q is a square, q ≡ 1 (mod 4)and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆O1 · qs+1pψ(2pn−1q ) 21+2tq2s+1pn+2 −28+6tq6s+3p2n+6O2 −22−2t · qs+1pψ(2pn−1q ) −22−2tq2s+1p2 213−6tq6s+3pn+6O1′ − · qs+1pψ(2pn−1q ) 21+2tq2s+1pn+2 −28+6tq6s+3p2n+6O2′ 22−2t · qs+1pψ(2pn−1q ) −22−2tq2s+1p2 213−6tq6s+3pn+616. There exist integers n ≥ 0 and s, t ∈ {0, 1} such that 2+pnq is a square and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆P1 2t+1qs+1pψ(2+pnq ) 21+2tq2s+1p2 28+6tq6s+3pn+6P2 −22−2t · qs+1pψ(2+pnq ) 22−2tq2s+1pn+2 213−6tq6s+3p2n+6P1′ −2t+1qs+1pψ(2+pnq ) 21+2tq2s+1p2 28+6tq6s+3pn+6P2′ 22−2t · qs+1pψ(2+pnq ) 22−2tq2s+1pn+2 213−6tq6s+3p2n+617. There exist integers n ≥ 0 and s, t ∈ {0, 1} such that pn−2q is a square and E is Q-isomorphic to one of the following elliptic curvea2 a4 ∆Q1 2t+1qs+1pψ(pn−2q ) −21+2tq2s+1p2 28+6tq6s+3pn+6Q2 −22−2t · qs+1pψ(pn−2q ) 22−2tq2s+1pn+2 −213−6tq6s+3p2n+6Q1′ −2t+1qs+1pψ(pn−2q ) −21+2tq2s+1p2 28+6tq6s+3pn+6Q2′ 22−2t · qs+1pψ(pn−2q ) 22−2tq2s+1pn+2 −213−6tq6s+3p2n+6Theorem 2.2.18. The elliptic curves E defined over Q of conductor 28q2p2 and having atleast one rational point of order 2 are the ones such that one of the following conditions issatisfied:1. There exist integers m ≥ 0 and n ≥ 0 such that qmpn+12 is a square, p, q ≡ 1, 7 (mod 8),t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆A1 2t+2qpψ( qmpn+12 ) 21+2tqm+2pn+2 −29+6tq2m+6p2n+6A2 −23−tqpψ( qmpn+12 ) −23−2tq2p2 215−6tqm+6pn+6A1′ −2t+2qpψ( qmpn+12 ) 21+2tqm+2pn+2 −29+6tq2m+6p2n+6A2′ 23−tqpψ( qmpn+12 ) −23−2tq2p2 215−6tqm+6pn+61252. There exist integers m ≥ 0 and n ≥ 0 such that qmpn−12 is a square, p, q ≡ 1, 3 (mod 8),t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆B1 2t+2qpψ( qmpn−12 ) 21+2tqm+2pn+2 −29+6tq2m+6p2n+6B2 −23−tqpψ( qmpn−12 ) −23−2tq2p2 215−6tqm+6pn+6B1′ −2t+2qpψ( qmpn−12 ) 21+2tqm+2pn+2 −29+6tq2m+6p2n+6B2′ 23−tqpψ( qmpn−12 ) −23−2tq2p2 215−6tqm+6pn+63. There exist integers m ≥ 0 and n ≥ 0 such that qm+pn2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆C1 2t+2qpψ( qm+pn2 ) 21+2tqm+2p2 29+6tq2m+6pn+6C2 −23−tqpψ( qm+pn2 ) 23−2tq2pn+2 215−6tqm+6p2n+6C1′ −2t+2qpψ( qm+pn2 ) 21+2tqm+2p2 29+6tq2m+6pn+6C2′ 23−tqpψ( qm+pn2 ) 23−2tq2pn+2 215−6tqm+6p2n+64. There exist integers m ≥ 0 and n ≥ 0 such that qm−pn2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆D1 2t+2qpψ( qm−pn2 ) 21+2tqm+2p2 −29+6tq2m+6pn+6D2 −23−tqpψ( qm−pn2 ) −23−2tq2pn+2 215−6tqm+6p2n+6D1′ −2t+2qpψ( qm−pn2 ) 21+2tqm+2p2 −29+6tq2m+6pn+6D2′ 23−tqpψ( qm−pn2 ) −23−2tq2pn+2 215−6tqm+6p2n+65. There exist integers m ≥ 0 and n ≥ 0 such that pn−qm2 is a square, t ∈ {0, 1} and E isQ-isomorphic to one of the following elliptic curvesa2 a4 ∆E1 2t+2qpψ(pn−qm2 ) 21+2tq2pn+2 −29+6tqm+6p2n+6E2 −23−tqpψ(pn−qm2 ) −23−2tqm+2p2 215−6tq2m+6pn+6E1′ −2t+2qpψ(pn−qm2 ) 21+2tq2pn+2 −29+6tqm+6p2n+6E2′ 23−tqpψ(pn−qm2 ) −23−2tqm+2p2 215−6tq2m+6pn+66. There exist integers m ≥ 0 and n ≥ 0 such that qm+12p is a square, qm ≡ 1 (mod 4),r, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curves126a2 a4 ∆F1 2t+2qpr+1ψ( qm+12p ) 21+2tqm+2p2r+1 29+6tq2m+6p6r+3F2 −23−tqpr+1ψ( qm+12p ) 23−2tq2p2r+1 215−6tqm+6p6r+3F1′ −2t+2qpr+1ψ( qm+12p ) 21+2tqm+2p2r+1 29+6tq2m+6p6r+3F2′ 23−tqpr+1ψ( qm+12p ) 23−2tq2p2r+1 215−6tqm+6p6r+37. There exist integers m ≥ 0 and n ≥ 0 such that qm−12p is a square, qm ≡ 3 (mod 4),r, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆G1 2t+2qpr+1ψ( qm−12p ) 21+2tqm+2p2r+1 −29+6tq2m+6p6r+3G2 −23−tqpr+1ψ( qm−12p ) −23−2tq2p2r+1 215−6tqm+6p6r+3G1′ −2t+2qpr+1ψ( qm−12p ) 21+2tqm+2p2r+1 −29+6tq2m+6p6r+3G2′ 23−tqpr+1ψ( qm−12p ) −23−2tq2p2r+1 215−6tqm+6p6r+38. There exist integers m ≥ 0 and n ≥ 0 such that pn+12q is a square, pn ≡ 1 (mod 4),s, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆H1 2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn+2 29+6tq6s+3p2n+6H2 −23−tqs+1pψ(pn+1q ) 23−2tq2s+1p2 215−6tq6s+3pn+6H1′ −2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn+2 29+6tq6s+3p2n+6H2′ 23−tqs+1pψ(pn+1q ) 23−2tq2s+1p2 215−6tq6s+3pn+69. There exist integers m ≥ 0 and n ≥ 0 such that pn−12q is a square, pn ≡ 3 (mod 4),s, t ∈ {0, 1} and E is Q-isomorphic to one of the following elliptic curvesa2 a4 ∆I1 2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn+2 −29+6tq6s+3p2n+6I2 −23−tqs+1pψ(pn+1q ) −23−2tq2s+1p2 215−6tq6s+3pn+6I1′ −2t+2qs+1pψ(pn+1q ) 21+2tq2s+1pn+2 −29+6tq6s+3p2n+6I2′ 23−tqs+1pψ(pn+1q ) −23−2tq2s+1p2 215−6tq6s+3pn+610. There exist integers r, s ∈ {0, 1} and E is Q-isomorphic to one of the following ellipticcurvesa2 a4 ∆J1 0 2q2s+1p2r+1 −29q6s+3p6r+3J2 0 −8q2s+1p2r+1 215q6s+3p6r+3127a2 a4 ∆K1 0 −2q2s+1p2r+1 29q6s+3p6r+3K2 0 8q2s+1p2r+1 −215q6s+3p6r+3Proof. The proof of this is completely analogous to [Mul06, p.146-149]. I will include thedetails here in the interest of correcting a few typos in the proof and to provide a referencefor the reader.We organize the curves from Theorem 2.1.1. Let A.I represent the curves labeled form1 to 9 in the two tables. Similarly, let A.II represent the curves labeled from 10 to 18 inthe tables and let A.III be the curves labeled from 19 to 27. Using the tables from [Mul06,p.15-16], we can easily compute the conductors of these curves just based on the valuationsof the values of a2, a4, and ∆. Keeping the notation from Theorem 2.1.1 (in particular, that` ≥ 1 unless it appears on the right hand side of an equation), we see thatA.I A.II A.IIIv2(a2) r1 = r1 + 1 if ` ≥ 1 ≥ r1 + 2> r1 + 1 if ` = 0v2(a4) `+ 2r1 − 2 2r1 2r1 + 1v2(∆) 2`+ 6r1 `+ 6r1 + 6 6r1 + 9In the first row, the greater sign in the second column comes from the fact that in thiscase, we can conclude that d must be even and so we get at least one additional exponent of2 coming from the d value in a2. Similarly, we do not know the exact contribution of v2(d)in the last column for a2 and so we have a greater than or equal to sign to account for this.The conductor computations yield for the curves in A.I thatv2(N) =0 if r1 = 0, ` = 6, a ≡ 1 (mod 4)1 if r1 = 0, ` ≥ 7, a ≡ 1 (mod 4)2 if r1 = 0, ` = 2, a ≡ 1 (mod 4), b ≡ −1 (mod 4)3 if r1 = 0, ` = 2, a ≡ −1 (mod 4), b ≡ 1 (mod 4)or r1 = 0, ` = 4, 5, a ≡ 1 (mod 4)4 if r1 = 0, ` = 2, a ≡ b ≡ ±1 (mod 4)or r1 = 0, ` ≥ 4, a ≡ −1 (mod 4)5 if r1 = 0, ` = 36 if r1 = 1, ` ≥ 27 if r1 = 1, 2, ` = 1.128For the curves in A.II, the conductor isv2(N) =0 if r1 = 0, ` = 6, a2 ≡ −1 (mod 4)1 if r1 = 0, ` ≥ 7, a2 ≡ −1 (mod 4)2 if r1 = 0, ` = 2, a− b ≡ 9 (mod 16)3 if r1 = 0, ` = 2, a− b ≡ 5 (mod 16)or r1 = 0, ` = 4, 5, a2 ≡ −1 (mod 4)4 if r1 = 0, ` = 2, a− b ≡ 1, 13 (mod 16)or r1 = 0, ` ≥ 4, a2 ≡ 1 (mod 4)5 if r1 = 0, ` = 0, b ≡ −1 (mod 4)or r1 = 0, ` = 3or r1 = 1, ` = 0, b4 ≡ 1 (mod 4)6 if r1 = 0, ` = 0, b ≡ 1 (mod 4)or r1 = 1, ` = 0, b4 ≡ −1 (mod 4)or r1 = 1, ` ≥ 1, b4 ≡ 1 (mod 4)7 if r1 = 0, ` = 1or r1 = 1, ` ≥ 1, b4 ≡ −1 (mod 4).Lastly, for the curves in A.III, we havev2(N) = 8.To compute vq(N), consider the condition(*) q appears as a coefficient of d2 in the associated Diophantine Equation from the tablesin Theorem 2.1.1.Then, we havevq(N) =0 if (*) does not hold and r2 = 0 and m = 01 if (*) does not hold and r2 = 0 and m 6= 02 if r2 = 1 or (*) holds.Similarly, let(**) p appears as a coefficient of d2 in the associated Diophantine Equation from the tablesin Theorem 2.1.1.129Then, we havevp(N) =0 if (**) does not hold and r3 = 0 and n = 01 if (**) does not hold and r3 = 0 and n 6= 02 if r3 = 1 or (**) holds.There is a key fact to point out in the above proof. To compute the conductor above, weused the tables from [Mul06, p.15-16]. In these tables, there is a different table when q = 3and for q ≥ 5. However, for our purposes, the only rows we are concerned with are the p andq-adic valuations of a2 and a4, the supplementary conditions and the conductor exponent.Looking only at these rows, we see that the q = 3 and the q = 5 cases coincide. From thesetables it also easy to see that the above values for vq(N) and vp(N) are indeed correct. It isthis fact here that shows us that the theorem from chapter 3 of [Mul06] can be generalizedalmost verbatim and give us the result stated above. Having finished the classification in this case, we proceed to specialize the above infor-mation in terms of setting q = 3 and q = 5. Most of the work for q = 3 was done in [Mul06]but as we will see, we will need to reorganize this data in a way that is suitable for thiswork. One thing we would like to now do is look at the table in Theorem 2.1.1 and solvethe listed Diophantine equations. It will turn out that for the most part we will not makeuse of the fact that we are dealing with primes p and q, but rather we will attempt to solvethe equations in greater generality. Doing this will require a range of techniques common toDiophantine equations including some recently developed results.130Chapter 3Diophantine EquationsThe goal of this chapter is to take the Diophantine equations that arose in Theorem 2.1.1and try to classify all integer solution to these equations. Throughout the course of solvingthese equations, we will require different techniques to solve equations with smaller exponentsand equations with larger exponents. This is because modularity methods, the techniquethat will end up being our primary tool, require that exponents be suitably large in orderto be applied. Thus, we will take a small detour and discuss integer, rational and {2, 5,∞}-solutions to the equations y2 = xq ± 2α5β for q ∈ {3, 5}. In our case we will find that ourtreatment of q = 5 will require a lot more work than was needed in the case y2 = x5±2α3β asin [Mul06]. There were complications with finding all rational solution to this equation and sowe reduce ourselves to finding integer solutions to these equations via Thue-Mahler equationsand an algorithm of [TdW92] with corrections in [TdW93] implemented by [Ham11]. All ofthe computer programs used in this section were implemented in MAGMA [BCP97].3.1 S-Integer Solutions to y2 = x3 ± 2α5βThe goal of this section is to find all integer solutions to y2 = x3±2α5β. In order to achievethis goal, it suffices to compute all S-integer solutions to y2 = x3 ± 2α5β for S = {2, 5,∞}and where 0 ≤ α, β ≤ 6. For this we use MAGMA to find the S-integer points and then usethis data to create the following theorems using their built in calculator. An explanation ofthis method can be found in [Mul06].The major idea here is to use bounds from Baker on linear forms in complex logarithms[Bak67] [Bak68] or the more recent updated work due to Matveev [Mat00]. In practice thesebounds can still be difficult to work with. In his 1987 thesis, de Weger [dW89] found a wayto lower the bound obtained via the LLL algorithm [LLL82]. Very briefly, let P1, ..., Pr be131generators for the free part of E(Q). Then any P ∈ E(Q) can be written of the formP = T + n1P1 + ...+ nrPrwhere T is an element of the torsion subgroup of E(Q). For a point (x, y) ∈ E(Q), one canshow similar to [Sil09, p. 292] that|x|−1/2p ≤ C2e−C3 max{|ni|}2where C2, C3 are constants depending on the elliptic curve, the generators Pi and S andp ∈ S is such that |x|p = max{|x|q : q ∈ S}. Finding a lower bound on this |x|p would inturn find a lower bound on max{|ni|} and thus would restrict the possible values of P . Theselower bounds can be obtained when p = ∞ in [Dav95] and in certain cases when the rankof E(Q) is bounded by at most 2 as in [RU96]. Fortunately, our elliptic curves of the formy2 = x3 ± 2α5β all have ranks bounded by at most 2. Hence we can apply these results andwe get the following tables as computed by MAGMA [BCP97].Theorem 3.1.1. The integer points ony2 = x3 + 2α5βwith x-coordinate pA, p 6= 2, 5, A ≥ 1 and α, β ≥ 0 integers are given by(α, β, x, y) = (5, 5, 41,±411).Theorem 3.1.2. The integer points ony2 = x3 + 2α5βwith x-coordinate 2ApB, p 6= 2, 5, B ≥ 1 and A,α, β ≥ 0 integers are given by132α β x y n0 + 6n 1 22n−4 · 41 ±23n−6 · 299 n ≥ 20 + 6n 4 22n+1 · 3 ±23n · 29 n ≥ 01 + 6n 0 22n−2 · 17 ±23n−3 · 71 n ≥ 11 + 6n 1 22n−2 · 32 ±23n−3 · 37 n ≥ 12 + 6n 2 22n+3 · 3 ±23n · 118 n ≥ 02 + 6n 3 −22n−2 · 31 ±23n−3 · 47 n ≥ 13 + 6n 0 −22n−2 · 7 ±23n−3 · 13 n ≥ 13 + 6n 0 22n+1 · 23 ±23n · 312 n ≥ 03 + 6n 1 22n+1 · 3 ±23n · 16 n ≥ 03 + 6n 3 −22n+1 · 3 ±23n · 28 n ≥ 03 + 6n 4 −22n+1 · 7 ±23n · 88 n ≥ 04 + 6n 1 22n+2 · 11 ±23n · 292 n ≥ 05 + 6n 5 22n · 41 ±23n · 411 n ≥ 0Table 3.1: Integer solutions to y2 = x3 + 2α5β with x-coordinate 2ApB, p 6= 2, 5 , B ≥ 1 andA,α, β ≥ 0.Theorem 3.1.3. The integer points ony2 = x3 + 2α5βwith x-coordinate 5ApB, p 6= 2, 5, B ≥ 1 and A,α, β ≥ 0 integers are given byα β x y m0 4 + 6m 52m+2 · 3 ±53m+2 · 26 m ≥ 03 3 + 6m 52m+1 · 13 ±53m+2 · 21 m ≥ 05 3 + 6m −52m+1 · 3 ±53m+2 m ≥ 05 5 + 6m 52m · 41 ±53m · 411 m ≥ 0Table 3.2: Integer solutions to y2 = x3 + 2α5β with x-coordinate 5ApB, p 6= 2, 5 , B ≥ 1 andA,α, β ≥ 0.Theorem 3.1.4. The integer points ony2 = x3 + 2α5βwith x-coordinate 2A5BpC, p 6= 2, 5, C ≥ 1 and A,B, α, β ≥ 0 integers are given by133α β x y n m0 + 6n 1 + 6m 22n−4 · 52m · 41 ±23n−6 · 53m · 299 n ≥ 2 m ≥ 00 + 6n 4 + 6m 22n+1 · 52m · 3 ±23n · 53m · 29 n ≥ 0 m ≥ 00 + 6n 4 + 6m 22n · 52m+2 · 3 ±23n · 53m+2 · 26 n ≥ 0 m ≥ 01 + 6n 0 + 6m 22n−2 · 52m · 17 ±23n−3 · 53m · 71 n ≥ 1 m ≥ 01 + 6n 1 + 6m 22n−2 · 52m · 32 ±23n−3 · 53m · 37 n ≥ 1 m ≥ 02 + 6n 2 + 6m 22n+3 · 52m · 3 ±23n · 53m · 118 n ≥ 0 m ≥ 02 + 6n 2 + 6m −22n−2 · 52m+1 · 3 ±23n · 53m+1 · 11 n ≥ 1 m ≥ 02 + 6n 3 + 6m −22n−2 · 52m · 31 ±23n−3 · 53m · 47 n ≥ 1 m ≥ 03 + 6n 0 + 6m −22n−2 · 52m · 7 ±23n−3 · 53m · 13 n ≥ 1 m ≥ 03 + 6n 0 + 6m 22n+1 · 52m · 23 ±23n · 53m · 312 n ≥ 0 m ≥ 03 + 6n 1 + 6m 22n+1 · 52m · 3 ±23n · 53m · 16 n ≥ 0 m ≥ 03 + 6n 3 + 6m −22n+1 · 52m · 3 ±23n · 53m · 28 n ≥ 0 m ≥ 03 + 6n 3 + 6m 22n · 52m+1 · 13 ±23n · 53m+2 · 21 n ≥ 0 m ≥ 03 + 6n 4 + 6m −22n+1 · 52m · 7 ±23n · 53m · 88 n ≥ 0 m ≥ 04 + 6n 1 + 6m 22n+2 · 52m · 11 ±23n · 53m · 292 n ≥ 0 m ≥ 04 + 6n 1 + 6m 22n+2 · 52m−1 · 11 ±23n · 53m−3 · 3927 n ≥ 0 m ≥ 15 + 6n 3 + 6m −22n · 52m+1 · 3 ±23n · 53m+2 n ≥ 0 m ≥ 05 + 6n 5 + 6m 22n · 52m · 41 ±23n · 53m · 411 n ≥ 0 m ≥ 0Table 3.3: Integer solutions to y2 = x3 + 2α5β with x-coordinate 2A5BpC , p 6= 2, 5, C ≥ 1and A,B, α, β ≥ 0.Theorem 3.1.5. The integer points ony2 = x3 − 2α5βwith x-coordinate pA, p 6= 2, 5, A ≥ 1 and α, β ≥ 0 integers are given byα β x y1 0 3 ±51 4 11 ±93 2 9 ±235 2 41 ±261Table 3.4: Integer solutions to y2 = x3 − 2α5β with x-coordinate pA, p 6= 2, 5 , A ≥ 1 andα, β ≥ 0.134Theorem 3.1.6. The integer points ony2 = x3 − 2α5βwith x-coordinate 2ApB, p 6= 2, 5, B ≥ 1 and A,α, β ≥ 0 integers are given byα β x y n1 + 6n 0 22n · 3 ±23n · 5 n ≥ 01 + 6n 4 22n · 11 ±23n · 9 n ≥ 02 + 6n 1 22n+1 · 3 ±23n+1 · 7 n ≥ 02 + 6n 2 22n+1 · 17 ±23n+1 · 99 n ≥ 03 + 6n 1 22n+1 · 7 ±23n+2 · 13 n ≥ 03 + 6n 2 22n+1 · 3 ±23n+2 n ≥ 03 + 6n 2 22n · 9 ±23n · 23 n ≥ 05 + 6n 2 22n · 41 ±23n · 261 n ≥ 0Table 3.5: Integer solutions to y2 = x3 − 2α5β with x-coordinate 2ApB, p 6= 2, 5 , B ≥ 1 andA,α, β ≥ 0.Theorem 3.1.7. The integer points ony2 = x3 − 2α5βwith x-coordinate 5ApB, p 6= 2, 5, B ≥ 1 and A,α, β ≥ 0 integers are given byα β x y m1 0 + 6m 52m · 3 ±53m · 5 m ≥ 01 4 + 6m 52m · 11 ±53m · 9 m ≥ 03 2 + 6m 52m · 9 ±53m · 23 m ≥ 05 2 + 6m 52m · 41 ±53m · 261 m ≥ 0Table 3.6: Integer solutions to y2 = x3 − 2α5β with x-coordinate 5ApB, p 6= 2, 5 , B ≥ 1 andA,α, β ≥ 0.Theorem 3.1.8. The integer points ony2 = x3 − 2α5βwith x-coordinate 2A5BpC, p 6= 2, 5, C ≥ 1 and A,B, α, β ≥ 0 integers are given by135α β x y n m0 + 6n 2 + 6m 22n−4 · 52m+1 · 13 ±23n−6 · 53m+1 · 83 n ≥ 2 m ≥ 01 + 6n 0 + 6m 22n · 52m · 3 ±23n · 53m · 5 n ≥ 0 m ≥ 01 + 6n 4 + 6m 22n · 52m · 11 ±23n · 53m · 9 n ≥ 0 m ≥ 02 + 6n 1 + 6m 22n+1 · 52m · 3 ±23n+1 · 53m · 7 n ≥ 0 m ≥ 02 + 6n 2 + 6m 22n+1 · 52m · 17 ±23n+1 · 53m · 99 n ≥ 0 m ≥ 02 + 6n 2 + 6m 22n−2 · 52m+1 · 37 ±23n−3 · 53m+1 · 503 n ≥ 1 m ≥ 03 + 6n 1 + 6m 22n+1 · 52m · 7 ±23n+2 · 53m · 13 n ≥ 0 m ≥ 03 + 6n 2 + 6m 22n+1 · 52m · 3 ±23n+2 · 53m n ≥ 0 m ≥ 03 + 6n 2 + 6m 22n · 52m · 9 ±23n · 53m · 23 n ≥ 0 m ≥ 04 + 6n 4 + 6m 22n−2 · 52m+2 · 17 ±23n−3 · 53m+2 · 349 n ≥ 1 m ≥ 05 + 6n 2 + 6m 22n · 52m · 41 ±23n · 53m · 261 n ≥ 0 m ≥ 0Table 3.7: Integer solutions to y2 = x3 − 2α5β with x-coordinate 2A5BpC , p 6= 2, 5, C ≥ 1and A,B, α, β ≥ 0.3.2 Integer and Rational Solutions to y2 = x5 ± 2α5βThe goal of this section is to compute solutions to y2 = x5 ± 2α5β for α, β ≥ 0. Thisequation is much more difficult to solve in comparison to the curve y2 = x5 ± 2α3β as wasconsidered by my predecessor in [Mul06]. To proceed to solve this question, we will beginwith attempting to solve this equation using Chabauty’s method. Chabauty’s method asimplemented here depends on two key factors. One is knowledge of the Mordell-Weil groupattached to the hyperelliptic curve and the other is ensuring that the aforementioned grouphas rank strictly less than the genus of our curves, which in our case is 2. Failing this, weapply the techniques of Elliptic Curve Chabauty to a covering of the remaining curves. Thismethod however also suffers from the Mordell-Weil group computations being difficult incertain cases. I will then briefly mention the work of Gallegos-Ruiz [GR11] on S-integralsolutions to Diophantine equations which is a continuation of the work of [BMS+08]. Inparticular, I will discuss how one verifies the ranks of certain genus 2 curves and then outlinehow an approach using these papers would proceed. Upon exhausting these techniques, wewill use the work of [Ham11] on Thue-Mahler equations to help solve the remaining cases.3.2.1 Chabauty’s MethodWe begin by using Chabauty’s method [Cha41] to figure out the rational solutions to theequation y2 = x5± 2α5β where (x, y) ∈ Q2 and α, β ∈ Z+. A hypothesized set of solutions isgiven as follows by searching for points with small heights.136Conjecture 3.2.1. Rational solutions to y2 = x5 +2α5β with integers 0 ≤ α, β ≤ 9 are givenby1. If α, β ∈ 2Z, then (x, y) = (0,±2α/25β/2) is a set of solutions.2. If α, β ∈ 5Z, then (x, y) = (−2α/55β/5, 0) is a set of solutions.3. In addition to the above solutions, we have solutions given byα β 2α5β (x, y)0 1 5 (−1,±2)1 0 2 (−1,±1)1 1 10 (−1,±3)1 2 50 (−1,±7)1 5 6250 (15,±875), (−154 ,±237532 )2 0 4 (2,±6)2 4 2500 (5,±75)2 5 12500 (5,±125)3 0 8 (1,±3)4 1 80 (−1,±9)5 0 32 (2,±8)6 2 1600 (−4,±24)6 4 40000 (20,±1800)8 1 1280 (4,±48), (−4,±16)8 3 3200 (−4,±176)8 5 800000 (20,±2000)Table 3.8: Rational solutions to y2 = x5 + 2α5β.Conjecture 3.2.2. Rational solutions to y2 = x5−2α5β with integers 0 ≤ α, β ≤ 9 are givenby1. If α, β ∈ 5Z, then (x, y) = (2α/55β/5, 0) is a set of solutions.2. In addition to the above solutions, we have solutions given by137α β 2α5β (x, y)0 4 625 (5,±50)1 3 250 (11,±401)2 2 100 (5,±55)2 4 2500 (5,±25)4 0 16 (2,±4)4 3 2000 (6,±76), (14,±732)4 4 10000 (10,±300)4 8 6250000 (25,±1875), (50,±17500)5 0 32 (6,±88)Table 3.9: Rational solutions to y2 = x5 − 2α5β.Despite inabilities to prove these conjectures, we will be able to prove the followingtheorem about the integer solutions to the above curves.Theorem 3.2.3. All integer solutions to y2 = x5±2α5β are given by taking the above integersolutions when (0 ≤ α, β ≤ 9 and if α, β ≥ 10, additional integer solutions are given bytranslations via the following. Take rα ≡ α (mod 10) for rα ∈ {0, 1, ..., 9} and similarly forrβ. Then take any integer solution (x, y) for C = 2rα5rβ above and notice that(X, Y ) := (x · 2(α−rα)/5 · 5(β−rβ)/5, y · 2(α−rα)/2 · 5(β−rβ)/2)is a solution to Y 2 = X5 ± 2α5β.As a final note, notice that in all of the x-coordinates above, only the primes 2, 3, 5, 7 and11 divide the numerator of the x-coordinate of a solution to y2 = x5 ± 2α5β.To prove this theorem, we will use a myriad of techniques. The first of which is Chabauty’smethod. The key to Chabauty’s method is to notice that the rational points on a hyperellipticcurve can be embedded into its Jacobian. It can be shown that the set of rational pointson the Jacobian form a finitely generated abelian group known as the Mordell-Weil group.When one can show that the Mordell-Weil group has rank strictly less than the genus, thenChabauty’s method can give us a way to determine the rational points on our curve. Onecan associate to the curve a p-adic power series and use Strassman’s Theorem to give anupper bound on the number of p-adic solutions to the power series in terms of its coefficients.This corresponds to an upper bound on the number of rational solutions to the originalequation and if the bound can be obtained, one can conclude that these are all the rationalpoints on your curve. Given that the prime above can be any prime outside of a finite set,there are many such options one can use to find these upper bounds. In fact, one can also138use information at many different primes to also help obtain tight bounds on the numberof solutions. For an excellent primer I refer the reader to [Mul06]. I will include some ofthe finer details missing from that excerpt here in this thesis as well as present Chabauty’smethod from the viewpoint of differentials as in [Col85]. For a different viewpoint see [CF96].Let C be a hyperelliptic curve defined byC : y2 = f(x)where disc(f) 6= 0 over a number field K. It does no harm for the reader to think of K as Qsince that is where our main application will be.An important concept for us will be the Jacobian of a hyperelliptic curve, denoted by Jthroughout. For an in depth discussion of the Jacobian, I refer the reader to the primer in[Mul06, Chapter 5]. For us, we will need a few crucial facts about the Jacobian that I willoutline here. Firstly, if C/K is a curve and C(K) 6= ∅ then we have thatJ(K) ∼= Div0(C/K)/Prin(C/K)where the Div0(C/K) is the set of rational degree zero divisors and Prin(C/K) is the set ofprincipal divisors over C (elements corresponding to div (f) for some f ∈ k(C)∗). In fact,one can define a map called the Abel-Jacobi map associated to P0 ∈ C(K) defined byC → JP 7→ [P − P0].The hyperelliptic curve has an invariant called the genus and is given by g =⌊deg(f)−12⌋[HS00, p.86-87] (this follows from the Riemann-Hurwitz formula [DS05, p.66]). Let Ω be thespace of regular differentials. The Riemann-Roch Theorem [Sil09, p.35] states that this spaceis a K-vector space and has dimension g. In fact, a basis for this space, as shown in [DR11,p.55-56] can be given by {dxy,xdxy, ...,xg−1dxy}.Further, letting p be an odd prime not dividing the discriminant of f , we have a bilinearpairing given byΩ× J(Qp)→ Qp〈ω,∑i[Pi −Qi]〉 7→∑i∫ PiQiω.139This pairing has the properties that it is Qp linear on the left, Z-linear on the right and thatthe kernel on the right is the torsion subgroup of J(Qp) [Sik13]. The key assumption thatthe rank r of the Jacobian J(Q) ≤ g − 1 will come into play in the following proposition.Proposition 3.2.4. With the notation as above and assuming that r ≤ g − 1, there exists adifferential ω called an annihilating differential such that〈ω,D〉 = 0holds for all D ∈ J(Q).The proof of the above is relatively straightforward as well. Since the dimension of Ω isg and the rank of J(Qp) is g − 1, one can simply take each of the basis elements ωg, takean arbitrary∑ri=1 niRi ∈ J(Qp) and solve the linear system created from these values. Anexplicit example of this method can be found in [MP12]. I will say a few words about thisintegral. These integrals above are called tiny integrals and can be evaluated like normalintegrals. Use a power series representation to write 1/y subject to y2 = f(x) as an elementof Zp[[x]] and then integrate term by term substituting the endpoints into the x-coordinateby the Fundamental Theorem of Calculus. Fix some point say P0 ∈ C(Q). Then using theAbel-Jacobi map, we see that if another point P ∈ C(Q) were to exist, then [P −P0] ∈ J(Q)implies that∫ PP0ω = 0 where ω is an annihilating differential. The integral becomes a powerseries in Zp[[x]] and thus our problem reduces to finding a root of a polynomial in this ring.At first glance this problem may seem harder, however a theorem of Strassman can help us.Theorem 3.2.5. (Strassman’s Theorem) Suppose thatf(x) =∞∑i=0aixi ∈ Zp[[x]]where Zp is the p-adic integers. Further suppose that limi→∞ai = 0. Let k = minivp(ai) and letN = max{i : vp(ai) = k}. Then the number of roots of f(x) in Zp is at most N .The last ingredient in this method is to show that all known points form all the possibleneighbourhoods where potential points on C(Q) can live. This can be accomplished by aMordell-Weil Sieve. Suppose that our curve C/Q has rank 1 and that J(Q) = 〈D〉. Thenwith the Abel-Jacobi map, we have that if P ∈ C(Q), then [P − P0] = nD for any fixed P0.We would like to know what possible values of n there are. Use the known points of C(Q) togive an initial set of these n values and then for different values of p, compute the order of thereduced divisor D¯ ∈ J(Fp). Doing this for multiple values of p can help narrow the possiblevalues of n and these can be translated to find a small collection of neighbourhoods wherepoints of C(Q) can live. When we can show that all points must live in the neighbourhoods140as discovered from Chabauty’s method, we are done after including possible torsion pointsof J(Q).With the above approach in mind, I will divide our curves into groups of curves withrank 0, 1, ≥ 2 and those whose rank we cannot determine. The strategy for determiningthe Mordell-Weil rank of Jacobians is to use descent arguments to get an upper bound onthe rank and then attempt to find linearly independent points on the Jacobian to establishthis rank. There is no currently known algorithm to compute this number exactly and thisfact causes many problems for us in determining the exact structure of these Mordell-Weilgroups.Case 1: Known rank 0 curvesOf the 100 cases of curves with a positive sign, there are 36 which can by proven to haverank exactly 0. These correspond to(α, β) ∈{(0, 0), (0, 2), (0, 3), (0, 5), (0, 7), (0, 8), (1, 3), (1, 8), (2, 2), (2, 3), (2, 7), (2, 8), (3, 2),(3, 3), (3, 7), (3, 8), (4, 0), (4, 2), (4, 5), (4, 7), (5, 2), (5, 7), (6, 0), (6, 3), (6, 5), (6, 8),(7, 2), (7, 7), (8, 0), (8, 4), (8, 5), (8, 9), (9, 0), (9, 2), (9, 5), (9, 7)}and with a negative sign, we see that there are 28 cases given by(α, β) ∈{(1, 0), (1, 5), (2, 0), (2, 1), (2, 5), (2, 6), (3, 2), (3, 3), (3, 7), (3, 8), (4, 1), (4, 6), (5, 2),(5, 7), (6, 1), (6, 4), (6, 6), (6, 9), (7, 2), (7, 7), (8, 0), (8, 4), (8, 5), (8, 9), (9, 0), (9, 2),(9, 5), (9, 7)}.Of these curves, we can run the commands for each (α, β) in the above list_<x> := PolynomialRing(Rationals());C := HyperellipticCurve(x^5+2^a * 5^b);Chabauty0(Jacobian(C));and find that rational points occur on the following curves in the positive case141α β 2α5β (x, y)0 0 1 (−1, 0)0 0 1 (0,±1)0 2 25 (0,±5)0 5 3125 (−5, 0)0 8 390625 (0,±625)2 2 100 (0,±10)2 8 1562500 (0,±1250)4 0 16 (0,±4)4 2 400 (0,±20)6 0 64 (0,±8)6 8 25000000 (0,±5000)8 0 256 (0,±16)8 4 160000 (0,±400)8 5 800000 (20,±2000)Table 3.10: Rational solutions to y2 = x5 + 2α5β for rank 0 curvesand in the negative case none of the curves have any rational points other than the pointat infinity.Case 2: Known rank 1 curvesIn this case, we discuss the situation where we know the generator of a curve with rank1. The following list of coordinates give rise to such curves in the positive sign case(α, β) ∈{(0, 1), (0, 4), (1, 1), (1, 4), (1, 6), (2, 0), (2, 1), (2, 4), (2, 5), (3, 0), (3, 1), (3, 4), (4, 1),(4, 4), (5, 0), (5, 1), (5, 4), (6, 1), (6, 4), (7, 0), (7, 1), (7, 4), (8, 2), (8, 3), (9, 1)}and in the negative sign case, we have(α, β) ∈{(0, 0), (0, 2), (1, 4), (2, 2), (3, 0), (3, 1), (4, 0), (4, 2), (5, 0), (5, 1), (6, 0), (6, 2), (8, 2)}.There are also curves where the computation for the given model of our hyperellipticcurves above go beyond what MAGMA can do. To handle these cases, we can change themodel to use smaller coefficients and this helps limit the search space to something moremanageable. There were six other cases in the positive sign setting given by(α, β) ∈{(0, 6), (2, 6), (3, 5), (5, 5), (7, 5), (8, 7)}142and ten others in the negative sign setting given by(α, β) ∈{(0, 5), (0, 7), (0, 8), (2, 8), (4, 7), (6, 5), (7, 0), (7, 5), (8, 8), (9, 1)}where we can use the trick of changing the model to help us compute the Mordell-Weil groupof the Jacobian by reducing the size of the coefficients involved. It will turn out that we canverify a generator in each of these cases as well.To determine the Jacobian in these cases, we use MAGMA. To illustrate the methodsused above, I will show them explicitly using the curves y2 = x5 + 5 and y2 = x5 + 56. Tobegin with the first curve, we set up the polynomial ring, the hyperelliptic curve and theJacobian.> _<x> := PolynomialRing(Rationals());> C := HyperellipticCurve(x^5 + 5);> J := Jacobian(C);> J;Jacobian of Hyperelliptic Curve defined by y^2 = x^5 + 5 over Rational FieldWe would like to know what the rank of this Jacobian is and for this we can use“RankBounds”.> RankBounds(J);1 1There is also a command called “RankBound” which just gives the upper bound. From thiswe see that the lower and upper bounds of the Jacobian is indeed 1. Thus our curve hasrank 1. Next, we search for rational point on our curve up to some bound. We use the LLLalgorithm [LLL82] to help determine a basis of this group.> ratPoints := RationalPoints(J:Bound:=5000);> B := ReducedBasis(ratPoints);> B;[ (x + 1, 2, 1) ]The point P = (x + 1, 2, 1) is on the Jacobian and MAGMA suggests that this could bea possible generator. Remember that this point is in the Mumford representation (for moredetails, see [Mul06]). To determine if this is the generator, we need to check the following.Suppose that this was not the generator. Then there would be some point P0 such thatP = mP0 for some integer m. However, this m is an integer at least of size 2 and so using143the canonical height of the Jacobian (denoted by hˆ) [HS00, p. 199] [Sto02b, Chapter 6], wesee thathˆ(P ) = hˆ(mP0) = m2hˆ(P0) ≤ 4hˆ(P0).This gives us that hˆ(P0) ≤ 14 hˆ(P ) which in turn gives us a way to bound how far we need tocheck for rational points. The command RationalPoints(J:Bound:=N) will look for rationalpoints up to naive height N on the associated Kummer Surface to our Jacobian (for moredetails, see [CF96]). We can compute the maximum difference between the canonical heightand naive height using MAGMA and thus, to verify the bound, we need to check for rationalpoints up to the boundexp(hˆ(P )4+HC)where HC is the height constant, this maximal height difference between canonical and naiveheights. For more details see [Sto99] [Sto02b]. Computing gives us that> P := B[1];> HC := HeightConstant(J:Effort:=2);> hP := Height(P);> boundToCheck := Exp(hP/4 + HC);> boundToCheck;11.1994504369602782170687264455As we have already checked up to 5000, we have successfully determined that P is agenerator. In some cases, the computations are too large to be performed. For example,consider the curve y2 = x5 + 56:> C1 := HyperellipticCurve(x^5 + 5^6);> J1 := Jacobian(C1);> ratPoints1 := RationalPoints(J1:Bound:=15000);> B1 := ReducedBasis(ratPoints1);> B1;[ (x^2 - 105/16*x - 525/16, 2205/64*x - 575/64, 2) ]> HC1 := HeightConstant(J1:Effort:=2);> hPoint := Height(B1[1]);> Exp(hPoint/4+HC1);57732.5711881962379928792966209This computation just exceeds what MAGMA can do (the max bound rational points willallow is 46340). We can reduce our model and perform this computation again. MAGMA hasa command that will give us a smaller model (though this is not an exact science - sometimesother models can do better).144> C2, mapC1toC2 := ReducedMinimalWeierstrassModel(C1);> C2;Hyperelliptic Curve defined by y^2 = 5*x^5 + 25 over Rational Field> J2 := Jacobian(C2);> ratPoints2 := RationalPoints(J2:Bound:=1000);> B2 := ReducedBasis(ratPoints2);> HC2 := HeightConstant(J2: Effort:=2);> hPoint2 := Height(B2[1]);> Exp(hPoint2/4+HC2);595.986306378025334312831604667This computation is manageable. In our specific setting, we can also find a different modelby hand by using say (x, y) 7→ (2x, 22y) or (x, y) 7→ (5x, 52y) (or a combination of thesemaps) provided that the power of α, β ≥ 4.Next, we can execute Chabauty’s method on all these curves that we have computeda Mordell-Weil basis for their respective Jacobians and come up with the following set ofsolutions to the curves y2 = x5 ± 2α5β. To use the following command, we need to tellMAGMA which prime specifically to check. This method returns the number of points thatare roots of the equation over Zp[[x]] discussed earlier modulo a prime power of p and takingthe cardinality of this set returns a bound on half the number of non-Weierstrass points. Bydefinition, Weierstrass points are the points (x, 0) and the point at infinity. We also need tocheck the torsion subgroup for any other points not coming from the free part of J(Q). Inour example y2 = x5 + 5, we have// Recall C := HyperellipticCurve(x^5 + 5)> RationalPoints(C:Bound:=1000000);{@ (1 : 0 : 0), (-1 : -2 : 1), (-1 : 2 : 1) @}> TorsionSubgroup(J);Abelian Group of order 1Mapping from: Abelian Group of order 1 to JacHyp:J given by a rule [no inverse]We can see that we do not have any points coming from the torsion subgroup and weonly have one non-Weierstrass point given by (−1,±2). Since we know two non-Weierstrasspoints on our curve, we are done if our Chabauty method returns the value 1.//Recall P := B[1];> #Chabauty(P,7);2145> #Chabauty(P,11);5> #Chabauty(P,13);1We reach our bound using the prime p = 13. Thus we can conclude that the number ofrational points on y2 = x5 + 5 is exactly 2 given by (−1 ± 2). We repeat this computationon all the curves y2 = x5 + 2α5β that have rank 1 to illustrate this method. The columndenoted by p denotes the prime used in Chabauty.α β C(Q)\∞ p0 1 (−1,±2) 130 4 (0,±25) 110 6 (0,±125) 71 1 (−1,±3) 111 4 111 6 112 0 (0,±2), (2,±6) 192 1 292 4 (0,±50), (5,±75) 612 5 (5,±125) 192 6 (0,±250) 113 0 (1,±3) 133 1 113 4 313 5 174 1 (1,±9) 19α β C(Q)\∞ p4 4 (0,±100) 115 0 (−2, 0), (2,±8) 1095 1 175 4 195 5 (−10, 0) 116 1 176 4 (0,±200), (20,−1800) 2297 0 117 1 297 4 597 5 118 2 (0,±80) 118 3 (−4,±176) 118 7 179 1 19Table 3.11: Rational points on y2 = x5 + 2α5β for known rank 1 curves.In the later versions of MAGMA, we can also perform Chabauty, the Mordell-Weil sievingand the torsion computation without explicitly giving the primes needed to check. This canbe done using> Chabauty(P); //Recall P is the generator of J.{ (-1 : -2 : 1), (1 : 0 : 0), (-1 : 2 : 1) }{ 3, 19, 29, 59, 79 }[ 5, 2, 3, 2 ]146The first line is the list of rational points. The next two are Mordell-Weil sieving datawhich we will not discuss here. This helps to take care of ghost solutions as in [Mul06, p.191]and [BC06, p.63-92]. Using this command, we do likewise for the curve y2 = x5 − 2α5β andsee thatα β C(Q)\∞0 0 (1, 0)0 20 5 (5, 0)0 70 81 42 2 (5,±55)2 83 03 14 0 (2,±4)4 2α β C(Q)\∞4 54 75 0 (2, 0), (6,±88)5 16 06 26 57 07 58 28 89 1Table 3.12: Rational points on y2 = x5 − 2α5β for known rank 1 curves.3.2.2 Elliptic Curve ChabautyWe now address the question of what happens when Chabauty’s method cannot be appliedeither because the rank of J(Q) is greater than 1 or because we cannot specifically computethe exact size of the rank. In this case, there are some more advanced techniques that onecan try to use (see also [Mul06], [Sto98] and [Sto02a] for other tricks, but the ones mentionedthere do not work here). The first of which is Elliptic Curve Chabauty. The idea behindthis method is to pass our curves to a covering collection of curves that map down to ellipticcurves over a relatively small number field K. Over this number field, we try to find allpoints where the x component is rational. These give a superset of all the rational pointson our elliptic curve. One can then map these points back to the original curve and verifywhether or not they are rational points for our original curve. For more details, see [Bru02]and [Bru03].Our first step is to get this covering collection. A common way to do this is to use a1472-cover descent. The main idea here is to write our hyperelliiptic curve C/K asC : y2 = f(x) = g(x)h(x) = fndeg(f)∏i=1(x− θi)where h(x) is monic (for us it will be a monic linear term), g(x) is of even degree andg(x), h(x) ∈ L[x] for some finite extension L of K and fn is the coefficient of xn in f(x).For our applications here our f(x) is a degree five monic polynomial and so we make thissimplifying assumption. For more details on any of these methods in full generality, thereader is encouraged to see [BS08], [BS09] and [BS10].Define HK := {δ ∈ A∗/A∗2 : NA/K(δ) ∈ k∗2} where A = K[x]/f(x) = K[θ] for θ = θ1.For δ ∈ HK , we can write down the cover given byDδ =y21 = δ1(x− θ1)...y25 = δ5(x− θ5)y = y1...y51 = δ1...δ5where here we are thinking of δ1, ..., δ5 as conjugates of δ and hence we think of y1, ..., y5 asconjugates as well. Condensing the above map and relabeling givesEγ : γy21 = g(x)E′γ : (1/γ)y22 = h(x)where this γ is associated to the δ from above as follows. Suppose that δ ∈ A and letL[Θ] = L[x]/h(x). Define the mapj : A→ L[Θ]θ 7→ Θand extend the above map in the natural way. Then, we have that γ = NL[Θ]/L(j(δ)). Denotethis γ by γ(δ) to remind ourselves of the association. The question now arises “How manyof these such δ (and consequently γ(δ)) values do we have to consider?” The smallest suchset we can hope to obtain by purely local means is given by the fake 2-Selmer group. This is148defined as follows. The map from A→ A⊗K Kv gives a commutative diagramC(K)µK> HKC(Kv)∨µKv> HKvρKv∨From here, we define the fake 2-Selmer group asSel(2)fake(C/K) = {δ ∈ HK : ρKv ∈ µKv(C(Kv)) for all places v ∈ K}.This group was first introduced by Poonen and Schaefer [PS97]. It is related to the Selmergroup by being equal to a quotient of the Selmer group. For more details on this, see[SvL13]. With all this terminology now introduced, we can note that there is an algorithm forcomputing the fake 2-Selmer group and this has been implemented in MAGMA. Proceedingwith the computation gives> P<x> := PolynomialRing(Rationals());< poly := x^5 + 2*5^5;> C := HyperellipticCurve(poly);> Hk,AtoHk := TwoCoverDescent(C);> Hk := [d @@ AtoHk : d in Hk];> Hk;[1,-1/5*theta - 1]> K<w> := NumberField(poly);> HK := [Evaluate(P!d, w) : d in Hk];> HK;[1,1/5*(-w - 5)]This means that every rational point on our curve C leads to a rational point with thesame Q-rational x-coordinate on one of the two genus 1 curves given byE1 : y2 = g(x)E2 : y2 = −1/5(w + 5)g(x)149where g(x) := x5+2·55x−w . We load this now into MAGMA> PK<X> := PolynomialRing(K);> g := ExactQuotient(Evaluate(PK!poly,X), X - w);We will run Elliptic Curve Chabauty on each elliptic curve. To do this, we iterate over allvalues of “HK” and use MAGMA’s built-in capabilities. First we set up our elliptic curve.> for d in HK dofor> C2 := HyperellipticCurve(d*g);for> E, mE := EllipticCurve(C2, RationalPoints(C2:Bound:=20)[1]);Here we might need some user intervention depending on if we cannot find any small rationalpoints. If you know a rational point on the curve, for example infinity when d = 1 and x = 15when d = −1/5(w + 5), then you can use the commandfor> Rep(Points(C2, Infinity()))instead of searching for rational points. With the elliptic curve built, we can now attemptto compute the rank. This command can take a bit of time depending on the curve. For thepurposes of showing sample output, I will assume that d = −1/5(w+ 5) and that we use thepoint at x = 15.for> E, mE := EllipticCurve(C2, Rep(Points(C2, 15)));for> bounds, gens := MordellWeilShaInformation(E);for> end for;Torsion Subgroup = Z/2The 2-Selmer group has rank 3Found a point of infinite order.Found 2 independent points.After 2-descent:2 <= Rank(E) <= 2Sha(E)[2] is trivial(Searched up to height 16 on the 2-coverings.)It is here that we should note we need the rank above to be exact. If we cannot computethe exact rank of E then we cannot guarantee that the next steps are valid. Here we knowrank(E) = 2 and so we can continue to apply Elliptic Curve Chabauty since this rank isless than the degree of the number field extension (which is 5). Next, we set up an abstractrepresentation of the Mordell-Weil group of E.150> T, mapTtoE := TorsionSubgroup(E);> A := AbelianGroup(Invariants(T) cat [0 : g in gens]);> gensE := [mapTtoE(T.j) : j in [1..Ngens(T)]] cat gens;> MWmap := map<A -> E | a :-> &+[e[i]*gensE[i] : i in [1..#e]]where e := Eltseq(a)>;> MWmap;Mapping from: GrpAb: A to CrvEll: E given by a rule [no inverse]Some notes on the above code. The code “T.j” refers to the elements of T (and in partic-ular the generators). The symbols “& +” will take the map on each of the generators and putthem together to form an element in E. Elliptic Curve Chabauty requires a representation ofthe Mordell-Weil group and a map from E to P1. The major idea is the following. Let C/Kbe an algebraic curve where K is a number field. Suppose that C covers an elliptic curveE defined over L where L is a finite extension of K. Next, consider following commutativediagramCφ> EP1pi∨Φ >where Φ is defined over K and φ, pi are defined over L. The map φ is the aforementionedcover of E. From this map, we can see thatΦ(C(K)) ⊆ φ(E(L)) ∩ P1(K)The right most set can sometimes be bounded using p-adic methods. One such situation iswhen rank(E) ≤ [L : K]. The details of this method can be found in [Bru02] and [Bru03]. Ifthe bound is sharp, then we have found all rational points on C. For our purposes, MAGMAhas the built in functionality and so we will not go into the details. We compute the necessaryvalues as follows.> P1 := ProjectiveSpace(Rationals(), 1);> piC2toP1 := map<C2 -> P1 | [FunctionField(C2).1,1]>;> piE := Inverse(mE)*piC2toP1;> set, n := Chabauty(MW, piE);> set, n;{0,$.1 + $.2 - $.3,$.1 + $.2 - 3*$.3,151-2*$.3}396902628105984000Lastly, we need to verify that the output is valid. This is the case if the index of A inE(K) is finite and coprime to n as computed in the above code. This can be done using the“Saturation” command.> primes := PrimeDivisors(n);> p := 1;> omitPrimes := [];> while(p lt (Max(primes)+1)) dowhile> p := NextPrime(p);while> if not(p in primes) thenwhile|if> omitPrimes := Append(omitPrimes,p);while|if> end if;while> end while;> sat := Saturation(gens, Max(primes) :TorsionFree, OmitPrimes := omitPrimes);> sat eq gens;TrueThe “Saturation” command takes in rational points on E and an integer n and returns asequence of points generating a subgroup of E(Q) such that the given points are containedin the subgroup and that the subgroup is p-saturated for all primes p up to n excluding thosegiven by the optional “OmitPrimes” parameter. Recall that a subgroup S is p-saturated in agroup G if there is no intermediate subgroup H for which the index in S is finite and divisibleby p. This is the exact condition we wish to verify. If “TorsionFree” is specified above to betrue, then the torsion points are omitted from the result. Checking this to be equal to thegenerators is enough.If this last check of equality with the generators turns out to be false, then use thesaturated group in place of the original generators (in our case given by “gens”) and repeatthe computation. Check that the new second value does not have new prime divisors or elserepeat the saturation step with the new primes and loop until you get an equality.Translating back to the curve C, we see that the rational points on C corresponding tod = −1/5(w + 5) are given by> xCoords := {piE(MW(s)) : s in set};> ptsOnC := &join{Points(C, pt[2] eq 0 select Infinity() else pt[1]/pt[2]) :152pt in xCoords};> xCoords; ptsOnC;{ (15 : 1), (-5 : 1) }{@ (15 : 875 : 1), (15 : -875 : 1) @}We can repeat this1 for the point d = 1 and see thatC(Q) = {∞, (15,−875), (15, 875), (−15/4,−2375/32), (−15, 2375/32)}.We can run this method on the remaining curves and we can verify the rational points ony2 = x5 ± 2α5β for the curvesα β ± 2α5β (x, y)0 1 − 5 None0 3 − 125 None0 4 − 625 (5,±50)0 6 − 15625 None0 9 + 1953125 None0 9 − 1953125 None1 5 + 6250 (15,±875),(−154 ,±237532 )2 9 + 7812500 None3 5 − 25000 None5 3 + 4000 None5 3 − 4000 None5 4 − 20000 None5 6 + 500000 Noneα β ± 2α5β (x, y)5 6 − 500000 None5 8 + 12500000 None5 8 − 12500000 None5 9 + 62500000 None5 9 − 62500000 None6 6 + 1000000 None8 3 − 32000 None8 7 + 20000000 None8 7 − 20000000 None8 8 − 100000000 None9 4 + 320000 None9 4 − 320000 None9 6 + 8000000 NoneTable 3.13: Rational solutions to y2 = x5 ± 2α5β solved using Elliptic Curve Chabauty.3.2.3 S-integer PointsYet another technique we can attempt to use is that of finding S-integer points on hyperel-liptic curves based on [GR11] and [BMS+08]. Currently, the code is not yet fully implementedin MAGMA. In lieu of applying this method by hand, we note that the calculation is madesubstantially easier if one already knows the Mordell-Weil group exactly and so I will givesome examples of how one verifies this in practice for higher ranked curves.1In reality, we get an error in this case caused by a MAGMA programming error. However in the version Iused, we can use a different model of our curve using the command “MinimalReducedWeierstrassModel(C)”and come up with the other two points in the same manner.153Contrary to the rank 1 case where we could do a simple computation involving the heightconstant and one fourth the value of the presumed generator, when we consider ranks 2or higher we need to use more sophisticated machinery [Sto02b]. The main idea here is tocompute a value called the covering radius % (or at least an upper bound on it). Once wehave this value, we use an upper bound similar to that in the rank 1 case to check that thereare no more rational points on the Jacobian of naive height at mostexp(%2 +HC)where again HC is the height constant. We illustrate this method using the curves y2 =x5 − 2453 and y2 = x5 − 2459. First we proceed with the second curve which has rank 2 asgiven by> _<x> := PolynomialRing(Rationals());> C := HyperellipticCurve(x^5 - 2^4*5^9);> C1, mapCtoC1 := ReducedMinimalWeierstrassModel(C); C1;Hyperelliptic Curve defined by y^2 = -5*x^6 + 50*x over Rational Field> J1 := Jacobian(C1);> B := ReducedBasis(RationalPoints(Jacobian(C1):Bound:=8000));> B;[ (x^2 + x + 18/7, 9/7*x + 12/7, 2),(x^2 + 17/3*x + 32/3, 1/9*x + 712/9, 2) ]Then to compute the covering radius, we use the formula found in [Sto02b, p. 178] asgiven by%2 =hˆ(P1)hˆ(P2)hˆ(P1 ± P2)4Reg(P1, P2)where Reg denotes the regulator and the sign above is chosen to give the smaller value ofhˆ(P1 ± P2). If we use the height pairing matrix M = (〈Pi, Pj〉)1≤i,j≤2 where〈Pi, Pj〉 =hˆ(P1 + P2)− hˆ(P1)− hˆ(P2)2we can also see that%2 =〈P1, P1〉〈P2, P2〉(〈P1, P1〉+ 〈P2, P2〉+ 2|〈P1, P2〉|)4 det(M)valid since hˆ(P1 + P2)− hˆ(P1 − P2) = 2(hˆ(P1) + hˆ(P2)). This gives> H1 := HeightPairingMatrix(B);> H1;154[5.70203132416513267250593814558 0.607244504928574524217534640890][0.607244504928574524217534640890 8.21959988070301859177146320080]> rsq := H1[1,1]*H1[2,2]*(H1[1,1]+H1[2,2]-2*Abs(H1[1,2])/(4*Determinant(H1));> rsq;3.20197769760017634737894478458> HC := HeightConstant(J1 : Effort := 2);> Exp(HC + rsq);7604.25353647160094586160852567This confirms that our generators for J(Q) are given by(x2 + x+ 18/7, 9x/7 + 12/7, 2), (x2 + 17x/3 + 32/3, x/9 + 712/9, 2).If we pull back, we see that the generators are> C5 := HyperellipticCurve(x^5 - 2^4*5^9);> C10, mapping := ReducedMinimalWeierstrassModel(C5);> basOrig := [Pullback(mapping,b) : b in B];> basOrig;[ (x^2 - 175/9*x + 8750/9, 1225/27*x - 16250/27, 2),(x^2 - 425/16*x + 1875/8, -47525/64*x + 314375/32, 2) ]With rank 3 and higher, there is no exact formula like in the r = 2 case. However MAGMAis still capable of computing this. We perform this computation for our only rank 3 curvegiven by y2 = x5 − 2453. As always, we produce potential candidates for our generators.> _<x> := PolynomialRing(Rationals());> C2 := HyperellipticCurve(x^5 - 2^4*5^3);> C3, mapC2toC3 := ReducedMinimalWeierstrassModel(C2); C3;Hyperelliptic Curve defined by y^2 = 2*x^5 - 125 over Rational Field> J3 := Jacobian(C3);//Can take a while> B3 := ReducedBasis(RationalPoints(Jacobian(C3):Bound:=24500));> B3;[(x - 3, 19, 1), (x - 7, 183, 1), (x^2 + 1/2*x + 3/2, 19/2*x + 5/2, 2) ]Next, we compute the covering radius. This is done by using the height pairing matrixas before, creating a lattice from the matrix and then computing the covering radius.> H3 := HeightPairingMatrix(B3);155> L3 := LatticeWithGram(H3);> r3 := CoveringRadius(L3);Runtime error in ’CoveringRadius’: Argument 1 must be over Z or QThe problem with this code is that the height pairing matrix is over a real field of precision30. In order to compute the covering radius, we must tell MAGMA to think of “H3” as arational matrix. To do this, we truncate the entries at a few different values for n and thenwe compute the covering radius. Note that the command “CoveringRadius” runs in timeapproximately factorial in the size of the matrix and so is really only feasible for small valuesof the size of the matrix.> seq := [];> n := 8; //Vary for precision>for a in [1,2,3] do> for b in [1,2,3] do//Trick to round values multiply by power of 10, floor, then divide.> seq := Append(seq,Floor((10^n)*H[a][b])/10^n);> end for;>end for;> H3 := Matrix(RationalField(),3,3,seq);> L3 := LatticeWithGram(H3);> r3 := CoveringRadius(L3);> r3;2.27285021881299If we vary the n values from say 1 to 10, we can see that %2 ≤ 2.3. This gives> HC := HeightConstant(Jacobian(C):Effort:=2);> Exp(2.3^2 + HC);24116.3154447130917357362220427and we have already verified past this bound so our basis is given by(x− 3, 19, 1), (x− 7, 183, 1), (x2 + x/2 + 3/2, 19x/2 + 5/2, 2).Now that one has these points, one could in principle use the techniques of [GR11] to findthe S-integral points on this curve. In order to use this technique, one need only two morefacts, namely a bound on the number of S-integral points of the curve and a variant of theMordell-Weil Sieve capable of searching up to the bounds. I will not do this here becausethe Mordell-Weil Sieving code still requires some in depth user knowledge before executing156properly which the author has yet to obtain. I will however state that there is no reason thatthis technique could not be done if the user had enough knowledge of the code (or enoughtime to write his or her own). We will see in the section on Thue-Mahler equations thatwork has already been done to classify the integer points on these two curves and that willbe enough for our purposes.3.2.4 Other TechniquesWe can also use elliptic curves in the rare case that the polynomial on the right factors.For example, fory2 = x5 − 2555 = (x− 10)(x4 + 10x3 + 100x2 + 1000x+ 10000),we have that the factors on the right are coprime and hence we can use techniques for finding{2, 5∞}-integral points on the elliptic curvey2 = x4 + 10x3 + 100x2 + 1000x+ 10000.We use MAGMA via the command//The following command includes integral points> SIntegralQuarticPoints([1,10,100,1000,10000], [2,5]);[[ 30, -1100 ],[ -10, -100 ],[ -55/4, 2525/16 ],[ 0, 100 ]]A quick check shows that none of these points leads to a {2, 5,∞}-integer point on y2 =x5− 2555. Thus the only such point must come when y = 0 and hence x = 10. This gives usthat the only {2, 5}-integer points on this curve is the point (x, y) = (10, 0).3.2.5 Integer Points Via Thue-Mahler EquationsWhen the previous methods fail one can always try to simplify the search by trying to findinteger solutions to your equations. The goal of this section is to compute integer solutionsto y2 = x5±2α5β with α, β ≥ 0. Our primary tools will be techniques from algebraic numbertheory as well as reductions to Thue-Mahler equations, that is, equations of the formf(x, y) = cpz11 ...pznn157where f(x, y) is an irreducible binary form in Z[x, y] of degree at least 3. From here, we applyan algorithm of [TdW92] as implemented by [Ham11] to solve these equations. Throughoutwe assume that xy 6= 0 as these solutions are easy to obtain. In this case, we know thatthe greatest common divisor of x and y has only powers of 2 and 5. Thus, removing all thepowers of 2 and 5 in x and y, lety = 2y15y2 yˆx = 2x15x2xˆ.Plugging this into our equation, we have22y152y2 yˆ2 = 25x155x2xˆ5 ± 2A5B.Notice that two of the powers of 2 must be equal and the third must be a larger power ofthe smaller two. This leads to the following cases:1. 2y1 = 5x1 and 2y2 = 5x22. 2y1 = 5x1 and 2y2 = B3. 2y1 = 5x1 and 5x2 = B4. 2y1 = A and 2y2 = 5x25. 2y1 = A and 2y2 = B6. 2y1 = A and 5x2 = B7. 5x1 = A and 2y2 = 5x28. 5x1 = A and 2y2 = B9. 5x1 = A and 5x2 = B.We proceed case by case. Note in all the cases below, the class number is coprime to 5 andso factoring in the ring of integers will be permitted.1. 2y1 = 5x1 and 2y2 = 5x2.This case is equivalent to gcd(x, y) = 1. For the duration of this case, to easy notation,we suppose that we have a solution to y2 = x5± 2α5β with gcd(x, y) = 1. The work forthe equation with a negative sign has been done as a consequence of the following.Theorem 3.2.6. [GLT08] (Goins, Luca, Togbe) The only solutions to y2 = x5 −2α5β for gcd(x, y) = 1 and α, β non-negative integers are given by (x, y, α, β) =(11,±401, 1, 3) or (x, y, α, β) = (1, 0, 0, 0).158This was done using work on Lucas sequences and S-integral points on elliptic curves.For the rest of this section, we attempt to apply a myriad of techniques to the equationy2 = x5 + 2α5β with gcd(x, y) = 1 to come up with a full solution set to the aboveproblem. We attempt to solve this first by factoring.Case 1: Suppose that α = 2a and β = 2b for some non-negative integers a and b.Then we have (y − 2a5b)(y + 2a5b) = x5. Now, let d = gcd(y − 2a5b, y + 2a5b). Addingthe two values shows that d | 2y. Now if d | y, then d | 2α5β and so d | x, contradictingthe coprimality of x and y. If d = 2 then we know that 2 - y from the above argumentand so we must have that a = 0 when d = 2. The argument to follow only works whend = 1 and so we assume that a 6= 0. The case where a = 0 was solved earlier usingChabauty’s method. Unique factorization tells us thaty − 2a5b = M5 and y + 2a5b = N5and so subtracting gives N5 + (−M)5 = 2a5b. Using Theorem 1.6.2 (valid since theright hand side is a square), we have that the only solution to x5 + y5 = Cz2 forC ∈ {1, 2, 5, 10} with x > y is given by C = 2 and (n, x, y, z) = (5, 3,−1,±11). Sincehere z is a power of 2 and 5 in our situation, we have a contradiction. SummarizinggivesLemma 3.2.7. There are no solutions to y2 = x5 + 2α5β with α, β even non-negativeintegers with α > 0 and x, y coprime integers.This technique worked because we could factor in a unique factorization domain. Wenow attempt to try this over the ring Z(√2) which is also a unique factorization domain.Case 2: We assume now that α = 2a + 1 and β = 2b. Performing a factorization inZ[√2] gives us (y−2a5b√2)(y+2a5b√2) = x5. Again, let d = gcd(y−2a5b√2, y+2a5b√2)for d ∈ Z[√2]. Adding the two values shows that d | 2y. Now if d | y, then d | 2α5βand so d | x thus d = 1. If d = 2 then as d | 2a5b√2, we have that d | y and the aboveagain shows that d | x, a contradiction. Unique factorization tells us thaty + 2a5b√2 = (1 +√2)k(m+ n√2)5 (3.1)where m,n ∈ Z, 1 +√2 is a fundamental unit for Z[√2] and we have absorbed rootsof unity into the 5th power. We may suppose without loss of generality that 0 ≤ k ≤ 4since we can absorb all the 5th powers into the (m + n√2)5 term if necessary. Usingthe first equation above, we can expand and compare coefficients of√2 to get a set of159equations for each value of k given by fk(m,n) = 2δ15δ2 given byk = 0 f0(m,n) := n(5m4 + 20m2n2 + 4n4)k = 1 f1(m,n) := m5 + 5m4n+ 20m3n2 + 20m2n3 + 20mn4 + 4n5k = 2 f2(m,n) := 2m5 + 15m4n+ 40m3n2 + 60m2n3 + 40mn4 + 12n5k = 3 f3(m,n) := 5m5 + 35m4n+ 100m3n2 + 140m2n3 + 100mn4 + 28n5k = 4 f4(m,n) := 12m5 + 85m4n+ 240m3n2 + 340m2n3 + 240mn4 + 68n5.To solve this, we use the Thue-Mahler solver. The program takes as an input anirreducible binary form of degree at least 3, a parameter and a set of primes given bya tuple ([c0, ..., cn], [p1, ..., pk], A) wherec0XN + c1XN−1Y + ...+ cNYN = ±Apδ11 · · · pδkkand outputs coprime integers X, Y with gcd(c0, Y ) = 1 satisfying the above equation.First, let’s see why solving for solutions with gcd(X, Y ) = 1 is not an issue. Notice thatfor a solution (m0, n0) to our Thue-Mahler equation above implies that gcd(m0, n0) | yby Equation 3.1. As gcd(m0, n0) | 2a5b, we must also have that gcd(m0, n0) | x as well.So gcd(m0, n0) | gcd(x, y) = 1 showing that gcd(m0, n0) = 1 since we assumed that xand y were coprime. Thus solving the Thue-Mahler equation for coprime m and n issufficient.For our purposes, the restriction that gcd(c0, Y ) = 1 is unwanted so to deal with this,we solve the following set of related Thue-Mahler equations gj(m,n) = ±2δ15δ2 wheregj(m,n) are given by [Ham11, p.6]k = 0 g1(m,n) := 5m4 + 20m2n2 + 4n4k = 0 g2(m,n) := m4 + 100m2n2 + 500n2k = 1 g3(m,n) := m5 + 5m4n+ 20m3n2 + 20m2n3 + 20mn4 + 4n5k = 2 g4(m,n) := 2m5 + 15m4n+ 40m3n2 + 60m2n3 + 40mn4 + 12n5k = 2 g5(m,n) := m5 + 15m4n+ 80m3n2 + 240m2n3 + 320mn4 + 192n5k = 3 g6(m,n) := 5m5 + 35m4n+ 100m3n2 + 140m2n3 + 100mn4 + 28n5k = 3 g7(m,n) := m5 + 35m4n+ 500m3n2 + 3500m2n3 + 12500mn4 + 17500n5k = 4 g8(m,n) := 12m5 + 85m4n+ 240m3n2 + 340m2n3 + 240mn4 + 68n5k = 4 g9(m,n) := 6m5 + 85m4n+ 480m3n2 + 1360m2n3 + 1920mn4 + 1088n5k = 4 g10(m,n) := 3m5 + 85m4n+ 960m3n2 + 5440m2n3 + 15360mn4 + 17408n5.160To give an idea of the above collection and where they come from, take the case withk = 0 above. To go to the second k = 0 case, multiply every term by 53 and thenrelabel mnew = 5m. Each time you do this you will find solutions with d | y where d isthe base of the number we are multiplying by. For example, solving with j = 9 above,we see that we will find solutions to our original Thue-Mahler equation with 2 | n andsolving with j = 10 above, we see that we will find solutions to our original Thue-Mahler equation with 4 | n. Solving the above sets of equations gives the followingsolutions.j Solutions to gj(m,n) = ±2δ15δ2 in the form (m,n, δ1, δ2)1 (1, 0, 0, 1), (−1, 0, 0, 1)2 (1, 0, 0, 0), (−1, 0, 0, 0)3 (1, 0, 0, 0), (−1, 0, 0, 0)4 (−1, 1, 0, 1), (1,−1, 0, 1)5 (−2, 1, 4, 1), (2,−1, 4, 1), (1, 0, 0, 0), (−1, 0, 0, 0)6 (−1, 1, 1, 0), (1,−1, 1, 0)7 (−1, 0, 0, 0), (1, 0, 0, 0), (−5, 1, 1, 4), (5,−1, 1, 4)8 (−1, 1, 0, 0), (1,−1, 0, 0), (−2, 1, 2, 0), (2,−1, 2, 0)9 (2,−1, 4, 0), (−2, 1, 4, 0), (−4, 1, 6, 0), (4,−1, 6, 0)10 (4,−1, 8, 0), (−4, 1, 8, 0), (−8, 1, 10, 0), (8,−1, 10, 0)Table 3.14: Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0).Notice that in the cases of j = 4 and j = 6 above, we get only one more additionalsolution to our original equation with gcd(c0, Y ) > 1 and these come when Y = 0. Tosatisfy gcd(X, Y ) = 1, we have that only (m,n) = (±1, 0) are added solutions givenfrom this reduction process. Our k = 0 situation above does not give us a valid solutionsince the coefficient of√2 on the left is nonzero. Thus our final list of solutions forfk(m,n) = 2α5β is as follows:161k Solutions to fk(m,n) = 2δ15δ2 in the form (m,n, δ1, δ2) Corresponding (x, y, α, β)0 None None1 (1, 0, 0, 0) (−1, 1, 1, 0)2 (−1, 1, 0, 1) (−1,−7, 1, 2)2 (1, 0, 0, 0) (1, 3, 3, 0)3 (1,−1, 1, 0) (1,−3, 3, 0)3 (1, 0, 0, 0) (−1, 7, 1, 2)4 (−1, 1, 0, 0) (−1,−1, 1, 0)4 (−2, 1, 2, 0) (2,−8, 5, 0)Table 3.15: Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to the coefficientof gj(m, 0).To compute the last column above, we plug back in (m,n) into the Equation 3.1and solve for the corresponding variables. Notice that the last solution, we obtain asolution with gcd(x, y) > 1. Notice that finding solutions with gcd(m,n) = 1 does notnecessarily guarantee finding solutions with gcd(x, y) = 1 but it does ensure that wefind all such solutions with gcd(x, y) = 1. Summarizing the above, we get the followingLemma 3.2.8. Solutions to y2 = x5 + 2α5β with α, β non-negative integers with α oddand β even and x, y coprime integers are given by the following(x, y, α, β) ∈ {(−1,±1, 1, 0), (−1,±7, 1, 2), (1,±3, 3, 0)}Case 3: We assume now that α = 2a and β = 2b + 1. Notice that Z[√5] is not aunique factorization domain, however Z[1+√52 ] is. Performing a factorization in Z[1+√52 ]gives us (y − 2a5b√5)(y + 2a5b√5) = x5. Again, let d = gcd(y − 2a5b√5, y + 2a5b√5)for d ∈ Z[1+√52 ]. Adding the two values shows that d | 2y. Now if d | y, then d | 2α5βand so d | x thus d = 1. If d = 2 then we know from above that d - y and so α = 0 andy is odd. Thus x is even and modulo 8, we have thaty2 = x5 + 2α5β ≡ 0 + 52b+1 ≡ 5 (mod 8)which is a contradiction since odd squares modulo 8 are congruent to 1. Thus we havethat d = 1 and so unique factorization tells us thaty + 2a5b√5 =(1 +√52)k(m+ n√52)5(3.2)162where m,n ∈ Z, 1+√52 is a fundamental unit for Z[1+√52 ] and we have absorbed rootsof unity into the 5th power. As before, we may suppose without loss of generality that0 ≤ k ≤ 4.In contrast to the previous situation where we had leading coefficients that mighthave shared prime divisors with the n value, we do not have coefficients that are nolonger necessarily integers. To simplify this we’ll multiply through to eliminate thedenominators (notice that our set of primes already contains 2 so this introduces noadditional concerns). This can give us additional solutions depending on how big thepower of 2 is. Evaluating the powers of the fundamental unit, we see that{1 +√52}k=4k=0={1,1 +√52,3 +√52, 2 +√5,7 + 3√52}.From this, we see when k = 0, 3, then we can multiply by 25 to clear denominatorsand in the other three k cases, we multiply by 26 to clear denominators. This gives thefollowing quintic forms fk(m,n) = 2δ15δ2 after expanding and comparing coefficients of√5.k = 0 g0(m,n) := 25f0(m,n)/(5n) := m4 + 10m2n2 + 5n4k = 1 g1(m,n) := 26f1(m,n) := m5 + 5m4n+ 50m3n2 + 50m2n3 + 125mn4 + 25n5k = 2 g2(m,n) := 26f2(m,n) := m5 + 15m4n+ 50m3n2 + 150m2n3 + 125mn4 + 75n5k = 3 g3(m,n) := 25f3(m,n) := m5 + 10m4n+ 50m3n2 + 100m2n3 + 125mn4 + 50n5k = 4 g4(m,n) := 26f4(m,n) := 3m5 + 35m4n+ 150m3n2 + 350m2n3+ 375mn4 + 175n5.Solving using the Thue-Mahler solver, we havej Solutions to gj(m,n) = ±2δ15δ2 in the form (m,n, δ1, δ2)0 (−1, 0, 0, 1), (1, 0, 0, 1), (1,−1, 4, 1), (−1, 1, 4, 1), (1, 1, 4, 1)(−1,−1, 4, 1)1 (−1, 0, 0, 0), (1, 0, 0, 0), (0, 1, 0, 2), (0,−1, 0, 2), (1, 1, 8, 0), (−1,−1, 8, 0)2 (−1, 0, 0, 0), (1, 0, 0, 0), (−1, 1, 6, 0), (1,−1, 6, 0), (−5, 1, 7, 2), (5,−1, 7, 2)3 (−1, 0, 0, 0), (1, 0, 0, 0), (0,−1, 1, 2), (0, 1, 1, 2)4 (1,−1, 5, 0), (−1, 1, 5, 0), (3,−1, 8, 0), (−3, 1, 8, 0), (−5, 1, 5, 2), (5,−1, 5, 2)Table 3.16: Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0).163Now, notice that in the k = 0, 3 cases, there are no solutions with the power of 2 atleast 5. In the other three cases, there are solutions with the power of 2 at least 6 andthe parity of m and n are the same. In these cases, we get two solutions given bym+ n√5 andm+ n√52Call solutions relating to the first equation to be type 1 and those of the second type2. This gives the following table of solutions.k Type Sol’ns to fk(m,n) = 2δ15δ2 of form (m,n, δ1, δ2) Corresponding (x, y, α, β)0 1 (−1, 1, 4, 1) (−4,−176, 8, 3)0 1 (1, 1, 4, 1) (−4, 176, 8, 3)1 1 (1, 1, 8, 0) (288, 4, 14, 1)1 2 (1, 1, 8, 0) (1, 9, 4, 1)2 1 (−1, 1, 6, 0) (−4, 64, 10, 1)2 2 (−1, 1, 6, 0) (−1,−2, 0, 1)2 1 (−5, 1, 7, 2) (20,−4000, 12, 5)2 2 (−5, 1, 7, 2) (5,−125, 2, 5)3 1 (1, 0, 0, 0) (−1, 2, 0, 1)3 1 (0, 1, 1, 2) (5, 125, 2, 5)4 1 (−1, 1, 5, 0) (−4, 16, 8, 1)4 1 (−3, 1, 8, 0) (4, 128, 14, 5)4 2 (−3, 1, 8, 0) (1,−9, 4, 1)4 1 (−5, 1, 5, 2) (20, 2000, 8, 5)Table 3.17: Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to the coefficientof gj(m, 0).Lemma 3.2.9. Solutions to y2 = x5 +2α5β with α, β non-negative integers with α evenand β odd and x, y coprime integers are given by the following(x, y, α, β) ∈ {(−1,±2, 0, 1), (−1,±9, 4, 1)}.Case 4: We assume now that α = 2a + 1 and β = 2b + 1. Notice that both x and ymust be odd since the power of 2 is nonzero. Now we have a bigger issue to deal with.The class group of Q(√10) has order 2 so we have to argue using ideals in the ring of164integers OQ(√10) = Z[√10]. Factoring into ideals gives〈y − 2a5b√10〉〈y + 2a5b√10〉 = 〈x〉5Now suppose there exists a proper prime ideal p containing both ideals on the right.Notice that summing the two generators shows that 2y ∈ p. Since p is prime either2 ∈ p or y ∈ p. If 2 ∈ p, Then notice that p must also contain x5 and hence mustcontain x. As x is odd, we get that 1 = gcd(x, 2) ∈ p which is a contradiction. Nowsuppose that y ∈ p. Then again as p = 〈y − 2a5b√5〉, we see that x ∈ p and hence1 = gcd(x, y) ∈ p which is again a contradiction. Thus the two ideals are coprime.Hence we can factor as〈y − 2a5b√10〉 = I5for some ideal I. Now the class group of Z[√10] is of size 2 so we must have that I isprincipal since 2 and 5 are coprime. Thus, setting I = (m+ n√10), we have thaty − 2a5b√10 = (3 +√10)k(m+ n√10)5for k ∈ {0, 1, 2, 3, 4} where 3 +√10 is a fundamental unit and we use the usual trick ofabsorbing roots of unity to the 5th power. As in case 2, we can expand the equationabove and compare coefficients of√10 to get a set of equations for each value of k givenby fk(m,n) = 2δ15δ2 given byk = 0 f0(m,n) := 5n(m4 + 20m2n2 + 20n4)k = 1 f1(m,n) := m5 + 15m4n+ 100m3n2 + 300m2n3 + 500mn4 + 300n5k = 2 f2(m,n) := 6m5 + 95m4n+ 600m3n2 + 1900m2n3 + 3000mn4 + 1900n5k = 3 f3(m,n) := 37m5 + 585m4n+ 3700m3n2 + 11700m2n3+ 18500mn4 + 11700n5k = 4 f4(m,n) := 228m5 + 3605m4n+ 22800m3n2 + 72100m2n3+ 114000mn4 + 72100n5.Again we mimic case 2. We need to account for the cases when the leading coefficientshares a divisor with an element of {2, 5}. To do this, we solve the following set of165related Thue-Mahler equations gj(m,n) = ±2δ15δ2 where gj(m,n) are given byk = 0 g1(m,n) := m4 + 20m2n2 + 20n4k = 1 g2(m,n) := m5 + 15m4n+ 100m3n2 + 300m2n3 + 500mn4 + 300n5k = 2 g3(m,n) := 6m5 + 95m4n+ 600m3n2 + 1900m2n3 + 3000mn4 + 1900n5k = 2 g4(m,n) := 3m5 + 95m4n+ 1200m3n2 + 7600m2n3 + 24000mn4 + 30400n5k = 3 g5(m,n) := 37m5 + 585m4n+ 3700m3n2 + 11700m2n3 + 18500mn4 + 11700n5k = 4 g6(m,n) := 228m5 + 3605m4n+ 22800m3n2 + 72100m2n3+ 114000mn4 + 72100n5k = 4 g7(m,n) := 114m5 + 3605m4n+ 45600m3n2 + 288400m2n3+ 912000mn4 + 1153600n5k = 4 g8(m,n) := 57m5 + 3605m4n+ 91200m3n2 + 1153600m2n3+ 7296000mn4 + 18457600n5.Solving the above sets of equations gives the following solutions.j Solutions to gj(m,n) = ±2δ15δ2 in the form (m,n, δ1, δ2)1 (1, 0, 0, 0), (−1, 0, 0, 0)2 (1, 0, 0, 0), (−1, 0, 0, 0)3 (−5, 1, 0, 2), (5,−1, 0, 2)4 (−5, 1, 4, 2), (5,−1, 4, 2), (−10, 1, 4, 2), (10,−1, 4, 2)5 (−5, 2, 0, 2), (5,−2, 0, 2), (5,−1, 5, 2), (−5, 1, 5, 2)6 (3,−1, 0, 0), (−3, 1, 0, 0)7 (6,−1, 4, 0), (−6, 1, 4, 0)8 (12,−1, 8, 0), (−12, 1, 8, 0)Table 3.18: Solutions to gj(m,n) = ±2δ15δ2 with gcd(m,n) = 1 and n coprime to thecoefficient of gj(m, 0).Here we see we gain additional solutions when we allow gcd(c0, Y ) > 1 only in the casewhen k = 2 and this gives the solutions (−5, 2, 5, 2) and (5,−2, 5, 2) with the secondsolution corresponding to a negative sign in fk(m,n) = ±2δ15δ2 . Thus our final list ofsolutions for fk(m,n) = 2α5β is as follows:166k Solutions to fk(m,n) = 2δ15δ2 in the form (m,n, δ1, δ2) Corresponding (x, y, α, β)0 None None1 (1, 0, 0, 0) (−1, 3, 1, 1)2 (−5, 1, 0, 2) (15,−875, 1, 5)2 (−5, 2, 5, 2) (−15,−2375, 11, 5)3 (−5, 2, 0, 2) (15, 875, 1, 5)3 (5,−1, 5, 2) (−15, 2375, 11, 5)4 (−3, 1, 0, 0) (−1,−3, 1, 1)Table 3.19: Solutions to fk(m,n) = 2δ15δ2 with gcd(m,n) = 1 and n coprime to the coefficientof gj(m, 0).Summarizing the above, we get the followingLemma 3.2.10. The only solution to y2 = x5 + 2α5β with α, β non-negative integerswith α and β odd and x, y coprime integers is given by (x, y, α, β) = (−1,±3, 1, 1).Thus, collecting all the above lemmas gives rise to the solutions with the powers of 2and 5 in x and y are equal as(x, y,±, A,B) ∈ {(2(A−1)/55B/5,±2(A−1)/25B/2,+, 1 + 10A1, 10B1),(2(A−3)/55B/5,±2(A−3)/25B/2(3),+, 3 + 10A1, 10B1),(−2(A−1)/55(B−1)/5,±2(A−1)/25(B−1)/2(3),+, 1 + 10A1, 1 + 10B1),(−2(A−1)/55(B−2)/5,±2(A−1)/25(B−2)/2(7),+, 1 + 10A1, 2 + 10B1),(−2(A−4)/55(B−1)/5,±2(A−4)/25(B−1)/2(9),+, 4 + 10A1, 1 + 10B1),(2(A−1)/55(B−3)/5(11),±2(A−1)/25(B−3)/2(401),−, 1 + 10A1, 3 + 10B1)}.2. 2y1 = 5x1 and 2y2 = B. Cancelling out the equal terms and relabelling termsyˆ2 = 5xˆ2xˆ5 ± 2Aˆ.We have solved the case where 5 | x2 via Chabauty methods so without loss of generality,we may suppose that x2 > 0. Modulo 5 considerations shows us that A is even. Thus,bringing over and factoring gives(yˆ − 2A/2)(yˆ + 2A/2) = 5xˆ2xˆ5or that(yˆ − 2A/2i)(yˆ + 2A/2i) = 5xˆ2xˆ5167As the terms are coprime in each case, we have thatyˆ ∓ 2A/2 = 5xˆ2K5yˆ ± 2A/2 = L5where K,L ∈ Z or we have thatyˆ + 2A/2i = γxˆ2α5where α = (a + bi) ∈ Z[i] and γ = 1 ± 2i. Without loss of generality again, we mayassume that γ = 1 + 2i since if α = a+ bi is a solution to2A/2i = =((1 + 2i)xˆ2α5)then we have that α0 = −a+ bi is a solution to2A/2i = =((1− 2i)xˆ2α50).In the first case, we get the Thue-Mahler equations corresponding to5sx5 − y5 = ±2r for s ∈ {0, ..4} and where gcd(y, 5) = 1.where we also note that gcd(y, 5) = 1 or we get a Thue-Mahler equation correspondingto one of the following curves (denote the left hand sides by g0, g1, g2 and g3 respectively)2y5 + 5y4x− 20y3x2 − 10y2x3 + 10yx4 + x5 = ±2r4y5 − 15y4x− 40y3x2 + 30y2x3 + 20yx4 − 3x5 = ±2r−2y5 − 55y4x+ 20y3x2 + 110y2x3 − 10yx4 − 11x5 = ±2r−24y5 − 35y4x+ 240y3x2 + 70y2x3 − 120yx4 − 7x5 = ±2r.We enter these cases into the Thue-Mahler equation solver and come up with thefollowing solutions5s(0)5 − (1)5 = −205s(0)5 − (−1)5 = 2051(1)5 − (1)5 = 2251(−1)5 − (−1)5 = −22and for the other 4 equations (where below we keep the same sign for ± when we make168a choice)2 solutions to g0(x, y) = ±2r are given by(x, y, r,±) ∈ {(±1, 0, 0,±), (−1, 1, 2,−), (1,−1, 2,+), (0,±1, 1,±)}.Solutions to g1(x, y) = ±2r(x, y, r,±) ∈ {(0,±1, 2,±), (−1,−1, 2,+), (1, 1, 2,−)}.Solutions to g2(x, y) = ±2r(x, y, r,±) ∈ {(0,±1, 1,±)}.There are no solutions to g3(x, y) = ±2r.Transferring back to the original equation gives rise to the solutions(x, y,±, A,B) ∈ {(2(A−1)/55B/5,±2(A−1)/25B/2,+, 1 + 10A1, 10B1),(2(A−3)/55B/5,±2(A−3)/25B/2(3),+, 3 + 10A1, 10B1),(−2(A−2)/55(B+1)/5,±2(A−2)/25B/2(3),+, 2 + 10A1, 4 + 10B1),(−2(A−2)/55(B+3)/5,±2(A−2)/25B/2(11),−, 2 + 10A1, 2 + 10B1),(−2(A−2)/55(B+1)/5,±2(A−2)/25B/2,−, 2 + 10A1, 4 + 10B1),(2(A−4)/55(B+2)/5,±2(A−4)/25B/2(3),−, 4 + 10A1, 8 + 10B1)}.3. 2y1 = 5x1 and 5x2 = B. Canceling out the equal terms and relabeling terms5yˆ2 yˆ2 = xˆ5 ± 2Aˆ.If yˆ2 is even, then we get the equationY 2 = X5 ± 2Aˆ.Otherwise, if yˆ2 is odd, multiplying through by 55 gives the equationY 2 = X5 ± 2Aˆ55.Here we can use the results from Chabauty and Elliptic Curve Chabauty methods to2Notice that above we reversed the roles of x and y. This is since the Thue-Mahler equation solver hasthe additional constraint that it only returns solutions with gcd(c0, y) = 1 where c0 is the coefficient of x5.In the form above, we note that any solution must necessarily have gcd(c0, y) = 1.169see that the only solutions to Y 2 = X5 ± 2Aˆ with Aˆ ≤ 9 are given by(X, Y, Aˆ,±) ∈ {(−1, 0, 0,+), (0,±1, 0,+), (1, 0, 0,−), (−1,±1, 1,+), (0,±2, 2,+),(2,±6, 2,+), (1,±3, 3,+), (0,±4, 4,+), (2,±4, 4,−), (−2, 0, 5,+),(2,±8, 5,+), (2, 0, 5,−), (6,±88, 5,−), (0,±8, 6,+), (0,±16, 8,+)}and the only solutions to Y 2 = X5 ± 2Aˆ55 with Aˆ ≤ 9 are given by(X, Y, Aˆ,±) ∈ {(−5, 0, 0,+), (5, 0, 0,−), (15,±875, 1,+), (−15/4,±2375/32, 1,+),(5,±125, 2,+), (−10, 0, 5,+), (10, 0, 5,−), (20,±2000, 8,+)}.Of the above solutions, in this case, we only consider the ones where 2 - gcd(x, y)(otherwise we violate the condition that 2y1 = 5x1). The remaining 5 cases give rise tothe solutions(x, y,±, A,B) ∈ {(2(A−1)/55B/5,±2(A−1)/25B/2,+, 1 + 10A1, 10B1),(2(A−3)/55B/5,±2(A−3)/25B/2(3),+, 3 + 10A1, 10B1),(2(A−1)/55B/5(3),±2(A−1)/25(1+B)/2(7),+, 1 + 10A1, 5 + 10B1),(2(A−1)/55B/5(3),±2(A−1)/25(1+B)/2(19),+, 11 + 10A1, 5 + 10B1),(2(A−2)/55B/5,±2(A−2)/25(1+B)/2,+, 2 + 10A1, 5 + 10B1)}.4. 2y1 = A and 2y2 = 5x2. Canceling out the equal terms and relabeling termsyˆ2 = 2xˆ1xˆ5 ± 5Bˆ.In the case where yˆ2 = 2xˆ1xˆ5 − 5Bˆ, if x1 ≥ 2, then reducing modulo 4 shows us thatyˆ2 = 2xˆ1xˆ5 − 5Bˆ has no solutions. Thus we are left only with the caseyˆ2 = 2xˆ5 − 5Bˆwith gcd(xˆ, yˆ) = 1. Here we may use [AMLST09] to see that the only solutions to thisequation are given by(xˆ, yˆ, Bˆ) ∈ {(3,±19, 3), (7,±183, 3)}170and these translate to(x, y,±, A,B) ∈ {(2(A+1)/55(B−3)/5(3),±2A/25(B−3)/2(19),−, 4 + 10A1, 3 + 10B1),(2(A+1)/55(B−3)/5(7),±2A/25(B−3)/2(3)(61),−, 4 + 10A1, 3 + 10B1))}.We finish this case off by considering when yˆ2 = 2xˆ1xˆ5 + 5Bˆ. Notice here that reducingmodulo 4 yields that xˆ1 ≥ 2. Considerations modulo 8 thus reveal that Bˆ is odd whenyˆ2 = 4xˆ5 + 5Bˆ and that Bˆ is even when yˆ2 = 2xˆ1xˆ5 + 5Bˆ for xˆ1 ≥ 3.For the case where yˆ2 = 4xˆ5 +5Bˆ with Bˆ is odd, we may factor over the ring of integersof Q(√5) (namely Z[(1 +√5)/2]) to see that(yˆ − 5(Bˆ−1)/2√5)(yˆ + 5(Bˆ−1)/2√5) = 4xˆ5.The greatest common divisor of the terms on the left is 2 and so we have thatyˆ + 5(Bˆ−1)/2√52= k(a+ b√52)5where = (1+√5)/2. Expanding and simplifying gives a subset of the cases consideredin case 1. In particular, we are looking for solutions there with 24+k in the Thue-Mahlerequation. This gives desired solutions when k = 0, 4 (see the table from the first casein this section) and in these cases, we have the solutions given by(a, b, k, yˆ, Bˆ) ∈ {(−1, 1, 0, 11, 3), (1, 1, 0,−11, 3), (−1, 1, 4, 1, 1), (−5, 1, 4, 125, 5)}and also solutions coming just from units as given by(a, b, k, yˆ, Bˆ) ∈ {(1, 0, 1, 1, 1), (1, 0, 3, 3, 1)}.When we translate these solutions to the equation yˆ2 = 4xˆ5 + 5Bˆ, we get(xˆ, yˆ, Bˆ) ∈ {(−1,±11, 3), (−1,±1, 1), (5,±125, 5), (1,±3, 1)}.In this case, we need that 2y1 = A and 2y2 = 5x2 which eliminates the third solutionabove. Summarizing, we get the solutions to the original equation given by(x, y,±, A,B) ∈ {(2(A+2)/55(B−1)/5,±2A/25(B−1)/2(3),+, 8 + 10A1, 1 + 10B1),(2(A+2)/55(B−1)/5,±2A/25(B−1)/2,+, 8 + 10A1, 1 + 10B1),(2(A+2)/55(B−3)/5,±2A/25(B−3)/2(11),+, 8 + 10A1, 3 + 10B1)}.171For the case where yˆ2 = 2xˆ1xˆ5 + 5Bˆ for xˆ1 ≥ 3 and Bˆ is even, we isolate and factor overZ to see that(yˆ − 5Bˆ/2)(yˆ + 5Bˆ/2) = 2xˆ1xˆ5.The left hand side has only a common factor of 2. This gives rise to the pair of equationsyˆ ∓ 5Bˆ/2 = 2K5yˆ ± 5Bˆ/2 = 2xˆ1−1L5.Subtracting yields ±5Bˆ/2 = 2xˆ1−2L5−K5. We solve this using our Thue-Mahler solverto see that the only solutions with KL 6= 0 are given when K,L ∈ {±1}. This gives(yˆ, Bˆ, xˆ1, K, L) ∈ {(±3, 0, 3, 1, 1), (±3, 0, 4,−1, 1)}.Translating to the general solution of our equation gives(x, y,±, A,B) ∈ {(2(A+3)/55B/5,±2A/25B/2(3),+, 2 + 10A1, 10B1),(2(A+4)/55(B−2)/5,±2A/25(B−2)/2(3),+, 6 + 10A1, 2 + 10B1)}.Summarizing, this case gives the following solutions:(x, y,±, A,B) ∈ {(2A/55B/5,±2A/25B/2(3),+, 2 + 10A1, 10B1),(2(A+1)/55B/5,±2A/25B/2,−, 4 + 10A1, 10B1),(2(A+4)/55(B−2)/5,±2A/25(B−2)/2(3),+, 6 + 10A1, 2 + 10B1),(2(A+2)/55(B−1)/5,±2A/25(B−1)/2(3),+, 8 + 10A1, 1 + 10B1),(2(A+2)/55(B−1)/5,±2A/25(B−1)/2,+, 8 + 10A1, 1 + 10B1),(2(A+2)/55(B−3)/5,±2A/25(B−3)/2(11),+, 8 + 10A1, 3 + 10B1),(2(A+1)/55(B−3)/5(3),±2A/25(B−3)/2(19),−, 4 + 10A1, 3 + 10B1),(2(A+1)/55(B−3)/5(7),±2A/25(B−3)/2(3)(61),−, 4 + 10A1, 3 + 10B1)}.5. 2y1 = A and 2y2 = B. Canceling out the equal terms and relabeling termsyˆ2 = 2xˆ15xˆ2xˆ5 ± 1.In this case, again if x1 > 1, then we see that the −1 case gives a contradiction viamodulo 4 considerations. This leaves us withyˆ2 = 2 · 5xˆ2xˆ5 − 1172and the positive 1 case which we defer until later. Factoring gives(yˆ + i)(yˆ − i) = 2 · 5xˆ2xˆ5.As the two terms on the left are coprime outside of 2, we see thatyˆ + i = (1± i)(1± 2i)xˆ2(a+ bi)5.We use the Thue-Mahler equation solver on these equations and see that the onlysolutions come when xˆ2 = 0, 1 or 2 giving solutions as(xˆ, yˆ, xˆ1, xˆ2) ∈ {(0,±1, 1, 0), (1,±3, 1, 1), (1,±7, 1, 2)}.These in turn give rise to solutions to the original equation of the form(x, y,±, A,B) ∈ {(2(A+1)/55B/5,±2A/25B/2,−, 4 + 10A1, 10B1),(2(A+1)/55(B+1)/5(3),±2A/25B/2,−, 4 + 10A1, 4 + 10B1),(2(A+1)/55(B+2)/5(7),±2A/25B/2,−, 4 + 10A1, 8 + 10B1)}.Now, to handle the case whereyˆ2 = 2xˆ15xˆ2xˆ5 + 1we first note that if xˆ1 ≤ 2, we have that modulo 8 considerations give a contradiction.Hence xˆ1 ≥ 3. Factoring gives(yˆ + 1)(yˆ − 1) = 2xˆ15xˆ2xˆ5.The left hand side is coprime outside of 2 and hence we have one ofyˆ ∓ 1 = 2xˆ1−15xˆ2K5yˆ ± 1 = 2L5yˆ ∓ 1 = 2 · 5xˆ2K5yˆ ± 1 = 2xˆ1−1L5.Subtracting the equations and dividing by 2 in each case gives∓1 = 2xˆ1−25xˆ2K5 − L5 ∓ 1 = 5xˆ2K5 − 2xˆ1−2L5.Both of these can be solved using the Thue-Mahler equation solver. The solutions with173KL 6= 0 are given by(xˆ1, xˆ2, K, L) ∈ {(3, 0, 1, 1), (3, 0,−1,−1)}in the first case and in the second case by(xˆ1, xˆ2, K, L) ∈ {(3, 0, 1, 1), (3, 0,−1,−1), (4, 1, 1, 1), (4, 1,−1,−1)}.Converting to solutions of the original equation gives(x, y,±, A,B) ∈ {(2A/55B/5,±2A/25B/2(3),+, 2 + 10A1, 10B1),(2(A+4)/55(B+1)/5(9),±2A/25B/2,+, 6 + 10A1, 4 + 10B1)}Collecting all the solutions in this case yields the following list:(x, y,±, A,B) ∈ {(2A/55B/5,±2A/25B/2(3),+, 2 + 10A1, 10B1),(2(A+1)/55B/5,±2A/25B/2,−, 4 + 10A1, 10B1),(2(A+4)/55(B+1)/5(9),±2A/25B/2,+, 6 + 10A1, 4 + 10B1),(2(A+1)/55(B+1)/5(3),±2A/25B/2,−, 4 + 10A1, 4 + 10B1),(2(A+1)/55(B+2)/5(7),±2A/25B/2,−, 4 + 10A1, 8 + 10B1)}.6. 2y1 = A and 5x2 = B. Canceling out the equal terms and relabeling terms5yˆ2 yˆ2 = 2xˆ1xˆ5 ± 1.We can reduce this to exactly case (iii), which we solved using Chabauty and EllipticCurve Chabauty techniques, by removing the power of 5 on the left hand side bymultiplying through by 55 if yˆ2 is odd and a suitable power of 2 which gives rise to theequationsY 2 = X5 ± 2Aˆ or Y 2 = X5 ± 2Aˆ55.Here we are interested in the cases where 2 | gcd(X, Y ) and thus are left with the listgiven by(X, Y, Aˆ,±) ∈ {(2,±6, 2,+), (2,±4, 4,−), (2,±8, 5,+), (6,±88, 5,−)}174for the first equation and for the second equation,(X, Y, Aˆ,±) ∈ {(20,±2000, 8,+)}.The cases where (X, Y, Aˆ,±) = (2,±8, 5,+) or (6,±88, 5,−) are inadmissible since thepower of 2 in Y 2 does not match the value of Aˆ. This leaves us only three solutionswhich in terms of our original equation are given by(x, y,±, A,B) ∈ {(2A/55B/5,±2A/25B/2(3),+, 2 + 10A1, 10B1),(2(A+1)/55B/5,±2A/25B/2,−, 4 + 10A1, 10B1),(2(A+2)/55B/5,±2(A+12)/25(B+1)/2,+, 8 + 10A1, 5 + 10B1)}.7. 5x1 = A and 2y2 = 5x2. In this case, we have2yˆ1 yˆ2 = xˆ5 ± 5Bˆ.We can proceed similar to case 3. If yˆ1 is even, then we look at solutions to the equationY 2 = X5 ± 5Bˆand if yˆ1 is odd, multiply through by 25 and look at solutions to the equationY 2 = X5 ± 255Bˆ.Using Chabauty and Elliptic Curve Chabauty, we have the solutions given in the caseof Y 2 = X5 ± 5Bˆ by(X, Y, Bˆ,±) ∈ {(−1, 0, 0,+), (0,±1, 0,+), (1, 0, 0,−), (−1,±2, 1,+), (0,±5, 2,+),(0,±25, 4,+), (5,±50, 4,−), (−5, 0, 5,+),(5, 0, 5,−), (0,±125, 6,+), (0,±625, 8,+)}and in the second case by(X, Y, Bˆ,±) ∈ {(−2, 0, 0,+), (2,±8, 0,+), (2, 0, 0,−),(6,±88, 0,+), (−10, 0, 5,+), (10, 0, 5,−)}.In the first case, there are only two solutions with non-zero entries and these are givenby(X, Y,C,±) ∈ {(−1,±2, 1,+), (5,±50, 4,−)}175The second solution above does not satisfy 2y2 = 5x2 (it satisfies 2y2 = B). Theremaining case corresponds to the solution given by(x, y,±, A,B) ∈ {(−2A/55(B−1)/5,±2(2+A)/25(B−1)/2,+, 10A1, 1 + 10B1)}In the second case, we see that the only solutions correspond to when Bˆ = 0. Theseare precisely the solutions discussed in case (ix) below and are given by(x, y,±, A,B) ∈ {(2A/55B/5(3),±2(1+A)/25B/2(11),−, 5 + 5A1, 10B1),(2A/55B/5,±2(1+A)/25B/2,+, 5 + 5A1, 10B1)}.Hence all solutions in this case are given by(x, y,±, A,B) ∈ {(−2A/55(B−1)/5,±2(2+A)/25(B−1)/2,+, 10A1, 1 + 10B1),(2A/55B/5(3),±2(1+A)/25B/2(11),−, 5 + 5A1, 10B1),(2A/55B/5,±2(1+A)/25B/2,+, 5 + 5A1, 10B1)}.8. 5x1 = A and 2y2 = B. In this case, we have2yˆ1 yˆ2 = 5xˆ2xˆ5 ± 1.In the case where xˆ2 > 0, we see that modulo 5, we must have thatyˆ2 = ±2yˆ1which is a contradiction unless yˆ1 is even. When this is even, this case gives rise tosolutions ofY 2 = X5 ± 5Cfor some constant 0 ≤ C ≤ 9. These solutions were discussed in case (vii) aboveand solved by Chabauty and Elliptic Curve Chabauty techniques. There are only twosolutions with non-zero entries and these are given by(X, Y,C,±) ∈ {(−1,±2, 1,+), (5,±50, 4,−)}.The case when C = 1 above does not correspond to a solution in this case as it is toosmall. Hence we are given solutions only by the second case above which correspondsto(x, y,±, A,B) ∈ {(2A/55(1+B)/5,±2(2+A)/25B/2,−, 10A1, 4 + 10B1)}.176If xˆ2 = 0, then we see as in case (ix) below that the solutions are given by(x, y,±, A,B) ∈ {(2A/55B/5(3),±2(1+A)/25B/2(11),−, 5 + 5A1, 10B1),(2A/55B/5,±2(1+A)/25B/2,+, 5 + 5A1, 10B1)}.Hence all solutions in this case are given by(x, y,±, A,B) ∈ {(2A/55(1+B)/5,±2(2+A)/25B/2,−, 10A1, 4 + 10B1),(2A/55B/5(3),±2(1+A)/25B/2(11),−, 5 + 5A1, 10B1),(2A/55B/5,±2(1+A)/25B/2,+, 5 + 5A1, 10B1)}.9. 5x1 = A and 5x2 = B. In this case, we have2yˆ15yˆ2 yˆ2 = xˆ5 ± 1.By work done by Bennett and Skinner as stated in Theorem 1.6.2, this gives rise to asolution toX5 + Y 5 = CZ2with C ∈ {1, 2, 5, 10}. In their work, they show that this case has no solutions wheny = ±1 except for(X, Y, Z, C) ∈ {(3,−1,±11, 2), (1,−1, 0, C), (−1, 1, 0, C), (1, 1, 1, 2)}.We drop the cases where z = 0 as these have been considered already. Translating backto the original problem, we have that(x, y,±, A,B) = {(2A/55B/5(3),±2(1+A)/25B/2(11),−, 5 + 5A1, 10B1),(2A/55B/5,±2(1+A)/25B/2,+, 5 + 5A1, 10B1)}are the solutions that arise from this case.We finish off with a theorem that will be used heavily later.Theorem 3.2.11. Let d, `,m ≥ 0 be integers and let p 6= 5 be an odd prime number. Thensolutions to the equationsd2 = 2`5mp5 ± 1 d2 = ±2`p5 ± 5m d2 = ±2`5m ± p5d2 = 2` ± 5mp5 5d2 = ±2` ± p5 5d2 = ±2`p5 ± 1177are given by192 = 2 · 35 − 53 (183)2 = 2 · 75 − 534012 = −2 · 53 + 115 5(7)2 = 35 + 25(19)2 = −35 + 211.In particular, the equationsd2 = 2`5mp5 ± 1 d2 = 2` ± 5mp5 5d2 = ±2`p5 ± 1have no solutions.3.3 Other ResultsThe following are results used from Wilfrid Ivorra [Ivo04][p.38-45]. Definef(n) :=18 + 2 log2 n if n < 296435 + 10 log2 n if n ≥ 296.Theorem 3.3.1. Solutions to d2 − 1 = 2`pn with d ≥ 1 and d, `, n ∈ Z nonnegative integersare given by1. n = 0 and (d, `, p) = (3, 3, p).2. n = 1 and (d, `, p) ∈ {(2, 0, 3), (5, 3, 3), (7, 4, 3), (9, 4, 5), (2p − 1, `, 2`−2 + 1), (2p +1, `, 2`−2 − 1)} with ` ≥ 5 in the last two cases.3. n = 2 and (d, `, p) = (17, 5, 3).Theorem 3.3.2. Solutions to d2 + 1 = 2`pn with d ≥ 1 and d, `, n ∈ Z nonnegative integersare given by1. n = 0 and (d, `, p) = (1, 1, p).2. n = 1 and (d, `, p) ∈ {(5, 1, 13), (d, 0, 1), (d, 1, 1)}.3. n = 2 and (d, `, p) = (d, 1, p) with p ≡ 1 (mod 4) and p 6= 13.4. n = 4 and (d, `, p) = (239, 1, 13).Theorem 3.3.3. Solutions to d2 + 2` = pn with d ≥ 1 and d, `, n ∈ Z nonnegative integersare given by1781. n = 1 and ` < log2 p.2. n = 2 and (d, `, p) ∈ {(3, 4, 5), (1, 3, 3), (p− 2, `, 2`−2 + 1)} where in the last case, ` ≥ 5.3. n = 3 and (d, `, p) ∈ {(5, 1, 3), (11, 2, 5)}.4. n = 4 and (d, `, p) = (7, 5, 3).Theorem 3.3.4. Solutions to d2 − 2` = pn with d ≥ 1 and d, `, n ∈ Z nonnegative integersare given by1. n = 0 and ` odd.2. n = 1 and ` < f(p).3. n = 2 and (d, `, p) = (p+ 2, `, 2`−2 − 1) with ` ≥ 4.4. n = 3 and (d, `, p) = (71, 7, 17).Theorem 3.3.5. Solutions to d2 + pn = 2` with d ≥ 1 and d, `, n ∈ Z nonnegative integersare given by1. n = 0 and (d, `, p) = (1, 1, p).2. n = 1 and ` < f(p).3. n = 3 and (d, `, p) ∈ {(13, 9, 7)}.Theorem 3.3.6. Solutions to 2d2 + 1 = pn with d ≥ 1 and d, n ∈ Z nonnegative integers aregiven by1. n = 1.2. n = 2.3. n = 5 and (d, p) = (11, 3).Theorem 3.3.7. Solutions to 2d2− 1 = pn with d ≥ 1 and d, n ∈ Z nonnegative integers aregiven by1. n = 0 and d = 1.2. n = 1.3. n = 2.4. n = 3 and (d, p) = (78, 23).This final result on Catalan’s conjecture is due to Miha˘ilescu [Mih04].Theorem 3.3.8. (Miha˘ilescu’s Theorem) The only solution to the Diophantine equationxm − yn = 1 in positive integers x, y, n,m with n,m ≥ 2 is given by 32 − 23 = 1.1793.4 Diophantine Equations Relating to Elliptic CurvesIn this section, we solve the following Diophantine equations where d ≥ 1, `,m, n ≥ 0,p 6= 2, 5 and we do not have (+,−) in the two ± signs:1. d2 − 2`5mpn = ±12. d2 ± 2`pn = ±5m3. d2 ± 2`5m = ±pn4. d2 ± 2` = ±5mpn5. 5d2 ± 2` = ±pn6. 5d2 ± 2`pn = ±1.This will help simplify the classification of primes such that there is an elliptic curve withrational two torsion and conductor in the set 50p, 200p and 400p. Throughout let Pmin(n)denote the minimum prime dividing n.Theorem 3.4.1. Let p 6= 2, 5 be a prime number with `,m, n ≥ 0 and d ≥ 1.1. Solutions to d2 − 2`5mpn = 1 are given by(a) n = 0 and (d, `,m) ∈ {(3, 3, 0), (9, 4, 1)}.(b) n = 1 anda) p = 3 and (d, `,m) ∈ {(2, 0, 0), (5, 3, 0), (7, 4, 0), (4, 0, 1), (11, 3, 1), (49, 5, 2)}.b) (d, `,m, p) ∈ {(2p− 1, `, 0, 2`−2 + 1), (2p+ 1, `, 0, 2`−2 − 1)} with ` ≥ 5.c) (d, `,m, p) ∈ {(2`−1 + 1, `,m, 2`−2+15m )} with m ≥ 1 and ` ≥ 4 even.d) (d, `,m, p) ∈ {(p+ 1, 0,m, 5m − 2), (p− 1, 0,m, 5m + 2)} with m ≥ 1.e) (d, `,m, p) ∈ {(4p− 1, 3,m, 5m+12 )} with m ≥ 1.f) (d, `,m, p) ∈ {(2p− 1, `,m, 2`−25m + 1), (2p+ 1, `,m, 2`−25m − 1)} with m ≥ 1and ` ≥ 3.g) (d, `,m, p) = (2 · 5m − 1, 4,m, 5m−14 ) and m odd.(c) n = 2 and (d, `,m, p) ∈ {(17, 5, 2, 3), (19, 3, 1, 3), (99, 3, 2, 7)}.(d) n = 3 and (d, `,m, p) = (26, 0, 2, 3).(e) n = 4 and (d, `,m, p) = (161, 6, 1, 3).(f) Pmin(n) ≥ 7, d = 2 · 5m − 1 and 4pn = 5m − 1 with m odd and ` = 4.2. Solutions to d2 − 2`5mpn = −1 with `,m, n ≥ 0 and d ≥ 1 are given by180(a) m = 0 andi. p ≥ 5 and (d, `, n) = (1, 1, 0).ii. p = 13 and (d, `, n) ∈ {(5, 1, 1), (239, 1, 4)}.iii. p 6= 13, p ≡ 1 (mod 4) and (`, n) = (1, 2).iv. (`, n) ∈ {(0, 1), (1, 1)}(b) n = 0 and (d, `,m) ∈ {(2, 0, 1), (3, 1, 1), (7, 1, 2)}.(c) m,n ≥ 1, ` = 0, 1 and p ≡ 1 (mod 4).1. (a) Case 1: m = 0 or n = 0. This is done in Ivorra’s thesis (see Theorem 3.3.1).Note that the (d, `, n, p) = (9, 4, 1, 5) solution above is given as (d, `,m, n, p) =(9, 4, 1, 0, p). Thus throughout, we assume that m > 0 and that n > 0.Case 2: p = 3 This case was done in [Mul06] Gives the solutions(d, `,m, n) ∈ {(161, 6, 1, 4), (9, 4, 1, 0), (26, 0, 2, 3), (49, 5, 2, 1),(4, 0, 1, 1), (19, 3, 1, 2), (11, 3, 1, 1)}and a potential solution given from 5 · 3` = 2m−2 + 1 with ` ≥ 1 and m ≥ 5. Forthis case, modulo 5 shows us that 2m−2 ≡ −1 (mod 5) and so m ≡ 0 (mod 4).Modulo 3 gives 2m−2 ≡ −1 (mod 3) and so m is odd which gives a contradiction.Thus this last case does not occur and the other cases are listed in the statement.Hence throughout we suppose that p ≥ 7.Case 3: ` = 0. Here we have that d2 − 1 = 5mpn. As d is even, we have thatmodulo 4, we see that p ≡ 3 (mod 4) and that n is odd. Factoring the left handside gives (d− 1)(d+ 1) = 5mpn which givesd± 1 = 5md∓ 1 = pnwhere the signs above match up3 In either case, taking the difference gives±2 = 5m − pn.As we have already shown that n is odd, we begin breaking this into mini cases.3A positive sign for the first term means you must use a negative sign in the second term. This conventionwill be consistent throughout the paper, especially in this section.181Case 3a: n = 1. This gives p = 5m ∓ 2 and d = p ± 1 where again the signsmatch up.Case 3b: 3 | n. This gives (pn/3)3 = 5m ∓ 2. If m is even, then looking atthis as an integer solution to an elliptic curve of the form y2 = x3 ± 2. Weuse Theorem 3.1.4 to show that this has no solution for x prime when we havey2 = x3 + 2 and Theorem 3.1.8 to see when y2 = x3 − 2 that the solutions aregiven by (x, y) = (3,±5) (a solution we have already found). Now, assume thatm is odd. Since we know that p ≥ 7 a quick check locally at 3 shows that(pn/3)3 = 5m ∓ 2 ≡ 2∓ 2 (mod 3)so pn ≡ 0 (mod 3) or pn ≡ 1 (mod 3). The first case gives p = 3, a case we havealready considered above, and so we must have that (pn/3)3 = 5m + 2 with modd. Multiplying by 53 gives (5pn/3)3 = (5(m+3)/2)2 + 2 · 53, that is, a solution toy2 = x3 − 250. A check with Theorem 3.1.8 shows that there are no {2, 5,∞}solutions.Case 3c: Pmin(n) ≥ 5. This gives pn = 5m ∓ 2. If we think of this as a solution(x, y, z) = (1,∓1, p) to 5mxn + 2yn = zn for a prime n ≥ 3, then we can useTheorem 1.6.1 to show that we have no solutions.Case 4: ` = 1. Here we have locally at 8 thatd2 ≡ 1 + 2`5mpn ≡3 (mod 8) if either n is even or p ≡ 1, 5 (mod 8)7 (mod 8) if n is odd and p ≡ 3, 7 (mod 8)holding since 2 · 5m ≡ 2 (mod 8) always. This is a contradiction since odd squaresmodulo 8 are congruent to 1.Case 5: ` = 2. If ` = 2, then notice that 4 · 5m ≡ 4 (mod 8) and that 4pn ≡4 (mod 8). Hence we have that 1 + 2l5mpn ≡ 5 (mod 8) always, a contradiction.Case 6: ` ≥ 6, Pmin(n) ≥ 7. If ` ≥ 6, then using Theorem 1.6.4, we see thatthere are no solutions in this case.Case 7: 3 | n. Here, we have the equation d2 = 2`5m(pn/3)3+1. Let ` = 3L+λ andm = 3M + µ where λ, µ ∈ {0, 1, 2} and L,M are nonnegative integers. Rewritingthe equation givesd2 = 2λ5µ(2L5Mpn/3)3 + 1.182Multiplying through by 22λ and 52µ gives(2λ5µd)2 = (2L+λ5M+µpn/3)3 + 22λ52µThus it suffices to look up all {2, 5,∞}-integer points on elliptic curves with theabove form. This is done in Section 3.1. A quick check on the tables shows thatthe only entry that has an x-coordinate of the form x = 2L+λ5M+µpn/3 and theconstant term of the form 22λ52µ corresponds to the solution(25 · 26)2 = (25 · 3)3 + 20 · 54.This corresponds to(d, `,m, n, p) = (26, 0, 2, 3, 3)which is a solution we already have found since p = 3.Case 8: 5 | n. Checking with Theorem 3.2.11, we see that this case gives nosolutions.Case 9: ` ≥ 3, m,n ≥ 1 and one of 2 | n or ` = 3, 4, 5 with Pmin(n) ≥ 7 orn = 1. We rearrange the equation to d2 − 1 = 2`5mpn and factor the left handside to see that(d− 1)(d+ 1) = d2 − 1 = 2`5mpn.This leads to the following possible situationsd∓ 1 = 2`−1d± 1 = 2 · 5mpnd∓ 1 = 2`−1pnd± 1 = 2 · 5md∓ 1 = 2`−15md± 1 = 2 · pnd∓ 1 = 2`−15mpnd± 1 = 2.The last case has no such solutions as m and p are positive. Eliminating d in eachof the first 3 cases gives the equationsi. ±1 = 5mpn − 2`−2.ii. ±1 = 5m − 2`−2pn.iii. ±1 = pn − 2`−25m.Now we divide into subcasesCase 9a: n is even.i. For this case, rewrite the equation as 2`−2 ± 1 = 5m(pn/2)2. Using Theorem1.6.2, we get a contradiction when ` ≥ 6. Hence, we have that ` = 3, 4, 5.183In these cases, the only solution with n ≥ 2 even is given by (p, `,m, d, n) =(3, 5, 0, 17, 2).ii. For this case, rewrite the equation as 2`−2pn = 5m ∓ 1 which gives a solutionto the equation Cz2 = xm + ym where C ∈ {1, 2} (possible grouping extrapowers of 2 inside z). Theorem 1.6.2 implies that m < 4. Since m > 0, and nis even, a quick check of the 3 possible values of n shows that we have n = 1in all admissible cases, contradicting this case.iii. For this case, if pn = 2`−25m − 1, then if ` − 2 ≥ 3, we have that pn ≡−1 (mod 8), a contradiction since n is even. If ` = 4, then we get thatpn ≡ 3 (mod 8) which is also a contradiction. This leaves the case when` = 3. In this case, a result of Cohn [Coh96] shows us that we have nosolutions when m > 2. For m ≤ 2, we get solutions given by(d, `,m, n, p) ∈ {(19, 3, 1, 2, 3), (99, 3, 2, 2, 7)}.Now, if pn = 2`−25m + 1, we move the 1 to the other side and factor to seethat(pn/2 − 1)(pn/2 + 1) = 2`−25m.Since the greatest common divisor on the left is 2, we see that when ` = 3 wehave a contradiction. Thus ` ≥ 4 and further, we are in one of the followingcases pn/2 ∓ 1 = 2 · 5mpn/2 ± 1 = 2`−3pn/2 − 1 = 2pn/2 + 1 = 2`−35m.In the second case, it is clear that the only solution comes when (d, `,m, n, p) =(17, 5, 0, 2, 3). In the first case, eliminating the p factor gives 5m − 2`−4 = ±1and Miha˘ilescu’s Theorem (Theorem 3.3.8) implies that either m = 1 and` = 6 or that m = 0 and ` = 5. In these cases, we have the solutions given by(d, `,m, n, p) ∈ {(161, 6, 1, 4, 3), (17, 5, 0, 2, 3)}.Case 9b: ` = 3, 4, 5 and Pmin(n) ≥ 7.i. For this case, plugging in ` = 3, 4, 5 immediately shows that there are nosolutions with n ≥ 3.ii. In this case, notice that ±1 = 5m − 2pn can be handled by Theorem 1.6.1 toshow that the only solution to this equation is given by −1 = 50− 2p0. When` = 4 or 5, if −1 = 5m − 2`pn, then we have modulo 4 that −1 ≡ 1 (mod 4)184which is a contradiction. So we may suppose that we are in the case where1 = 5m − 4pn or 1 = 5m − 8pn. For the latter note that modulo 8, wehave that 1 ≡ 5m (mod 8) and so we get that m is even. Factoring gives8pn = (5m/2 − 1)(5m/2 + 1) where the factors on the right have a greatestcommon divisor of 2 and so we have one of the following cases5m/2 ± 1 = 2pn5m/2 ∓ 1 = 45m/2 ± 1 = 4pn5m/2 ∓ 1 = 2.Subtracting pairwise shows that either ±1 = pn − 3 or ±1 = 2pn − 1 givingthat either p is even, a contradiction or that n = 0 in the latter case whichalso leads to a contradiction.This leaves the case 1 = 5m − 4pn. Modulo 8 gives 5 ≡ 5m (mod 8) and thisshows that m is odd. This leaves the situation mentioned in the statement.iii. As above, notice that ±1 = pn − 2 · 5m can be handled by Theorem 1.6.1 toshow that the only solutions to this equation are given when Pmin(n) < 5. Theequation ±1 = pn−4 ·5m can be made to look like a solution to xn+yn = Cz2with C = 1 or 5. Thus by Theorem 1.6.2, we see that this has no solutions.Lastly, the equation ±1 = pn−8 ·5m can be made to look like xn +yn = 5mz3which also has no solutions via Theorem 1.6.6.Case 9c: n = 1.i. In this case, we have that ±1 = 5mp − 2`−2. Modulo 5 considerations gives∓1 = 2`−2 (mod 5) and this shows that ` is even. In the −1 case, we have that5mp = (2(`−2)/2 − 1)(2(`−2)/2 + 1) and so factoring gives 2(`−2)/2 ± 1 = 5m and2(`−2)/2 ∓ 1 = 5m which leads to p = 5m ∓ 2. The other case gives p = 2`−2+15mand d = 2`−1 + 1.ii. In the case where 1 = 5m−2`−2p, if ` = 3, then modulo 4 considerations gives1 ≡ 1 + 2 (mod 4) a contradiction. If ` = 4, then we get solutions given byp = 5m−14 . Lastly, if ` ≥ 5, then modulo 8 considerations show that m is even.Rearranging the equation and factoring gives2`−2p = (5m/2 − 1)(5m/2 + 1).This gives the cases5m/2 ± 1 = 2p5m/2 ∓ 1 = 2`−35m/2 ± 1 = 2`−3p5m/2 ∓ 1 = 2.185Considerations modulo 4 reduce the above to5m/2 + 1 = 2p5m/2 − 1 = 2`−35m/2 − 1 = 2`−3p5m/2 + 1 = 2.The second case above is impossible since m > 0. In the remaining case, wesee that the second equation reads 1 = 5m/2 − 2`−3 and so by Miha˘ilescu’sTheorem (Theorem 3.3.8) we have that m = 2, ` = 5 and p = 3.In the case where −1 = 5m − 2`−2p, if ` ≥ 4, then we get a contradictionby considerations modulo 4. If ` = 3, then we get the solutions given byp = 5m+12 , d = 4p− 1.iii. Here we have p = 2`−25m ± 1 with d = 2p∓ 1.(b) Case 1: m = 0 or n = 0. This is done in Ivorra’s thesis (see Theorem 3.3.2) togive the solutions as stated.Case 2: ` ≥ 3 Modulo 8 gives d2 ≡ −1 (mod 8) which has no solutions.Case 3: ` = 2. Locally at 8, we have−1 + 2`5mpn ≡ −1 + 4 · 5mpn ≡ 3 (mod 8),a contradiction. Thus there is no solution when ` = 2.Case 4: ` ≤ 1. There is little we can say in this case other than that solutionsdo in fact exist.Theorem 3.4.2. Let p 6= 2, 5 be a prime number with `,m, n ≥ 0 and d ≥ 1.1. Solutions to d2 − 2`pn = 5m are given by(a) n = 0 and (d, `,m, p) ∈ {(3, 3, 0, p), (3, 2, 1, p)}.(b) n = 1 anda) m = 0 with (d, `, p) ∈ {(2, 0, 3), (5, 3, 3), (7, 4, 3), (2p − 1, `, 2`−2 + 1), (2p +1, `, 2`−2 − 1)} with ` ≥ 5 in the last two cases.b) m > 0, ` = 0 and p ≡ 3 (mod 4).c) m > 0 is odd, ` = 2 and p = d2−5m4 .d) m > 0 even, ` = 3 and p = 5m/2+12 .186e) m > 0 even, ` ≥ 3 and p = 2`−2± 5m/2 with d = 2`−1± 5m/2 and both signs thesame.(c) n = 2 andi. (d, `,m, p) = (5m+12 , 2,m,5m−14 ) and m is odd.ii. (d, `,m, p) ∈ {(17, 5, 0, 3)}.(d) 2 ‖ n and 4pn/2 = 5m − 1 with m odd, ` = 2 and either n = 2 or Pmin(n/2) ≥ 7.(e) n = 3 and (d, `,m, p) ∈ {(29, 3, 4, 3), (59, 7, 2, 3), (73, 2, 1, 11)}.(f) n = 4 and (d, `,m, p) = (129, 4, 3, 11).(g) Pmin(n) ≥ 7, m ≥ 1 anda ` = 0 with p ≡ 3 (mod 4).b ` = 2 with pn = d2−5m4 and m odd.c ` = 3 with pn = 5m/2+12 and m even.d ` ∈ {4, 5} with pn = 2`−2 ± 5m/2 and m even.2. Solutions to d2 + 2`pn = 5m are given by(a) n = 0 and (d, `,m, p) ∈ {(11, 2, 3, p), (3, 4, 2, p)}.(b) n = 1 anda) (d, `,m, p) = (23, 5, 4, 3).b) p = 5m − d2 and ` = 0.c) p = 5m−d24 and ` = 2.d) (d, `,m, p) = (±(5m/2 − 2`−1), `,m, 5m/2 − 2`−2) ` ≥ 3 and m ≥ 2 even.e) p = 5m/2−14 , d = 5m/2 − 2 and ` = 4 with m ≥ 2 even.(c) n ≥ 2 is even and ` = 0 with m even and pn = 2 · 5m/2 − 1.(d) n ≥ 2 is even and ` ∈ {0, 2} with m odd.(e) n = 2 and (d, `,m, p) = (117, 4, 6, 11).(f) Pmin(n) ≥ 7 and 0 ≤ ` ≤ 5 and m ≥ 1.3. Solutions to d2 + 5m = 2`pn are(a) n = 0 and (d, `,m, p) = (1, 1, 0, p).(b) n = 1, ` = 0 and p ≡ 1 (mod 4).(c) n = 1 and ` = 1.(d) n = 2, ` = 0, p = 5m+12 , d = p− 1.187(e) n = 2, ` = 1 and m = 0 with p ≡ 1 (mod 4) and p 6= 13.(f) n = 3 and (d, `,m, p) ∈ {(7, 1, 1, 3), (99, 1, 2, 17)}.(g) n = 4 and (d, `,m, p) = (239, 1, 0, 13).(h) n = 5 and (d, `,m, p) ∈ {(19, 1, 3, 3), (183, 1, 3, 7)}.Proof. 1. Case 1: n = 0 or m = 0. This is done in Ivorra’s thesis. If n = 0, we lookup Theorem 3.3.4 for solutions to d2 − 2` = 5m. This gives solutions when m = 0 orm = 1. When m = 0, we use Miha˘ilescu’s Theorem (Theorem 3.3.8) to get the onlysolution of (d, `,m) = (3, 3, 0). When m = 1, we see that ` must be even via modulo 5considerations and so the left hand side factors to give the equationsd− 2`/2 = 1d+ 2`/2 = 5.Subtracting these two equations gives 2`/2+1 = 4 and so ` = 2. This gives anothersolution given by (d, `,m) = (3, 2, 1).When m = 0 and n > 0, via Theorem 3.3.1 (recalling that p 6= 5), we get the solutionsmentioned in the theorem. For the duration of this proof, we suppose that m,n > 0.Case 2: 3 | n. If ` ≡ 0 (mod 3) then (x, y) = (2`/3pn/3, d) is a solution to y2 = x3 +5m.Using Theorem 3.1.2, we have that the only solution of this shape that arises is when(x, y) = (6,±29) and this gives (d, `,m, n, p) = (29, 3, 4, 3, 3).If ` ≡ 1 (mod 3) then (x, y) = (2(`+2)/3pn/3, 2d) is a solution to y2 = x3 + 225m.Using Theorem 3.1.2, we have that the only solution of this shape that arises is when(x, y) = (24,±118) and this gives (d, `,m, n, p) = (59, 7, 2, 3, 3).If ` ≡ 2 (mod 3) then (x, y) = (2(`+4)/3pn/3, 4d) is a solution to y2 = x3 + 245m.Using Theorem 3.1.2, we have that the only solution of this shape that arises is when(x, y) = (44,±292) and this gives (d, `,m, n, p) = (73, 2, 1, 3, 11).Case 3: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Case 4: Pmin(n) ≥ 7 and ` ≥ 6. Using the modular method of Theorem 1.6.5, we canshow that this equation has no solutions.Case 5: n even. We break this into subcases.188Case 5a: ` = 0. In this case, we have d2 − pn = 5m. Notice that d must be even sincep is odd and so considerations modulo 4 give −1 ≡ 1 (mod 4), a contradiction.Case 5b: ` = 1. In this case, we have d2 − 2pn = 5m. Considerations modulo 5 gived2 ≡ 2pn (mod 5). Since n is even, we know that 2 must be a square modulo 5, acontradiction.Case 5c: ` = 2. Local considerations at 8 give us thatd2 = 2`pn + 5m ≡ 4 + 5m (mod 8)and this value is either 1 or 5 depending on if m is odd or even. If m is even, then weget d2 ≡ 5 (mod 8) which is a contradiction. Hence m is odd. Thus, let m = 2k + 1and j = n/2. Factoring gives5m = d2 − 4p2j = (d− 2pj)(d+ 2pj).As gcd(d− 2pj, d+ 2pj) = 1, we have thatd+ 2pj = 5md− 2pj = 1.Subtracting these two equations gives 4pj = 5m − 1. If 4 | n, then j is also even andhence the above gives rise to a solution to z2 = xm + ym. By Theorem 1.6.2, we knowthat this implies that m < 4. Checking values of m = 1 and m = 3 shows that m = 3gives rise to p = 31 and j = 1 contradicting that j was even.If 2 ‖ n, then we get 4pn/2 = 5m − 1. Modulo 8 considerations shows that m is odd.Case 5d: ` ≥ 3. Local considerations at 8 give us that1 ≡ d2 = 2`pn + 5m ≡ 5m (mod 8)and so, we see that m must be even. Now, considerations at 3 show us thatd2 = 2`pn + 5m ≡ 2` + 1 (mod 3).Thus, ` has to be odd (as 2 is a quadratic non-residue modulo 3). Next, modulo 5 givesd2 = 2`pn ≡ 2 · (2(`−1)/2pn/2)2 (mod 5)189and so once again, we see that 2 is a square modulo 5 which is a contradiction.Case 6: n = 1 or Pmin(n) ≥ 7. When ` = 0, d is even and so modulo 4 considerationsshow that p ≡ 3 (mod 4). When ` = 1, we see that d2 − 2pn = 5m. Reducing modulo4 yields that d2 ≡ 3 (mod 4) and this is a contradiction. When ` = 2, then we haved2 = 5m + 4pn ≡ 5m + 4 (mod 8) showing that m is odd since 5 is not a square modulo8. Lastly, if ` ≥ 3, then m is even by reducing modulo 8 and thus we can factor giving(d− 5m/2)(d+ 5m/2) = 2`pn. This gives the following casesd− 5m/2 = 2d+ 5m/2 = 2`−1pnd∓ 5m/2 = 2pnd± 5m/2 = 2`−1.In the first case, eliminating d yields 5m/2 = 2`−2pn − 1. When ` = 3, This givespn = 5m/2+12 which is a potential solution. When ` ≥ 4, modulo 4 considerations give acontradiction. In the second case, eliminating gives ±5m/2 = 2`−2− pn. The case when` = 3 is solved via Theorem 1.6.1 and the other cases are included in the theorem.2. Case 1: m = 0 or n = 0 If m = 0 then we immediately see we have no solution sinced is positive. If n = 0, then from Ivorra’s thesis (see Theorem 3.3.3), we see that theonly solutions are given by(d, `,m) ∈ {(3, 4, 2), (11, 2, 3)}.From now on, assume that m > 0 and n > 0.Case 2: 3 | nIf ` ≡ 0 (mod 3) then (x, y) = (−2`/3pn/3, d) is a solution to y2 = x3 + 5m.If ` ≡ 1 (mod 3) then (x, y) = (−2(`+2)/3pn/3, 2d) is a solution to y2 = x3 + 225m.If ` ≡ 2 (mod 3) then (x, y) = (−2(`+4)/3pn/3, 4d) is a solution to y2 = x3 + 245m.Using Theorem 3.1.2, we see that all the solutions which have a negative x-coordinateare not integer solutions (there is one {2, 5,∞}-integer solution with constant term22+6n · 53 but this does not translate to a solution) and so these cases do not give asolution.Case 3: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.190Case 4: Pmin(n) ≥ 7 and ` ≥ 6. This is shown to have no solutions via Theorem1.6.5.Case 5: n even. We break this up into subcasesCase 5a: ` odd. Here modulo 5 considerations gives d2 = −2`pn (mod 5). Since−2 ≡ 3 (mod 5) is a non-quadratic residue, this gives a contradiction and hence nosolutions for this case.Case 5b: ` even. If ` = 0, then if m is even, we can factor to get thatd∓ 5m/2 = −1d± 5m/2 = pnand eliminating d yields 2 · 5m/2 = pn − 1 which is included in the theorem. If ` = 2,then modulo 8 considerations shows that m is odd. In this situation, there are manypossible solutions.Now, if ` ≥ 4, modulo 8 considerations gives m even. Isolating and factoring gives thecasesd− 5m/2 = −2d+ 5m/2 = 2`−1pnd− 5m/2 = −2`−1d+ 5m/2 = 2pnd− 5m/2 = −2pnd+ 5m/2 = 2`−1.Eliminating for d in the first case gives 5m/2 = 2`−2pn+1 and in the last two cases gives5m/2 = pn + 2`−2.In the case where 5m/2 = 2`−2pn + 1 = (2(`−2)/2pn/2)2 + 1, we have no solutions viaMiha˘ilescu’s Theorem (Theorem 3.3.8).In the case where 5m/2 = pn + 2`−2, when ` = 4, we get the only admissible solutionvia [Coh03] to be (d, `, n,m, p) = (117, 4, 2, 6, 11). When ` ≥ 6, we get that by lookingmodulo 8, we see that m/2 is also even. Hence factoring gives(5m/4 − pn/2)(5m/4 + pn/2) = 2`−2191which in turn gives the cases5m/4 − pn/2 = 25m/4 + pn/2 = 2`−3.Eliminating for p gives 5m/4 = 2`−4 − 1 and so 1 = 2`−4 − 5m/4. Miha˘ilescu’s Theorem(Theorem 3.3.8) implies that `− 4 = 0 (since ` is even) and this gives a contradiction.Thus there are no solutions in this case.Case 6: n = 1. If ` = 0, then we can only say p = 5m − d2. If ` = 1 then modulo4 considerations gives d2 ≡ 5m − 2p ≡ 1 − 2 ≡ −1 (mod 4) which is a contradiction.When ` = 2, then we can only say that d2 = 5m − 4p. When ` ≥ 3, modulo 8considerations show that m is even. Just as in the case 5b above, we can factor to give(d− 5m/2)(d+ 5m/2) = 2`p leading tod− 5m/2 = −2d+ 5m/2 = 2`−1pd− 5m/2 = −2`−1d+ 5m/2 = 2pd− 5m/2 = −2pd+ 5m/2 = 2`−1.Eliminating for d in the first case gives 5m/2 = 2`−2p + 1 and in the last two casesgives 5m/2 = p + 2`−2. The second case gives p = 5m/2 − 2`−2. The first case needs tobe broken into more cases. When ` = 3, we get a contradiction by looking modulo 4.When ` = 4, we get p = 5m/2−14 and we can say nothing more. When ` ≥ 5, modulo8 considerations give m/2 is even and so factoring gives (5m/4 − 1)(5m/4 + 1) = 2`−2p.This leads to 5m/4 ∓ 1 = 2p5m/4 ± 1 = 2`−3.Looking at the second equation, Miha˘ilescu’s Theorem (Theorem 3.3.8) implies that ofthe equations5m/4 + 1 = 2`−35m/4 − 1 = 2`−3only the second admits a solution given by m = 4 and ` = 5. This gives p = 3 in thefirst equation and thus the solution given by(d, `, n,m, p) = (23, 5, 1, 4, 3).This completes this case.1923. Case 1: m = 0 or n = 0 If m = 0, then Theorem 3.3.2 shows that either (d, `, n, p) =(1, 1, 0, p), p = 13 and (d, `, n) ∈ {(5, 1, 1), (239, 1, 4)} or that p ≡ 1 (mod 4), p 6= 13and (`, n) ∈ {(0, 1), (1, 1), (1, 2)}.If n = 0 and m > 0, then a check with Theorem 3.3.5 shows that m = 1 and ` <435 + 10 log2(5) ≤ 500. Checking with MAGMA for squares of the form 2` − 5 quicklyyields no results for ` ∈ {1, 2, ..., 500} (we could also refer to [Coh03] to see this has nosolutions). Thus, throughout suppose that n > 0 and m > 0.Case 2: ` ≥ 3Here, we haved2 = −5m + 2lpn ≡−5 (mod 8) if m ≡ 1 (mod 2)−1 (mod 8) if m ≡ 0 (mod 2)which is a contradiction since −1 and −5 are not squares modulo 8.Case 3: ` = 2If ` = 2, thend2 = −5m + 2lpn ≡−5 (mod 8) if m ≡ 1 (mod 2)−1 (mod 8) if m ≡ 0 (mod 2)+4p (mod 8) if n ≡ 1 (mod 2)4 (mod 8) if n ≡ 0 (mod 2)≡−1 (mod 8) if m ≡ 1 (mod 2)3 (mod 8) if m ≡ 0 (mod 2).Since 4p ≡ 4 (mod 8) for any odd prime. Again we have a contradiction since neither−1 nor 3 are squares modulo 8.Case 4: ` = 1When n ≥ 3, the solutions are given by theorems 1 and 3 of [AMLST09] (valid sincegcd(d, p) = 1 resulting from p 6= 5) to be(d, `,m, n, p) ∈ {(7, 1, 1, 3, 3), (99, 1, 2, 3, 17), (239, 1, 0, 4, 13), (19, 1, 3, 5, 3), (183, 1, 3, 5, 7)}.If n = 2, then since m > 0, we have that modulo 5 considerations give d2 ≡ 2p2 (mod 5)which is a contradiction since 2 is a quadratic non-residue.193If ` = 1 and n = 1, there are many solutions. If ` = 1 and n = 0, then the equationbecomes d2 + 5m = 2 which only has the solution d = 1 and m = 0.Case 5: ` = 0When ` = 0, the solutions are given by theorem 1 of [PR11] when n ≥ 3 and there areno such solutions. When n = 2, we have 5m = p2− d2 = (p− d)(p+ d). Since d is even,we know that gcd(p− d, p+ d) = gcd(p− d, 2p) = 1 and so this gives the equationsp− d = 1p+ d = 5mand hence p = 5m+12 , a case included in the theorem. When n = 1, this gives p = d2+5m.Hence d is even and so p ≡ 1 (mod 4).Theorem 3.4.3. Let p 6= 2, 5 be a prime number with `,m, n ≥ 0 and d ≥ 1.1. Solutions to d2 − 2`5m = pn are given by(a) n = 0 and (d, `,m) ∈ {(3, 3, 0), (9, 4, 1)}.(b) n = 1, m = 0 and ` < 435 + 10 log2(p).(c) n = 1 and m > 0.(d) n = 2 anda) (d, `, p) = (p+ 2, `, 2`−2 − 1) with ` ≥ 4.b) p = 2`−2 − 5m, d = 2`−1 − p and ` ≥ 3.c) p = 5m − 2`−2, d = 2`−1 + p and ` ≥ 3.d) p = 2`−2 · 5m − 1, d = p+ 2 and ` ≥ 3.e) (d, `,m, p) = (13, 5, 1, 3).(e) 2 ‖ n, m is odd, ` = 4 and p satisfies pn/2 = 5m − 4.(f) n = 3 and (d, `,m, p) ∈ {(299, 12, 1, 41), (71, 7, 0, 17), (411, 5, 5, 41)}.(g) n = 4 and (d, `,m, p) ∈ {(41, 6, 2, 3), (11, 3, 1, 3), (51, 3, 2, 7), (129, 4, 3, 11)}.(h) n = 6 and (d, `,m, p) = (37, 7, 1, 3).(i) Pmin(n) ≥ 7 and 0 ≤ ` ≤ 5 and m ≥ 1.2. Solutions to d2 + pn = 2`5m are given by194(a) n = 0 and (d, `,m, p) ∈ {(1, 1, 0, p), (13, 9, 3, 7)}.(b) n = 1.(c) 2 | n and ` = 0, 1.(d) n = 3 and (d, `,m, p) ∈ {(47, 8, 3, 31), (17, 9, 0, 7)}.(e) Pmin(n) ≥ 7 and 0 ≤ ` ≤ 5 and m ≥ 1.3. Solutions to d2 + 2`5m = pn are given by(a) n = 1 so p = d2 + 2`5m.(b) n = 2 anda) (d, `,m, p) = (p− 1, 0,m, 5m+12 ).b) (d, `,m, p) = (±(p− 2`−1), `,m, 5m + 2`−2) and ` ≥ 3.c) (d, `,m, p) = (p− 2, `,m, 2`−2 · 5m + 1) and ` ≥ 3.(c) n = 3 and (d, `,m, p) ∈ {(9, 1, 4, 11), (261, 5, 2, 41)}.(d) n = 4 and (d, `,m, p) = (1, 4, 1, 3).(e) n = 5 and (d, `,m, p) = (401, 1, 3, 11).(f) n = 6 and (d, `,m, p) = (23, 3, 2, 3).(g) n = 8 and (d, `,m, p) = (79, 6, 1, 3).Proof. 1. Case 1: m = 0 or n = 0. When m = 0, we have via Theorem 3.3.4 that(a) n = 0.(b) n = 1 and ` < f(p) < 435 + 10 log2(p).(c) n = 2 and (d, `, p) = (p+ 2, `, 2`−2 − 1) with ` ≥ 4.(d) n = 3 and (d, `, p) = (71, 7, 17).If both m = 0 and n = 0, then Miha˘ilescu’s Theorem (Theorem 3.3.8) implies that` = 3 and d = 3 is the only solution to d2 − 2` = 1.If only n = 0, then we get two solutions via Theorem 3.3.1, namely(d, `,m) ∈ {(3, 3, 0), (9, 4, 1)}and the first one is discussed above. Thus throughout, suppose that m,n > 0.195Case 2: 3 | n. Our equation is set up in the form d2 = (pn/3)3 + 2`5m. Thus wewant solutions to y2 = x3 + 2α5β with the x coordinate a positive prime. A check onTheorem 3.1.4 reveals that these solutions are given by(d, `,m, n, p) ∈ {(299, 12, 1, 3, 41), (71, 7, 0, 3, 17), (37, 7, 1, 6, 3), (411, 5, 5, 3, 41)}.Case 3: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Case 4: Pmin(n) ≥ 7 and ` ≥ 6. Using the modular method of Theorem 1.6.5, we canshow that this equation has no solutions.Case 5: n is even. Suppose n is even, so n = 2k. Then if ` = 0, we haved2 = p2k + 5m ≡ 1 +5 if m ≡ 1 (mod 2)1 if m ≡ 0 (mod 2)(mod 8)≡6 if m ≡ 1 (mod 2)2 if m ≡ 0 (mod 2)(mod 8)which is a contradiction since neither are squares modulo 8. If ` = 1, we haved2 = p2k + 2 · 5m ≡ 1 + 2 ≡ 3 (mod 8)which is a contradiction since 3 is not a square modulo 8. If ` = 2, we haved2 = p2k + 4 · 5m ≡ 1 + 4 ≡ 5 (mod 8)which is a contradiction since 5 is not a square modulo 8. So hence we can assume that` ≥ 3. Factoring gives(d+ pk)(d− pk) = 2`5m.Since gcd(d+ pk, d− pk) = gcd(d− pk, 2d) = 2, we have thatd− pk = 2d+ pk = 2`−15md± pk = 2 · 5md∓ pk = 2`−1where in the first case, we do not have the d+ pk = 2 case because this least to k = 0and d = 1 which does not yield a solution.Now in the first case, eliminating for d yields 2`−25m = pk + 1. Using the modularitymethod of Theorem 1.6.2 (where we think of 2`−25m = Cz2), we can see that k ≤ 3.196The cases when k = 0 or 3 were solved already and so k = 1 or k = 2.If k = 2, then ` = 3 since otherwise modulo 8 considerations give a contradiction (` = 4gives 3 is a square modulo 8 while ` ≥ 5 shows that −1 is a square modulo 8; bothare contradictions). Now, rewriting the equation gives −1 = p2 − 2 · 5m which has nosolutions when m > 2 by [Coh96]. When m = 1 or 2, we see that the only possiblesolutions come when (d, `,m, n, p) = (11, 3, 1, 4, 3) or (d, `,m, n, p) = (51, 3, 2, 4, 7). Ifk = 1, we get p = 2`−25m − 1.In the second case, we eliminate for d yielding 5m ∓ pk = 2`−2. We break this intosubcases.Case 5a: ` ≥ 6. By the previous cases, we know that k is a power of 2 since no otherprimes divide n. Hence, considerations modulo 8 yieldpk = ∓(2`−2 − 5m) ≡ ±5m (mod 8).Giving a contradiction unless m is even and the equation is 5m − pk = 2`−2. In thiscase, we factor the left hand side and see that(5m/2 − pk/2)(5m/2 + pk/2) = 2`−2.As gcd(5m/2 − pk/2, 5m/2 + pk/2) = gcd(5m/2 − pk/2, 2 · 5m/2) = 2, we must have that5m/2 + pk/2 = 2`−35m/2 − pk/2 = 2.Eliminating pk/2 in the above yields 5m/2−2`−4 = 1 and Miha˘ilescu’s Theorem (Theorem3.3.8) says that the only solutions are when m = 2 and ` = 6 giving (d, `,m, n, p) =(41, 6, 2, 4, 3).Case 5b: ` ≤ 5. In the case of 5m + pk = 2`−2, a quick inspection gives us thatthe only solution is when ` = 5 with m = k = 1 and p = 3. This gives the solution(d, `,m, n, p) = (13, 5, 1, 2, 3). In the case of 5m − pk = 2`−2, if ` = 3, then Theorem1.6.1 shows that only 2 and 3 can divide 5. We have already considered the case when3 | k and so we suppose that k is even. Then modulo 5 considerations give us that−pk ≡ 2 (mod 5) which is a contradiction since −2 is not a quadratic residue modulo5. If ` = 5 then Theorem 1.6.7 tells us that only 2, 3 or 5 may divide k. Since wehave considered the case when 3, 5 | k, we may suppose that k is even and the sameargument as above gives a contradiction.197If ` = 4 and k is even, the solutions to x2 + C = yn with C = 2, 4, or 8 are given by[Coh03, Section 5] to be (x, y, n, C) = (5, 3, 3, 2), (2, 2, 3, 4) and (11, 5, 3, 4) (this offersan alternative proof to the above). For our purposes, we require that y is a power of 5and so only the last solution is admissible in our setting.Now, suppose k is odd. Thus looking modulo 4 tells us that p ≡ 1 (mod 4). If m iseven, then factoring 5m − pk = 4 gives (5m/2 − 2)(5m/2 + 2) = pn/2 and so 5m/2 − 2 = 1which is a contradiction. Hence m is odd.2. Case 1: m = 0 or n = 0. If m = 0, then a check with Theorem 3.3.5 shows thatp = 7 and (d, `,m) = (13, 9, 3), n = 0 which contradicts [Coh03], or that n = 1 and `is bounded by a constant depending on p.If n = 0 and m > 0, then Theorem 3.3.2 shows that (`, n) ∈ {(0, 1), (1, 1), (1, 2)}. Weassume that n,m > 0 from now on.Case 2: 2 | n. If ` = 0 or ` = 1, then there is little to say. If ` ≥ 2, then modulo 4, wesee that1 + 1 ≡ d2 + pn ≡ 2`5m ≡ 0 (mod 4)and this is a contradiction.Case 3: 3 | n. Our equation becomes d2 = (−pn/3)3 + 2`5m. So we use Theorem 3.1.4to see that we get two solutions given by(d, `,m, p) ∈ {(47, 8, 3, 31), (17, 9, 0, 7)}.Case 4: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Case 5: Pmin(n) ≥ 7. In this case, using Theorem 1.6.4, we see that ` ≤ 5.3. Case 1: n ≥ 3 . This is a result of [LT08, Theorem 1.1] if either ` > 0 or m > 0 andgives solutions as stated in the following table.198α β x y n1 4 9 11 33 2 23 9 35 2 261 41 37 6 383 129 39 2 17771 681 34 1 1 3 46 1 79 9 41 3 401 11 53 2 23 3 66 1 79 3 8Table 3.20: Integer solutions to x2 + 2α5β = yn with x, y ≥ 1, gcd(x, y) = 1 and n ≥ 3 withα, β > 0.If ` = 0 then Theorem 1 of [PR11] tells us that the equation has no solutions. If m = 0,then papers by [Coh92],[AAM01] and/or [Sik03] give the solutions to x2 + 2k = yn as(x, y, k, n) ∈ {(5 · 23α, 3 · 22α, 6α + 1, 3), (7 · 22α, 3 · 2α, 4α + 5, 4),(11 · 25α+3, 3 · 2α+1, 10α + 5, 5), (2α, 22α+1, 6α + 1, n),(11 · 23α, 5 · 22α, 6α + 2, 3)}holding for all α ≥ 0. In our setting, namely that y a prime power, we have no solutions.Case 2: n = 2If ` = 0, then we factor to get (d− p)(d+ p) = −5m. Breaking into cases givesd− p = −1d+ p = 5md− p = −5md+ p = 1where the second case is inadmissible. For the first case, subtracting gives p = 5m+12and this is a stated solution.If ` = 1, then modulo 4 considerations gives us that0 ≡ 1− 1 ≡ d2 − p2 ≡ −2`5m ≡ 2 (mod 4)199a contradiction. If ` = 2, then modulo 8 considerations give us that d2 ≡ −3 (mod 8)another contradiction. So we assume that ` ≥ 3. Here we have−2`5m = d2 − p2 = (d− p)(d+ p).Since gcd(d− p, d+ p) = gcd(d− p, 2d) = 2, we have that we are in one of the followingthree casesd− p = −2 · 5md+ p = 2`−1d− p = −2`−1d+ p = 2 · 5md− p = −2 · 5md+ p = 2`−1 · 5m.In the first and second pairs of equations above, eliminating d via subtraction yieldsp = 5m + 2`−2.In the third pair of equations, Again subtracting to eliminate d yieldsp = 2`−2 · 5m + 1and all these cases are as stated in the lemma.Case 3: n = 0 . We get an immediate contradiction since d2 + 2`5m = 1 has nosolutions.Theorem 3.4.4. Let p 6= 2, 5 be a prime number with `,m, n ≥ 0 and d ≥ 1.1. Solutions to d2 − 2` = 5mpn are given by(a) n = m = 0 and ` = 3.(b) n = 1, m = 0 and ` < 435 + 10 log2 p.(c) n = 1, m > 0 and p = 5m ± 2`/2+1 and d = 5m ± 2`/2 and ` is even.(d) n = 2 and (d, `,m, p) = (123, 2, 3, 11).(e) n = 2, m = 0 and (d, `, p) = (p+ 2, `, 2`−2 − 1).(f) n = 3 and (d, `,m, p) ∈ {(71, 7, 0, 17), (26, 0, 2, 3, 3)}.(g) Pmin(n) ≥ 7 and pn = 5m ± 4 with ` = 2 and p ≡ 1 (mod 4).2. Solutions to d2 + 5mpn = 2` are given by(a) n = m = 0 and d = ` = 1.200(b) n = 1, m = 0 and ` < 435 + 10 log2 p.(c) n = 1 and p = 2`/2+1 − 5m.(d) n = 3 and (d, `,m, p) ∈ {(13, 9, 0, 7), (11, 8, 1, 3, 3)}.3. Solutions to d2 + 2` = 5mpn are given by(a) n = 0 and (d, `,m, p) ∈ {(3, 4, 2, 5), (11, 2, 3, 5)}.(b) n = 1, m = 0 and ` < log2 p.(c) n = 1 and p = d2+2`5m .(d) n even, 3, 5 - n and ` = 0 with pn = d2+15m and m odd.(e) 2 ‖ n, either n = 2 or Pmin(n/2) ≥ 7, ` = 2 with pn = d2+45m and m odd.(f) n = 2 anda) m = 0 and (d, `, p) ∈ {(1, 1, 3), (p − 2, `, 2`−2 + 1)} and in the second case wehave that ` ≥ 5.b) p = 2`−2+15m/2with `,m even.(g) n = 3 and (d, `,m, p) ∈ {(5, 1, 0, 3), (83, 12, 1, 13), (503, 8, 1, 37), (349, 10, 2, 17)}.(h) n = 4, m = 0 and (d, `, p) ∈ {(7, 5, 3)}.(i) Pmin(n) ≥ 7 and ` = 0, 2 or 4.Proof. In all cases, when m > 0, we know thatd2 ± 2` ≡ ±5mpn ≡ 0 (mod 5)and so d2 ≡ ∓2` (mod 5). If ` is odd, we get a contradiction since ∓2 is a non-quadraticresidue. Hence in all three parts to this theorem, we know that ` is even.1. Case 1: m = 0 or n = 0. In this setting if either m = 0 or n = 0, then Theorem3.3.4 gives the results as stated. For the case when both m = n = 0, we have that(d− 1)(d+ 1) = 2` and thus d− 1 = 2 and d+ 1 = 2`−1. Hence d = 3 and ` = 3.Case 2: 3 | n. If m ≡ 0 (mod 3) then (x, y) = (5m/3pn/3, d) is a solution to y2 = x3+2`.If m ≡ 1 (mod 3) then (x, y) = (5(m+2)/3pn/3, 5d) is a solution to y2 = x3 + 2`52. Ifm ≡ 2 (mod 3) then (x, y) = (5(m+4)/3pn/3, 25d) is a solution to y2 = x3 + 2`54.A check in Theorem 3.1.4 shows us that the only solutions to the above comes from thefirst case with (d, `,m, n, p) = (71, 7, 0, 3, 17) and from the third case with the solutiongiven by (d, `,m, n, p) = (26, 0, 2, 3, 3).201Case 3: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Case 4: Remaining cases. The sign of 2` on the left hand side of the equation isnegative and we will assume that our equation takes the shape d2−2` = 5mpn. We canfactor the equation to get(d− 2`/2)(d+ 2`/2) = 5mpnAs the elements on the left are coprime, we can break into factors givingd∓ 2`/2 = 5md± 2`/2 = pn.Eliminating d gives 5m ± 2`/2+1 = pn. We break into cases.Case 4a: n > 0 is even In this case, modulo 5 considerations show again that `/2 + 1must be even since ±2 are both non-quadratic residues modulo 5.Case 4a (i): pn = 5m + 2`/2+1. Let k = `/2 + 1. As k is even, factoring again gives(pn/2 − 2k/2)(pn/2 + 2k/2) = 5m and so we getpn/2 ∓ 2k/2 = −1pn/2 ± 2k/2 = 5m.Eliminating p gives ±2k/2+1 = 5m + 1. As k > 0 we know that k/2 + 1 > 1 and modulo4 considerations give a contradiction.Case 4a (ii): pn = 5m − 2`/2+1. In this case, either ` = 0 (impossible since `/2 + 1 iseven), ` = 2 or ` ≥ 4. In the case when ` ≥ 4, modulo 8 considerations give us thatpn ≡ 5m (mod 8) and as n is even, this implies that m is even as well. We factor theequation as (pn/2 − 5m/2)(pn/2 + 5m/2) = −2`/2+1. This givespn/2 − 5m/2 = −2pn/2 + 5m/2 = 2`/2pn/2 − 5m/2 = −2`/2pn/2 + 5m/2 = 2.Eliminating pn/2 in both cases gives 5m/2 = 1 + 2`/2−1 and Miha˘ilescu’s Theorem (The-orem 3.3.8) implies that there are no solutions. This leaves ` = 2 and pn = 5m − 4.Now solutions to x2 + 4 = yn are given in [Coh03] to be x = 2, 11 of which we onlyconsider x = 11 which gives us 112 + 4 = 53 and this is as stated in the theorem.202Case 4b: Pmax(n) ≥ 7 and `/2 + 1 6= 2, 3. In this case, we immediately see viaTheorem 1.6.1 that we have no solutions.Case 4c: Pmax(n) ≥ 7 and `/2+1 = 3. This gives pn = 5m±8 and by Theorem 1.6.7,we get a contradiction.Case 4d: Pmax(n) ≥ 7 and `/2 + 1 = 2. This gives pn = 5m ± 4.2. Case 1: ` ≤ 2. This case has only solutions when ` = 1 = d and n = m = 0.Throughout we suppose that ` ≥ 4 (recalling that ` is even).Case 2: m = 0 or n = 0. In this setting if either m = 0 or n = 0, then Theorem8.4 in [BS04] gives the results as stated when the other variable is at least 1. Whenn = 0 and m = 1, then [Coh03] gives us no solutions. When n = 1 and m = 0 then thesolutions are as stated.Case 3: 3 | n. If m ≡ 0 (mod 3) then (x, y) = (−5m/3pn/3, d) is a solution toy2 = x3 + 2`. If m ≡ 1 (mod 3) then (x, y) = (−5(m+2)/3pn/3, 5d) is a solution toy2 = x3 + 2`52. If m ≡ 2 (mod 3) then (x, y) = (−5(m+4)/3pn/3, 25d) is a solution toy2 = x3 + 2`54.A check in Theorem 3.1.4 shows us that the only solution to the above comes fromthe first case with (d, `,m, n, p) = (13, 9, 0, 3, 7) and the third case with (d, `,m, n, p) =(11, 8, 1, 3, 3).Case 4: n is even. In d2 − 2` = −5mpn, we know thatd2 ≡ d2 − 2` ≡ −5mpn ≡ −5m (mod 8)and this is a contradiction.Case 5: Remaining cases. As ` is even, factoring gives(d− 2`/2)(d+ 2`/2) = −5mpnand breaking into cases givesd− 2`/2 = −5md+ 2`/2 = pnd− 2`/2 = −pnd+ 2`/2 = 5m.203Eliminating d in either case gives 2`/2+1 = 5m + pn. When ` = 4 we get a solutionwith n = 1 and p = 3. From here out, we assume that ` ≥ 6 so that `/2 + 1 ≥ 4 andthat n is odd and Pmin(n) ≥ 5. Thus, we may apply Theorem 1.6.1 and see that theequation 2`/2+1 = 5m + pn has no solutions. This leaves only the case when n = 1 andp = 2`/2+1 − 5m.3. Case 1: m = 0 or n = 0. In this setting if either m = 0 or n = 0, then Theorem 3.3.3gives the results as stated.Case 2: 3 | n. If m ≡ 0 (mod 3) then (x, y) = (5m/3pn/3, d) is a solution to y2 =x3 − 2`. Theorem 3.1.8 tells us that the only solution is with (x, y) = (3, 5) and so(d, `,m, n, p) = (5, 1, 0, 3, 3). If m ≡ 1 (mod 3) then (x, y) = (5(m+2)/3pn/3, 5d) is asolution to y2 = x3 − 2`52. Theorem 3.1.8 tells us that the only solutions are with(x, y, `) ∈ {(5 · 13, 5 · 83, 12), (5 · 37, 5 · 503, 8)}and so(d, `,m, n, p) ∈ {(83, 12, 1, 3, 13), (503, 8, 1, 3, 37)}.If m ≡ 2 (mod 3) then (x, y) = (5(m+4)/3pn/3, 25d) is a solution to y2 = x3 − 2`54.Theorem 3.1.8 tells us that the only solution is with (x, y) = (25 · 17, 25 · 349) and so(d, `,m, n, p) = (349, 10, 2, 3, 17).Case 3: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Case 4: Pmin(n) ≥ 7, ` ≥ 6. Use Theorem 1.6.5 on the original equation to reach acontradiction. Hence ` = 0, 2, 4.Case 5: n even and ` ≥ 3. In the first case, modulo 8 arguments give thatd2 ≡ d2 + 2` ≡ 5mpn ≡ 5m (mod 8)and so m is even since 5 is not a square modulo 8. Factoring givesd− 5m/2pn/2 = −2d+ 5m/2pn/2 = 2`−1d− 5m/2pn/2 = −2`−1d+ 5m/2pn/2 = 2.Eliminating d gives 5m/2pn/2 = 2`−2 + 1. If ` = 4, we do not get any solutions since204both m and n are positive. So suppose that ` ≥ 6 (recall that ` is even). Now theother cases show that only powers of 2 divide n in this case. If n = 2 then we getthe solution p = 2`−2+15m/2. If 4 | n, then another modulo 8 argument with the equation5m/2pn/2 = 2`−2 + 1 shows that m/2 is even again. Thus, we can factor again to seethat5m/4pn/4 − 1 = 25m/4pn/4 + 1 = 2`−3and the first line gives a contradiction.Case 6: n is even and ` = 2. Then d2 + 4 = 5mpn so modulo 8 considerations givesthat m is odd. Now if in fact 4 | n, rearranging gives 5m(pn/4)4− d2 = 4. By absorbingpowers of 5 that are multiples of 4, we can examine the solutions to 5x4 − d2 = 4 and125x4− d2 = 4 to make a claim about solutions in this case. By [Lju67], we see that inboth of these cases (the smallest solution to 5x2− d2 = 4 is given by (x, d) = (1, 1) andthe smallest solution to 125x2 − d2 = 4 is given by (x, d) = (1, 11)) the only solutionsare given by n = 0 which was already considered in case 1. Hence we have that 2 ‖ n.Case 7: n is even 4 | d. This case can only occur if ` = 0 and so in this case we getthat 1 ≡ d2 + 1 ≡ 5mpn ≡ 5m (mod 8) and so again we see that m is even. Factoringhere givesd− 5m/2pn/2 = −1d+ 5m/2pn/2 = 1and this is a contradiction.Case 9: n is even 2 ‖ d. This case can only occur if ` = 0 giving 1 = 5mpn − d2.Now, if m is even, as in case 7 we reach a contradiction. Thus m is odd.Theorem 3.4.5. Let p 6= 2, 5 be a prime number with `, n ≥ 0 and d ≥ 1.1. Solutions to 5d2 − 2`pn = −1 are given by(a) n = 1, ` = 0 and p = 5d2 + 1.(b) n = 1, ` = 1 and p = 5d2+12 .(c) n = 2 and ` = 0.205(d) Pmin(n) ≥ 7 and ` = 1 and p ≡ 3 (mod 4).2. Solutions to 5d2 − 2`pn = 1 are given by(a) n = 0 and ` = 2 with d = 1.(b) n = 1 and ` = 0, 2.(c) ` = 2 with 2 ‖ n and either n = 2 or Pmin(n/2) ≥ 7, or 2 - n and Pmin(n) ≥ 7.Proof. We will start by considering both cases simultaneously then breaking down this the-orem into parts. Throughout, suppose that we have a solution to 5d2 − 2`pn = ±1.Case 1: ` ≥ 3. In this case, modulo 8 considerations gives d2 ≡ ±5 (mod 8) which is acontradiction.Case 2: ` = 0. We use Theorem 1.6.2 to see that n = 0, 1, 2, 3. For n = 3, notice thatsolutions to this equation are associated to solutions of (5d)2 = (5p)3±53 which the theoremsfrom Section 3.1 show is impossible. For n = 1, we get that p = 5d2 ± 1 which are as statedin the theorem. When n = 0, we have no solutions.1. Case 3: ` = 2. In this case, modulo 8 considerations gives 5d2 ≡ 4pn− 1 ≡ 3 (mod 8).Isolating gives d2 ≡ −1 (mod 8) which is a contradiction.Case 4: ` = 1. Here we have 5d2 = 2pn − 1. We have two casesCase 4a: 2 | n. Here, considerations modulo 5 give us that pn ≡ 3 (mod 5) which is acontradiction. Hence n is odd.Case 4b: n is odd Here we can break it up into subcases. First notice that if2pn ≡ 2 (mod 8) then 5d2 ≡ 2pn − 1 ≡ 1 (mod 8) and this is a contradiction since5 is not a square modulo 8. So we have that 2pn ≡ 6 (mod 8) which means thatpn ≡ 3 (mod 8) or pn ≡ 7 (mod 8). This translates to p ≡ 3 (mod 4).Case 4bi: 3 | n Transform the equation to (50d)2 = (10p)3 − 22 · 53 and then useSection 3.1 to see that there are no solutions.2. We proceed in cases.Case 3: ` = 1. Modulo 4 considerations gives d2 ≡ 2pn+1 ≡ 3 (mod 4) a contradictionsince 3 is not a square modulo 4.206Case 4: ` = 0 and n even. This gives the equation 5d2 = pn + 1. Note that d is even.Modulo 4 considerations shows that 0 ≡ pn + 1 ≡ 2 (mod 4) a contradiction.Case 5: ` = 0 or ` = 2 and 3 | n. In these cases, multiplying through by suitablepowers of 2 and 5 gives D2 − Xn = 2k15k2 where k1 and k2 are integers. Section 3.1shows that 3 does not divide n in these cases.Case 6: ` = 2 and 4 | n. In this case, using [Lju54], we see that the equation5x2−4y4 = 1 only has the solution (x, y) = (1, 1). Thus n = 0 in this case and we haved = 1.Case 7: 5 | n. Checking with Theorem 3.2.11, we see that this case gives no solutions.Theorem 3.4.6. Let p 6= 2, 5 be a prime number with `, n ≥ 0 and d ≥ 1.1. Solutions to 5d2 − 2` = pn are given by(a) n = 0 and (d, `) = (1, 2).(b) n = 1 and p = 5d2 − 2`.(c) 2 ‖ n, either n = 2 or Pmin(n/2) ≥ 7, and ` = 2.(d) n = 3 and (d, `, p) = (21, 3, 13).(e) n = 5 and (d, `, p) = (7, 1, 3).(f) Pmin(n) ≥ 7 and 1 ≤ ` ≤ 5.2. Solutions to 5d2 + pn = 2` are given by(a) n = 1.(b) n = 3 and (d, `, p) = (1, 5, 3).(c) n = 5 and (d, `, p) = (19, 11, 3).3. Solutions to 5d2 + 2` = pn are given by(a) n = 1 so p = 5d2 + 2`.(b) 2 ‖ n, n = 2 or Pmin(n/2) ≥ 7, and ` = 0, 2.(c) Pmin(n) ≥ 7 and 1 ≤ ` ≤ 5.207Proof. First, let’s consider the above cases simultaneously. Suppose n is even. If ` ≥ 3, thenlocal considerations at 8 reveal thatpn ≡ ±5d2 ≡ ±5 (mod 8)which is a contradiction. If instead ` = 1, then locally at 5 reveals that±2 ≡ 5d2 ± 2 ≡ ±pn ≡ ±1 (mod 5)which is a contradiction. Hence when n is even, ` = 0 or ` = 2 in each case. Now we lookindividually at each equation.1. Case 1: ` = 0. This implies that d is even. Considerations modulo 8 show thatpn = 5d2 − 1 ≡3 if 2 ‖ d−1 if 4 | d(mod 8)and this is a contradiction if n is even. Hence n is odd. By Theorem 1.6.2, there are nosolutions when n ≥ 4. Hence n = 1 or n = 3. For n = 3, the equation is 5d2 − 1 = p3has no solutions via theorem 2 of [Coh91]. Thus, we are left with the case when n = 1and the equation is p = 5d2 − 1 which has many solutions.Case 3: 3 | n. Rearrange the original equation to give (25d)2 = (5pn/3)3+2`53. We cansolve this using Theorem 3.1.4 to see that the only solution comes when (d, `, n, p) =(21, 3, 3, 13).Case 4: 5 | n. Checking with Theorem 3.2.11, we see that the only solution is givenby (d, `, n, p) = (7, 1, 53).Case 5: Pmin(n) ≥ 7 and ` ≥ 6. We can use Theorem 1.6.4 to reach a contradiction.Case 6: 4 | n and ` = 2. For this, we use [LY07] to see that the equation 5d2−(pn/4)4 =4 only has a solution when n = 0 and d = 1. Thus if n is even then 2 ‖ n.2. Case 1: n is even By above we know that ` = 0 or ` = 2 both of which quickly donot yield a solution.Case 2: ` ≤ 5. As d > 0, it is clear that the only solutions occur when(d, `, n, p) ∈ {(1, 3, 1, 3), (1, 4, 1, 11), (1, 5, 3, 3)}.208Hence we may suppose that ` ≥ 6. In fact, using the preliminary remark, we know thatn even gives no solutions.Case 3: 3 | n and ` ≥ 6. Rearrange the original equation to give (25d)2 = (−5pn/3)3 +2`53. We can solve this using Theorem 3.1.4 to see that the only solution comes when(d, `, n, p) = (1, 5, 3, 3).Case 4: 5 | n and ` ≥ 6 . Checking with Theorem 3.2.11, we see that the only solutionis given by (d, `, n, p) = (19, 11, 5, 3).Case 5: Pmin(n) ≥ 7 and ` ≥ 6. This follows from Theorem 1.6.4 that the associatedequation has no solutions.3. Case 1: 3 | n and ` ≥ 6. Rearrange the original equation to give (25d)2 = (5pn/3)3 −2`53. We can solve this using Theorem 3.1.8 to see that there are no solutions.Case 2: 5 | n and ` ≥ 6. Checking with Theorem 3.2.11, we see that this case givesno solutions.Case 3: Pmin(n) ≥ 7 and ` ≥ 6. We can use Theorem 1.6.4 to reach a contradiction.Case 4: 4 | n and ` = 0, 2. In this case, we rewrite the given equation as(pn/4)4 − 5d2 = 2`When ` = 0, results due to Cohn and Ljunggren (see [Coh07a], [Lju42], [Coh97]) showthat the only solution to x4 − 5y2 = 1 is given by (x, y) = (3, 4). When ` = 2, so theequation is x4−5d2 = 4, Ljunggren showed that there are no solutions to this equation(see [Lju67]). For a survey of these results, see [HPPT13]. Thus, we see that if n iseven, then 2 ‖ n.To complete this section, I will summarize the results above into a table. We will beparticularly interested in the cases when n ≥ 1, m ≥ 0 and ` = 2 or ` ≥ 4. We organize theabove by prime p. We will also include all sporadic cases for each prime and when we staterelevant theorems later in this thesis, we will check these primes by hand if necessary.209pN ` m n N 3,72`−2 ± 1 ≥ 5 0 1 1 12`−2+15m ≥ 2 even ≥ 1 1 1 12`−25m ± 1 ≥ 4 ≥ 1 1 1 15m−14 4 even ≥ 1 odd 1 or Pmin(n) ≥ 7 n 1Table 3.21: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + 2`5mpn = .Note: The above with = −1 has no solutions when ` ≥ 2.pN ` m n N δ 3,112`−2 ± 1 ≥ 5 0 1 1 −1 1d2−5m4 2 ≥ 1 odd 1 1 −1 12`−2 ± 5m/2 ≥ 4 ≥ 2 even 1 1 −1 15m−14 2 ≥ 1 odd 2 1 −1 1(5m−14)22 ≥ 1 odd 2 ‖ n and either n = 2 or Pmin(n/2) ≥ 7 n −1 1d2−5m4 2 ≥ 1 odd Pmin(n) ≥ 7 n −1 12`−2 ± 5m/2 4, 5 ≥ 2 even Pmin(n) ≥ 7 n −1 15m−d24 2 ≥ 1 odd 1 1 1 15m/2 − 2`−2 ≥ 4 ≥ 2 even 1 1 1 15m/2−14 4 ≥ 2 even 1 1 1 15m−d24 2 ≥ 1 odd ≥ 2 even n 1 15m−d24 2 ≥ 1 odd Pmin(n) ≥ 7 n 1 15m−d22` 4, 5 ≥ 1 even Pmin(n) ≥ 7 n 1 1Table 3.22: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2`pn = 5m.Note: The above with δ = = −1 has no solutions when ` ≥ 2.210pN ` m n N δ 3,7,11,17,31,41d2 − 2` · 5m ` m 1 1 −1 12`−2 − 5m ≥ 4 m 2 1 −1 15m − 2`−2 ≥ 4 m 2 1 −1 12`−2 · 5m − 1 ≥ 4 m 2 1 −1 1d2 − 2` · 5m 2, 4, 5 m Pmin(n) ≥ 7 n −1 1(5m − 4)2 4 ≥ 1 odd n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7 n −1 12` · 5m − d2 ` m 1 1 −1 −12` · 5m − d2 2, 4, 5 m Pmin(n) ≥ 7 n −1 −1d2 + 2` · 5m ` m 1 1 1 15m + 2`−2 ≥ 3 m 2 1 1 12` · 5m + 1 ≥ 3 m 2 1 1 1Table 3.23: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2`5m = pn.pN ` m n N δ 3,11,13,17,37d2 − 2` ` 0 1 1 −1 15m/2 ± 2`/2+1 ≥ 2 even ≥ 1 1 1 −1 12`−2 − 1 ≥ 4 even 0 2 1 −1 15m ± 4 2 ≥ 1 Pmin(n) ≥ 7 n −1 12` − d2 ≥ 4 0 1 1 −1 −12`/2+1 − 5m ≥ 2 even m 1 1 −1 −1d2+2`5m ` m 1 1 1 1d2+45m 2 m odd n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7 n 1 1d2+2`5m 2, 4 m Pmin(n) ≥ 7 n 1 12`−2+15m/2even m ≥ 0 even 2 1 1 1Table 3.24: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equationd2 + δ2` = 5mpn.211pN ` n N 5d2−14 2 One of n = 1, n = 2 or 2 ‖ n with Pmin(n/2) ≥ 7 n 1Table 3.25: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equation5d2 − 2`pn = .Note: The above with = −1 has no solutions when ` ≥ 2.pN ` n N δ 35d2 − 2` ` 1 1 −1 15d2 − 4 2 n = 2, or 2 ‖ n with Pmin(n/2) ≥ 7 n −1 15d2 − 2` 2, 4, 5 Pmin(n) ≥ 7 n −1 12` − 5d2 ` 1 1 −1 −15d2 + 2` ` 1 1 1 15d2 + 4 2 n = 2, or 2 ‖ n with Pmin(n/2) ≥ 7 n 1 15d2 + 2` 2, 4, 5 Pmin(n) ≥ 7 n 1 1Table 3.26: Summary of solutions with n ≥ 1, ` = 2 or ` ≥ 4 and m ≥ 0 for the equation5d2 − δ2` = pn.212Chapter 4Elliptic Curves With Rational TwoTorsion and Conductor 18p, 36p or 72pOrganized by PrimesIn this section we recall, simplify and summarize the results from [Mul06]. The majorityof the work was done in the aforementioned thesis however the theorems there were presenteddifferently and were not organized by the form of the prime p. In the following theorems, ifa case shows up degenerately, for example if some exponent is 0, then many cases of curvesmight give the same elliptic curves. I have made an attempt to not duplicate any ellipticcurves represented below.Additionally in the following theorems, much like what happened in the tables at the endof Section 3.4, we see that sometimes the n substituted into the elliptic curves is not the sameas the n in the prime decomposition. For an example, in the theorem for curves of conductorof 18p, we see a case where p = 2a3b + 1 which suggests that the exponent of n is 1 and so weare to plug 1 in the tables obtained from [Mul06]. However this is not necessarily the case.Somewhat annoyingly, some cases will have, for example, n = 2 in the Diophantine equationand after factoring yield a prime of the form p = 2a3b + 1 which has n = 1. Despite thispotential confusion, I have left the notation the same. This only becomes a problem whenn = 2. If you are simply reading the theorem and do not care about deriving it, then this isnot a problem and this note can be ignored without harm.Theorem 4.0.7. Let p be a prime distinct from 2 and 3. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 18p provided that p satisfies at least one ofthe following where d > 0, `1 ≥ 5, `2 ≥ 7 and m ≥ 0 are integers:1. p = 2`13m + 1.2. p = 2`13m − 1.2133. p = 3m + 2`1.4. p = 3m − 2`1.5. p = 2`1 − 3m.6. p = d2 + 2`23m.7. p = d2 − 2`23m.8. p = 2`23m − d2.9. p = d2+2`23m .10. p = 3d2 + 2`2.11. p = 3d2 − 2`2.12. p = 2`2 − 3d2.13. p ∈ {5, 7, 11, 17, 19, 23, 73}.Theorem 4.0.8. Let p be a prime distinct from 2 and 3. Then up to the finitely many primesinp ∈ {5, 7, 11, 17, 19, 23, 73}the following gives a complete list of the families of elliptic curves with nontrivial rationaltwo torsion of conductor 18p associated with the primes in the previous theorem:(1) The prime p has the form p = 2`−2 · 3m + 1 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 −3ψ(2`3mp+ 1) 2`−23m+2p 22`32m+6p2A2 2 · 3ψ(2`3mp+ 1) 32 2`+63m+6p•a2 a4 ∆D1 −3ψ(p2 − 2`3m) −2`−23m+2 22`32m+6p2D2 2 · 3ψ(p2 − 2`3m) 32p2 −2`+63m+6p4(2) The prime p has the form p = 2`−2 · 3m− 1 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves214•a2 a4 ∆A1 −3ψ(2`3mp+ 1) 2`−23m+2p 22`32m+6p2A2 2 · 3ψ(2`3mp+ 1) 32 2`+63m+6p•a2 a4 ∆B1 −3ψ(2`3m + p2) 2`−23m+2 22`32m+6p2B2 2 · 3ψ(2`3m + p2) 32p2 2`+63m+6p4(3) The prime p has the form p = 3m + 2`−2 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆D1 −3ψ(p2 − 2`3m) −2`−23m+2 22`32m+6p2D2 2 · 3ψ(p2 − 2`3m) 32p2 −2`+63m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 3(2m)/2+2`−2given bya2 a4 ∆E1 −3ψ(2`p+ 32m) 2`−232p 22`32m+6p2E2 2 · 3ψ(2`p+ 32m) 32m+2 2`+634m+6p• Rewriting `− 2 as ˆ`/2 + 1, we get solutions with p = 3m + 2ˆ`/2+1 given bya2 a4 ∆H1 −3ψ(2ˆ` + 3mp) 2ˆ`−232 22ˆ`3m+6pH2 2 · 3ψ(2ˆ` + 3mp) 3m+2p 2ˆ`+632m+6p2(4) The prime p has the form p = 3m − 2`−2 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆B1 −3ψ(2`3m + p) 2`−23m+2 22`32m+6p2B2 2 · 3ψ(2`3m + p) 32p2 2`+63m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 3(2m)/2−2`−2given bya2 a4 ∆G1 · 3ψ(32m − 2`p) −2`−232p 22`32m+6p2G2 − · 2 · 3ψ(32m − 2`p) 32m+2 −2`+634m+6p215• Rewriting `− 2 as ˆ`/2 + 1, we get solutions with p = 3m − 2ˆ`/2+1 given bya2 a4 ∆H1 −3ψ(2ˆ` + 3mp) 2ˆ`−232 22ˆ`3m+6pH2 2 · 3ψ(2ˆ` + 3mp) 3m+2p 2ˆ`+632m+6p2(5) The prime p has the form p = 2`−2 − 3m with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆B1 −3ψ(2`3m + p) 2`−23m+2 22`32m+6p2B2 2 · 3ψ(2`3m + p) 32p2 2`+63m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 2`−2−3(2m)/2given bya2 a4 ∆E1 −3ψ(2`p+ 32m) 2`−232p 22`32m+6p2E2 2 · 3ψ(2`p+ 32m) 32m+2 2`+634m+6p• Rewriting `− 2 as ˆ`/2 + 1, we get solutions with p = 2ˆ`/2+1 − 3m given bya2 a4 ∆I1 · 3ψ(2ˆ`− 3mp) 2ˆ`−232 −22ˆ`3m+6pI2 − · 2 · 3ψ(2ˆ`− 3mp) −3m+2p 2ˆ`+632m+6p2(6) The prime p has the form p = d2 + 2`3m with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆D1 −3ψ(p− 2`3m) −2`−23m+2 22`32m+6pD2 2 · 3ψ(p− 2`3m) 32p −2`+63m+6p2(7) The prime p has the form p = d2 − 2`3m with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆B1 −3ψ(2`3m + p) 2`−23m+2 22`32m+6pB2 2 · 3ψ(2`3m + p) 32p 2`+63m+6p2216(8) The prime p has the form p = 2`3m − d2 with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆C1 −3ψ(2`3m − p) 2`−23m+2 −22`32m+6pC2 2 · 3ψ(2`3m − p) −32p 2`+63m+6p2(9) The prime p has the form p = d2+2L3m with L ≥ 5, m ≥ 0 and d ∈ Z and corresponds tothe following elliptic curves• If L = `/2 + 1 then we get a curve as given bya2 a4 ∆H1 −3ψ(2` + 3mp) 2`−232 22`3m+6pH2 2 · 3ψ(2` + 3mp) 3m+2p 2`+632m+6p2• In the case where d = 1, we get a solution with L = `− 2 as given bya2 a4 ∆A1 −3ψ(2`3mp+ 1) 2`−23m+2p 22`32m+6p2A2 2 · 3ψ(2`3mp+ 1) 32 2`+63m+6p• In the case where d = 1, we get a solution with L = `− 2 and p = 2`3(2m)/2 − d2 asgiven bya2 a4 ∆J1 −3ψ(32mp− 2`) −2`−232 22`32m+6pJ2 2 · 3ψ(32mp− 2`) 32m+2p −2`+634m+6p2(10) The prime p has the form p = 3d2 + 2` with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves•a2 a4 ∆O1 · 3s+1ψ(p−2`3 ) −2`−232s+1 22`36s+3pO2 − · 2 · 3s+1ψ(p−2`3 ) 32s+1p −2`+636s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.(11) The prime p has the form p = 3d2 − 2` with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves217•a2 a4 ∆M1 · 3s+1ψ(2`+p3 ) 2`−232s+1 22`36s+3pM2 − · 2 · 3s+1ψ(2`+p3 ) 32s+1p 2`+636s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.(12) The prime p has the form p = 2` − 3d2 with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves•a2 a4 ∆N1 · 3s+1ψ(2`−p3 ) 2`−232s+1 −22`36s+3pN2 − · 2 · 3s+1ψ(2`−p3 ) −32s+1p 2`+636s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.Proof. To show this, we proceed in cases. We use the families in [Mul06, p.58-60] and simplifythem according to [Mul06, p.153-175]. In what follows, δ, ∈ {±1} not both negative. Alsoa, b ∈ Z are positive integers.Case 1: t2 = 2a3bpn + 1. This is case 1 in [Mul06, p.58-60]. We can use Lemma 4.7 in[Mul06, p.153] to see that the solutions with a ≥ 7 and b ≥ 0 arep a b t n Extra Information2a−2 · 3b ± 1 ≥ 3 ≥ 0 2p∓ 1 1 none2a−2 ± 1 ≥ 5 0 2p∓ 1 1 none2a−2+13b ≥ 5 ≥ 1 2a−1 + 1 1 3b|a− 217 7 2 577 2 noneAs this case only has solutions when n = 1 or p = 17, we can verify that these twosituations are covered by the finitely many primes listed above and by families (1), (2) and(9).Case 2: t2 = pn + δ2a3b. This case is a combination of cases 2 and 9 in [Mul06, p.58-60].We can use Lemma 4.9 in [Mul06, p.165] to see that the solutions with a ≥ 7 and b ≥ 0 are218p a b t n δt2 − 2a · 3b ≥ 2 ≥ 0 t 1 12a−2 · 3b − 1 ≥ 3 ≥ 0 p+ 2 2 13b − 2a−2 ≥ 3 ≥ 0 2a−1 + p 2 12a−2 − 3b ≥ 5 ≥ 0 2a−1 − p 2 117 7 0 71 3 173 15 2 827 3 17 7 4 113 4 15 9 1 131 6 1t2 + 2a · 3b ≥ 1 ≥ 0 t 1 −12a−2 · 3b + 1 ≥ 3 ≥ 0 p− 2 2 −13b + 2a−2 ≥ 3 ≥ 0 p− 2 · 3b 2 −117 7 2 287 4 −1As this case only has solutions when n = 1, n = 2 or p = 5, 7, 17 or 73, we can verify thatthese three situations are covered by the finitely many primes listed above and by families(1), (2), (3), (4), (5), (6) and (7).Case 3: t2 = 2a3b − pn. This case is case 3 in [Mul06, p.58-60]. We can use Lemma 4.9in [Mul06, p.165] to see that the solutions with a ≥ 7 and b ≥ 0 arep a b t n2a · 3b − t2 ≥ 1 ≥ 0 t 17 9 0 13 323 12 1 11 3As this case only has solutions when n = 1 or p = 7 or 23, we can verify that these twosituations are covered by the finitely many primes listed above and by family (8).Case 4: t2 = 3b + δ2apn. This case is a combination of cases 4 and 8 in [Mul06, p.58-60].We can use Lemma 4.8 in [Mul06, p.156-157] to see that the solutions with a ≥ 7 and b ≥ 0arep a b t n δ2a−2 ± 3b/2 ≥ 3 ≥ 0 (even) 2p∓ 3b/2 1 15 9 2 253 3 13b/2 − 2a−2 ≥ 3 ≥ 0 (even) ±(2p− 3b/2) 1 −17 7 8 17 2 −1219As this case only has solutions when n = 1 or p = 5 or 7, we can verify that these twosituations are covered by the finitely many primes listed above and by families (3), (4) and(5).Case 5: t2 = 2a + δ3bpn. This case is a combination of cases 5 and 6 in [Mul06, p.58-60].We can use Lemma 4.10 in [Mul06, p.171] to see that the solutions with a ≥ 7 and b ≥ 0 arep a b t n δt2 − 2a ≥ 1 (odd) 0 t 1 12a/2+1+13b ≥ 1 (even) ≥ 1 2a/2 + 1 1 13b ± 2a/2+1 ≥ 1 (even) ≥ 1 p± 2a/2 1 17 8 4 65 2 12a−2 − 1 ≥ 5 0 p+ 2 2 117 7 0 71 3 12a − t2 ≥ 5 (odd) 0 t 1 −12a/2+1 − 3b ≥ 4 (even) ≥ 1 ±(2a/2 − p) 1 −17 10 2 31 1 −15 12 1 61 3 −17 9 0 13 3 −1Firstly, all the cases above when n = 1 and b = 0 have been covered by cases 2 and 3so for this case we can assume that b ≥ 1 when n = 1. Under this additional restriction,we only have solutions when n = 1, n = 2 or p = 5, 7 or 17. We can verify that these threesituations are covered by the finitely many primes listed above and by families (1), (6) and(7).Case 6: t2 = 3bpn − 2a. This is case 7 in [Mul06, p.58-60]. We can use Lemma 4.10 in[Mul06, p.171] to see that the solutions with a ≥ 7 and b ≥ 0 arep a b t nt2+2a3b ≥ 2 ≥ 0 t 12a−2+13b/2≥ 3 (odd) ≥ 0 (even) 3b/2p− 2 211 9 1 59 317 15 2 107 319 7 1 143 3As a note for later, notice that in the second case above, setting a = 7 gives b = 2 andp = 11 as a solution which is already accounted for and so in the second case, we can assume220that a − 1 ≥ 7. This case only has solutions when n = 1, n = 2 or p = 11, 17 or 19, andthus we can verify that these three situations are covered by the finitely many primes listedabove and by family (9).Case 7: 3t2 = 2a + δp. This case is a combination of cases 10 to 12 in [Mul06, p.58-60].We can use Lemma 4.11 in [Mul06, p.174-175] to see that the solutions with a ≥ 7 and b ≥ 0arep a t n δ 3t2 − 2a ≥ 0 t 1 1 111 8 23 3 1 12a − 3t2 ≥ 0 t 1 −1 15 7 1 3 −1 13t2 + 2a ≥ 0 t 1 1 −1This case only has solutions when n = 1 or p = 5 or 11, and thus we can verify thatthese three situations are covered by the finitely many primes listed above and by families(10),(11) and (12).Theorem 4.0.9. Let p be a prime distinct from 2 and 3. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 36p provided that p satisfies at least one ofthe following where d > 0, n1 = 1 or Pmin(n1) ≥ 7, n2 ∈ {1, 2} and m1,m2 ≥ 0 are integerswith m1 odd and m2 even:1. p = d2 + 4 · 3m2.2. pn1 = d2 − 4 · 3m1.3. pn1 = 4 · 3m1 − d2.4. p = d2+3m14 and p ≡ −1 (mod 4).5. pn2 = 3d2+14 and p ≡ 1 (mod 4).6. p = 3d2 − 4.7. p ∈ {5, 13}.Theorem 4.0.10. Let p be a prime distinct from 2 and 3. Then up to the finitely manyprimes inp ∈ {5, 13}221the following gives a complete list of the families of elliptic curves with nontrivial rationaltwo torsion of conductor 36p associated with the primes in the previous theorem:(1) The prime p has the form p = d2 + 4 · 3m with m ≥ 0 even, d ∈ Z and corresponds to thefollowing elliptic curves•a2 a4 ∆C1 −3ψ(p− 4 · 3m) −3m+2 2432m+6pC2 2 · 3ψ(p− 4 · 3m) 32p −283m+6p2(2) The prime p has the form pn = d2− 4 · 3m with m ≥ 1 odd, n = 1 or Pmin(n) ≥ 7, d ∈ Zand corresponds to the following elliptic curves•a2 a4 ∆A1 −3ψ(4 · 3m + pn) 3m+2 2432m+6pnA2 2 · 3ψ(4 · 3m + pn) 32pn 283m+6p2n(3) The prime p has the form pn = 4 · 3m− d2 with m ≥ 1 odd, n = 1 or Pmin(n) ≥ 7, d ∈ Zand corresponds to the following elliptic curves•a2 a4 ∆B1 −3ψ(4 · 3m − pn) 3m+2 −2432m+6pnB2 2 · 3ψ(4 · 3m − pn) −32pn 283m+6p2n(4) The prime p has the form p = d2+3m4 with m ≥ 1 odd, p ≡ −1 (mod 4) and d ∈ Z andcorresponds to the following elliptic curves•a2 a4 ∆E1 −3ψ(4p− 3m) 32p −243m+6p2E2 2 · 3ψ(4p− 3m) −3m+2 2832m+6p(5) The prime p has the form pn = 3d2+14 with d an integer, n ∈ {1, 2}, s ∈ {0, 1}, p ≡1 (mod 4) and corresponds to the following elliptic curves•a2 a4 ∆I1 · 3s+1ψ(4pn−13 ) 32s+1pn −2436s+3p2nI2 − · 2 · 3s+1ψ(4pn−13 ) −32s+1 2836s+3pnwhere ∈ {±1} is the residue of 3s+1 modulo 4.222(6) The prime p has the form p = 3d2 − 4 with d an integer, s ∈ {0, 1} and corresponds tothe following elliptic curves•a2 a4 ∆J1 · 3s+1ψ(4+p3 ) 32s+1 2436s+3pJ2 − · 2 · 3s+1ψ(4+p3 ) 32s+1p 2836s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.Proof. To show this, we proceed in cases. We use the families in [Mul06, p.61-62] and simplifythem according to [Mul06, p.153-175].Case 1: t2 = pn− 4 · 3b. This is case 4 in [Mul06, p.61-62]. Here b even is given. We canuse Lemma 4.9 in [Mul06, p.165] to see that the solutions with b ≥ 0 arep b t nt2 + 4 · 3b ≥ 0 t 15 0 11 313 5 35 3n√t2 + 4 · 3b ≥ 1 t Pmin(n) ≥ 7According to [Luc02], we know that in the last case n < 5 and and thus this case givessolutions only when n = 1 or p = 5 or 13. We can verify that these two situations are coveredby the finitely many primes listed above and by (1).Case 2: t2 = 4 · 3b + δpn for δ ∈ {±1}. This case is a combination of cases 1 and 2 in[Mul06, p.61-62]. Here b odd is given. We can use Lemma 4.9 in [Mul06, p.165] to see thatthe solutions with b ≥ 0 arep b t n δt2 − 4 · 3b ≥ 0 t 1 113 1 47 3 1n√t2 − 4 · 3b ≥ 1 t Pmin(n) ≥ 7 14 · 3b − t2 ≥ 0 t 1 −1n√4 · 3b − t2 ≥ 1 t Pmin(n) ≥ 7 −1As this case only has solutions when n = 1, Pmin(n) ≥ 7 or p = 13, we can verify thatthese two situations are covered by the finitely many primes listed above by (2) and by (3).223Case 3: t2 = 4pn−3b. This is case 3 in [Mul06, p.61-62]. Here pn ≡ −1 (mod 4) is givenand so the case n even is impossible. We can use Lemma 4.8 in [Mul06, p.156-157] to seethat the solutions with b ≥ 0 and n odd arep b t n Extra Informationt2+3b4 ≥ 1 (odd) t 1 p ≡ 1 (mod 3)n√t2+3b4 ≥ 1 (odd) t Pmin(n) ≥ 7 p ≡ 1 (mod 3)As shown in [AAA02], the second case cannot occur for n ≥ 7. As this case only hassolutions when n = 1, we can verify that this situation is covered by family (4).Case 4: 3t2 = 4pn − 1. This is case 5 in [Mul06, p.61-62]. Here pn ≡ 1 (mod 4) is given.By [Coh07a, p.353] Corollary 6.3.15, we have when n = 2 thatp = (σ2 + τ 2 + στ) 1 = δ(σ2 + τ 2 + 4στ)with σ and τ of opposite parity and δ, ∈ {±1}. We assume without loss of generality thatσ is the even term. In the second equation above, notice that if δ = −1, then modulo 4considerations give us a contradiction. Hence δ = 1. Next assume towards a contradictionthat 2 ‖ σ. Then modulo 8, we have that 1 ≡ 4+1+0 ≡ 5 (mod 8) and that is a contradiction.Hence 4 | σ. Further, in the equation above involving p, we have that σ2 + τ 2 + στ > 0 forany integers σ and τ and so > 0. Combining this with 4 | σ shows us thatp = σ2 + τ 2 + στ ≡ 0 + 1 + 0 ≡ 1 (mod 4).Thus p ≡ 1 (mod 4) when n = 2 and p ≡ 1 (mod 4) when n = 1 by necessity of this case.We can use Lemma 4.11 in [Mul06, p.175] to see that the solutions arep b t n3t2+14 ≥ 0 t 1√3t2+14 ≥ 1 t 2As this case only has solutions when n = 1 or 2, we can verify that these two situationsare covered by family (5).Case 5: 3t2 = pn + 4. This is case 6 in [Mul06, p.61-62]. We can use Lemma 4.11 in[Mul06, p.174-175] to see that the only solutions are with p = 3t2 − 4. This situation iscovered by family (6).224Theorem 4.0.11. Let p be a prime distinct from 2 and 3. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 72p provided that p satisfies at least one ofthe following where d > 0, `1 ∈ {2, 3}, `2 ∈ {4, 5}, `3 ∈ {2, 4, 5}, n1 = 1 or Pmin(n1) ≥ 7,n2 ∈ {1, 2} and m,m1 ≥ 0 are integers with m1 odd:1. p = 2`13m + 1.2. p = 2`13m − 1.3. p = 3m + 2`1.4. p = 3m − 2`1.5. p = d2 + 2`33m.6. pn1 = d2 − 2`23m.7. pn1 = 2`23m − d2.8. p = 3m1+14 .9. p = 3m1+d24 with p ≡ 1 (mod 4).10. pn2 = 3d2+14 .11. 1 pn1 = d2+323m .12. p = 3d2 + 2`3.13. p = 3d2 − 2`2.14. p ∈ {5, 7, 11, 13, 17, 23, 29, 31, 37, 47, 67, 73, 107, 109, 193, 1153}.Theorem 4.0.12. Let p be a prime distinct from 2 and 3. Then up to the finitely manyprimes inp ∈ {5, 7, 11, 13, 17, 23, 29, 31, 37, 47, 67, 73, 107, 109, 193, 1153}the following gives a complete list of the families of elliptic curves with nontrivial rationaltwo torsion of conductor 72p associated with the primes in the previous theorem:(1) The prime p has the form p = 2`−2 · 3m + 1 with ` ∈ {4, 5}, m ≥ 0 and corresponds tothe following elliptic curves1In [BLM11, p.7], this case was erroneously claimed to be for odd m only. In fact cases such as m = 2and d = 11 do give valid elliptic curves.225•a2 a4 ∆A1 −3ψ(2`3mp+ 1) 2`−23m+2p 22`32m+6p2A2 2 · 3ψ(2`3mp+ 1) 32 2`+63m+6p•a2 a4 ∆G1 −3ψ(p2 − 2`3m) −2`−23m+2 22`32m+6p2G2 2 · 3ψ(p2 − 2`3m) 32p2 −2`+63m+6p4• In the additional case when ` − 2 = 2, we get solutions given by p = 2`3(2m)/2 + 1and the curve is given bya2 a4 ∆B1 3ψ(4 · 3m + p) 32m+2 2434m+6pB2 −2 · 3ψ(4 · 3m + p) 32p 2832m+6p2(2) The prime p has the form p = 2`−2 · 3m − 1 with ` ∈ {4, 5}, m ≥ 0 and corresponds tothe following elliptic curves•a2 a4 ∆A1 −3ψ(2`3mp+ 1) 2`−23m+2p 22`32m+6p2A2 2 · 3ψ(2`3mp+ 1) 32 2`+63m+6p•a2 a4 ∆E1 −3ψ(2`3m + p2) 2`−23m+2 22`32m+6p2E2 2 · 3ψ(2`3m + p2) 32p2 2`+63m+6p4• In the additional case when `− 2 = 2, we get solutions given by p = 4 · 3(2m)/2 − 1with m odd and the curve is given bya2 a4 ∆C1 3ψ(4 · 3m − pn) 32m+2 −2434m+6pC2 −2 · 3ψ(4 · 3m − pn) −32p 2832m+6p2(3) The prime p has the form p = 3m + 2`−2 with ` ∈ {4, 5}, m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆G1 −3ψ(p2 − 2`3m) −2`−23m+2 22`32m+6p2G2 2 · 3ψ(p2 − 2`3m) 32p2 −2`+63m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 3(2m)/2+2`−2given by226a2 a4 ∆K1 −3ψ(2`p+ 3m) 2`−232p 22`32m+6p2K2 2 · 3ψ(2`p+ 3m) 32m+2 2`+634m+6p• Rewriting `− 2 as ˆ`/2 + 1, we get solutions with p = 3m + 2ˆ`/2+1 given bya2 a4 ∆O1 −3ψ(2ˆ` + 3mp) 2ˆ`−232 22ˆ`3m+6pO2 2 · 3ψ(2ˆ` + 3mp) 3m+2p 2ˆ`+632m+6p2(4) The prime p has the form p = 3m − 2`−2 with ` ∈ {4, 5}, m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆E1 −3ψ(2`3m + p2) 2`−23m+2 22`32m+6p2E2 2 · 3ψ(2`3m + p2) 32p2 2`+63m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 3(2m)/2+2`−2given bya2 a4 ∆M1 −3ψ(3m − 2`pn) −2`−232p 22`32m+6p2M2 2 · 3ψ(3m − 2`pn) 32m+2 −2`+634m+6p• Rewriting `− 2 as ˆ`/2 + 1, we get solutions with p = 3m + 2ˆ`/2+1 given bya2 a4 ∆O1 −3ψ(2ˆ` + 3mp) 2ˆ`−232 22ˆ`3m+6pO2 2 · 3ψ(2ˆ` + 3mp) 3m+2p 2ˆ`+632m+6p2(5) The prime p has the form p = d2 +2`3m with ` ∈ {2, 4, 5}, m ≥ 0, d ∈ Z and correspondsto the following elliptic curves• In the case when ` = 2, we havea2 a4 ∆D1 3ψ(p− 4 · 3m) −3m+2 2432m+6pD2 −2 · 3ψ(p− 4 · 3m) 32p −283m+6p2• In the case when ` 6= 2, we have227a2 a4 ∆G1 −3ψ(p− 2`3m) −2`−23m+2 22`32m+6pG2 2 · 3ψ(p− 2`3m) 32p −2`+63m+6p2(6) The prime p has the form pn = d2 − 2`3m with ` ∈ {4, 5}, m ≥ 0, n = 1 or Pmin(n) ≥ 7,d ∈ Z and corresponds to the following elliptic curves•a2 a4 ∆E1 −3ψ(2`3m + pn) 2`−23m+2 22`32m+6pnE2 2 · 3ψ(2`3m + pn) 32pn 2`+63m+6p2n(7) The prime p has the form pn = 2`3m − d2 with ` ∈ {4, 5}, m ≥ 0, n = 1 or Pmin(n) ≥ 7,d ∈ Z and corresponds to the following elliptic curves•a2 a4 ∆F1 −3ψ(2`3m − pn) 2`−23m+2 −22`32m+6pnF2 2 · 3ψ(2`3m − pn) −32pn 2`+63m+6p2n(8) The prime p has the form p = 3m+14 with m ≥ 1 odd and corresponds to the followingelliptic curves•a2 a4 ∆A1 −3ψ(243mp+ 1) 223m+2p 2832m+6p2A2 2 · 3ψ(243mp+ 1) 32 2103m+6p• If p ≡ 1 (mod 4) then alsoa2 a4 ∆I1 3ψ(4p2 − 3m) 32p2 −243m+6p4I2 −2 · 3ψ(4p2 − 3m) −3m+2 2832m+6p2• In addition, replacing m above with 2m, then we get solutions with p = 3(2m)/2+14given bya2 a4 ∆K1 −3ψ(24p+ 32m) 2232p 2832m+6p2K2 2 · 3ψ(24p+ 32m) 32m+2 21034m+6p(9) The prime p has the form p = d2+3m4 with m ≥ 1 odd, p ≡ 1 (mod 4) and d ∈ Z andcorresponds to the following elliptic curves228•a2 a4 ∆I1 3ψ(4p− 3m) 32p −243m+6p2I2 −2 · 3ψ(4p− 3m) −3m+2 2832m+6p(10) The prime p has the form pn = 3d2+14 with d an integer, n ∈ {1, 2}, s ∈ {0, 1}, andcorresponds to the following elliptic curves•a2 a4 ∆S1 − · 3s+1ψ(4pn−13 ) 32s+1pn −2436s+3p2nS2 · 2 · 3s+1ψ(4pn−13 ) −32s+1 2836s+3pnwhere ∈ {±1} is the residue of 3s+1 modulo 4.(11) The prime p has the form pn = d2+323m with m ≥ 0, n = 1 or Pmin(n) ≥ 7, d ∈ Z andcorresponds to the following elliptic curves•a2 a4 ∆Q1 −3ψ(3mpn − 25) −2332 2103m+6pnQ2 2 · 3ψ(3mpn − 25) 3m+2pn −21132m+6p2n(12) The prime p has the form p = 3d2−2` with ` ∈ {4, 5}, s ∈ {0, 1}, d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆X1 · 3s+1ψ(2`+p3 ) 2`−232s+1 22`36s+3pX2 − · 2 · 3s+1ψ(2`+p3 ) 32s+1p 2`+636s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.(13) The prime p has the form p = 3d2 + 2` with ` ∈ {2, 4, 5}, s ∈ {0, 1}, d ∈ Z andcorresponds to the following elliptic curves• ` = 2 anda2 a4 ∆W1 − · 3s+1ψ(p−43 ) −32s+1 2436s+3pW2 · 2 · 3s+1ψ(p−43 ) 32s+1p −2836s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.• ` ∈ {4, 5} and229a2 a4 ∆Z1 · 3s+1ψ(p−2`3 ) −2`−232s+1 22`36s+3pZ2 − · 2 · 3s+1ψ(p−2`3 ) 32s+1p −2`+636s+3p2where ∈ {±1} is the residue of 3s+1 modulo 4.Proof. To show this, we proceed in cases. We use the families in [Mul06, p.58-60] and simplifythem according to [Mul06, p.153-175]. In what follows, δ, ∈ {±1} not both negative.Case 1: t2 = 2a3bpn + 1. This is case 1 in [Mul06, p.62-67]. We can use Lemma 4.7 in[Mul06, p.153] to see that the solutions with a ∈ {4, 5} and b ≥ 0 arep a b t n Extra Information3b+14 4 ≥ 0 odd 8p− 1 1 none2a−2 · 3b ± 1 ≥ 3 ≥ 0 2p∓ 1 1 none2a−2 ± 1 ≥ 5 0 2p∓ 1 1 none2a−2+13b ≥ 5 ≥ 1 2a−1 + 1 1 3b|a− 25 4 0 9 1 none5 5 1 49 2 noneSince we only care about cases when a = 4 or 5, in the fourth case above, since 3b|a− 2and a = 5, the only case that this applies is a = 5 and b = 1. Here 2a−2+13b = 3 is impossiblesince p > 3. Further, cases two and three in the table above can be merged together. As thiscase only has solutions when n = 1 or p = 5, we can verify that these situations are coveredby the finitely many primes listed above and families (1), (2) and (8).Case 2: t2 = 4 · 3b + δpn. This is a combination of cases 2 and 4 in [Mul06, p.62-67].Here b is given to be even. We can use Lemma 4.9 in [Mul06, p.165] to see that the solutionswith b ≥ 0 arep b t n δt2 − 4 · 3b ≥ 0 t 1 113 0 47 3 1n√t2 − 4 · 3b ≥ 1 t Pmin(n) ≥ 7 14 · 3b − t2 ≥ 0 t 1 −1n√4 · 3b − t2 ≥ 0 t Pmin(n) ≥ 7 −1In the cases above when p 6= 13, notice that we havepn = t2 − 4 · 3b = (t− 2 · 3b/2)(t+ 2 · 3b/2) or pn = 4 · 3b − t2 = (2 · 3b/2 − t)(2 · 3b/2 + t)230As the two factors are coprime, we have that t− 2 · 3b/2 = 1 or 2 · 3b/2− t = 1. Plugging thisinto the original equation givespn = t2 − 4 · 3b = (2 · 3b/2 + 1)2 − 4 · 3b = 4 · 3b/2 + 1which according to [Luc02] cannot occur when n ≥ 5 and so we can assume that n = 1 inthis case orpn = 4 · 3b − t2 = 4 · 3b − (2 · 3b/2 − 1)2 = 4 · 3b/2 − 1.Now, if b/2 is even, we can repeat the above procedure to getpn = (2 · 3b/4 − 1)(2 · 3b/4 + 1)and thus 2 · 3b/4 − 1 = 1 so b = 0 which gives p = 3, a contradiction. Hence in the secondcase we must have that b/2 is odd. If b/2 < 5, then we see that4 · 3b/2 + 1 ∈ {13, 37, 109, 325}(of which only 325 is not prime) or in the case with b/2 odd4 · 3b/2 − 1 ∈ {11, 107}and all of these cases are in the list of finitely many primes. Next, suppose that b/2 ≥ 5. Ineither case above, via [BVY04], we have that solutions of the formpn + 4 · 3b/2(−1)n = (±1)3correspond to newforms at level 6 which cannot occur. A summary of this result can befound in [Coh07b, p.526-527]. Hence, this case only has solutions when n = 1 or p is in thefinitely many primes above. We can verify that these situations are covered by the finitelymany primes listed above and families (1) and (2).Case 3: t2 = pn − 4 · 3b. This is case 13 in [Mul06, p.62-67]. Here b is given to be odd.We can use Lemma 4.9 in [Mul06, p.165] to see that the solutions with b ≥ 0 arep b t nt2 + 4 · 3b ≥ 0 t 15 0 11 313 5 35 3n√t2 + 4 · 3b ≥ 0 t Pmin(n) ≥ 7231According to [Luc02], we know that in the final case n < 5 and so this case only hassolutions when n = 1 or p = 5 or 13. We can verify that these situations are covered by thefinitely many primes listed above and family (5).Case 4: t2 = pn + δ2a3b. This is a combination of cases 3,5, and 14 in [Mul06, p.62-67].We can use Lemma 4.9 in [Mul06, p.165] to see that the solutions with a ∈ {4.5} and b ≥ 0arep a b t n δ t2 − 2a · 3b ≥ 2 ≥ 0 t 1 1 12a−2 · 3b − 1 ≥ 3 ≥ 0 2a−1 + p 2 1 13b − 2a−2 ≥ 3 ≥ 0 2a−1 − p 2 1 12a−2 − 3b ≥ 5 ≥ 1 p+ 2 2 1 1n√t2 − 2a · 3b ≤ 5 ≥ 0 t Pmin(n) ≥ 7 1 1t2 + 2a · 3b ≥ 1 ≥ 0 t 1 −1 12a−2 · 3b + 1 ≥ 3 ≥ 0 2a−1 + p 2 −1 13b + 2a−2 ≥ 3 ≥ 0 2a−1 − p 2 −1 173 4 7 595 3 −1 1193 4 4 2681 3 −1 11153 5 5 39151 3 −1 15 5 1 23 4 −1 1n√t2 + 2a · 3b ≤ 5 ≥ 0 t Pmin(n) ≥ 7 −1 12a · 3b − t2 ≥ 1 ≥ 0 t 1 1 −173 5 11 2359 3 1 −1n√2a · 3b − t2 ≤ 5 ≥ 0 t Pmin(n) ≥ 7 1 −1As before, we can use [Luc02], we know that n < 5 when pn = t2 + 2a3b. Also, the casewhere p = 2a−2 − 3b when a = 4 or 5 only gives the case p = 3, 5, or 7 and so this can besimplified. Summarizing, this case only has solutions when n = 1, n = 2, Pmin(n) ≥ 7 orp = 5, 7, 73, 193, or 1153, we can verify that these situations are covered by the finitely manyprimes listed above and families (1), (2), (3), (4), (5), (6), and (7).Case 5: t2 = 3b + δ2apn. This is a combination of cases 6 and 12 in [Mul06, p.62-67].We can use Lemma 4.8 in [Mul06, p.156-157] to see that the solutions with a ∈ {4, 5} andb ≥ 0 are232p a b t n δ3b/2+14 4 b/2 odd 3b/2 + 2 1 12a−2 ± 3b/2 ≥ 3 ≥ 0 even 2p∓ 3b/2 1 13b/2 − 2a−2 ≥ 3 ≥ 0 even ±(2p− 3b/2) 1 −1In the second case above, we only need the positive case as the negative case gives solutionsp = 5 and p = 7, both of which are in the first family. As this case only has solutions whenn = 1 or p = 5 or 7 under the simplification above, we can verify that these situations arecovered by the finitely many primes listed above and families (3), (4) and (8).Case 6: t2 = 4pn − 3b. This is case 7 in [Mul06, p.62-67]. We can use Lemma 4.8 in[Mul06, p.156-157] to see that the solutions with b ≥ 0 arep b t n Extra Informationt2+3b4 ≥ 0 odd t 1 p ≡ 1 (mod 3)3b+14 ≥ 0 odd 2p− 1 2 nonen√3b+14 ≥ 0 odd t Pmin(n) ≥ 7 p ≡ 1 (mod 3)As shown in [AAA02], the third case cannot occur for n ≥ 7. All of the situations aboveare covered in the family (8) and (9).Case 7: t2 = 2a+ δ3bpn. This is a combination of cases 9 and 10 in [Mul06, p.62-67]. Wecan use Lemma 4.10 in [Mul06, p.171-172] to see that the solutions with a ∈ {2, 4, 5} andb ≥ 1 (possible to assume since we covered b = 0 in case 4) arep a b t n δ3b ± 4 2 ≥ 1 p∓ 2 1 13b ± 23 4 ≥ 1 p∓ 4 1 123 − 3b 4 ≥ 1 ±(4− p) 1 −1The last case above only has positive solutions for b = 0 or b = 1 giving p = 7 or p = 5.As this case only has solutions when n = 1 and a = 2, 4 or p = 5, or 7, we can verify thatthese situations are covered by the finitely many primes listed above and families (3) and(4).Case 8: t2 = 3bpn − 2a. This is case 11 in [Mul06, p.62-67]. Now, we can assume thatb ≥ 1 since when b = 0, we can refer back to case 4. Notice that since b ≥ 1, we know that233a = 4 cannot happen by local considerations of the equation at 3. We can use Lemma 4.10in [Mul06, p.156-157] to see that the solutions with a ∈ {4, 5} and b ≥ 1 arep a b t nt2+2a3b 5 ≥ 1 t 12a−2+13b/25 ≥ 1 (even) 3b/2p− 2 267 5 3 8549 3n√t2+2a3b 5 ≥ 1 t Pmin(n) ≥ 7As this case only has solutions when n = 1, Pmin(n) ≥ 7 or p = 17, we can verify thatthese situations are covered by the finitely many primes listed above and family (11).Case 9: 3t2 = 2a + δpn. This is a combination of cases 15 and 16 in [Mul06, p.62-67].We can use Lemma 4.10 in [Mul06, p.174-175] to see that the solutions with a ∈ {4, 5} areof the form p = δ(3t2 − 2a). When δ = −1 then since we are only considering cases witha ∈ {4, 5}, we only get the primes p = 13, 29. This situation is covered by family (13).Case 10: 3t2 = pn − 2a. This is a combination of cases 18 and 19 in [Mul06, p.62-67].We can use Lemma 4.10 in [Mul06, p.174-175] to see that the solutions with a ∈ {2, 4, 5} areof the form p = 3t2 + 2a. This situation is covered by family (12).Case 11: 3t2 = 4pn − 1. This is case 17 in [Mul06, p.62-67]. We can use Lemma 4.10in [Mul06, p.174-175] to see that the solutions to this equation are of the form pn = 3t2 − 1with n = 1 or 2. This situation is covered by family (10).234Chapter 5Elliptic Curves With Rational TwoTorsion and Conductor 50p, 200p or400p Organized by PrimesIn this section, we continue our classification of elliptic curves with rational two torsionand given conductor. Before beginning this section, I would like to remind the reader of thenotes of Chapter 4. These same notes apply here. In this section I suppress the formal proofas well. These are done the same way as in the previous chapter except all of this work iscontained in this thesis.Theorem 5.0.13. Let p be a prime distinct from 2 and 5. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 50p provided that p satisfies at least one ofthe following where d > 0, `1 ≥ 5, `2 ≥ 7 and m ≥ 0 are integers:1. p ∈ {3, 7, 11, 13, 17, 31, 37, 41}2. p = 2`15m + 1.3. p = 2`15m − 1.4. p = 2`1+15m .5. p = 2`1 − 5m.6. p = 5m − 2`1.7. p = 2`2 + 5m.8. p = d2 − 2`25m.9. p = 2`25m − d2.23510. p = d2 + 2`25m.11. p = 5d2 − 2`2.12. p = 2`2 − 5d2.13. p = 5d2 + 2`2.Theorem 5.0.14. Let p be a prime distinct from 2 and 5. Then up to the finitely manyprimes inp ∈ {3, 7, 11, 13, 17, 31, 37, 41}the following gives a complete list of the elliptic curves with nontrivial rational two torsionof conductor 50p associated with the primes in the previous theorem:(1) The prime p has the form p = 2`−2 · 5m + 1 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 −2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆D1 5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2D2 −2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4(2) The prime p has the form p = 2`−2 · 5m− 1 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 −2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆B1 5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2B2 −2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4(3) The prime p has the form p = 2`−2+15m with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 −2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p236• In addition, replacing m above with 2m, then we get solutions via p = 2`−2+15(2m)/2givenby,a2 a4 ∆J1 5ψ(52mp2 − 2`) −2`−252 22`52m+6p2J2 −2 · 5ψ(52mp2 − 2`) 52m+2p2 −2`+654m+6p4(4) The prime p has the form p = 2`−2 − 5m with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆B1 5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2B2 −2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4• In addition, replacing m above with 2m, then we get solutions via p = 2`−2−5(2m)/2given by,a2 a4 ∆E1 5ψ(2`p+ 52m) 2`−252p 22`52m+6p2E2 −2 · 5ψ(2`p+ 52m) 52m+2 2`+654m+6p• In the additional case that `−2 can be written as ˆ`/2+1 with ˆ`even, we get solutionswith p = 2`/2+1 − 5m given bya2 a4 ∆I1 5ψ(2ˆ`− 5mp) 2ˆ`−252 −22ˆ`5m+6pI2 −2 · 5ψ(2ˆ`− 5mp) −5m+2p 2ˆ`+652m+6p2for this new `(5) The prime p has the form p = 5m − 2`−2 with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆B1 5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2B2 −2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 5(2m)/2−2`−2given bya2 a4 ∆G1 5ψ(52m − 2`p) −2`−252p 22`52m+6p2G2 −2 · 5ψ(52m − 2`p) 52m+2 −2`+654m+6p237• In the additional case that `−2 can be written as ˆ`/2+1 with ˆ`even, we get solutionswith p = 5m − 2ˆ`/2+1 given bya2 a4 ∆H1 5ψ(2ˆ` + 5mp) 2ˆ`−252 22ˆ`5m+6pH2 −2 · 5ψ(2ˆ` + 5mp) 5m+2p 2ˆ`+652m+6p2(6) The prime p has the form p = 2`−2 + 5m with ` ≥ 7 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆D1 5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2D2 −2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4• In addition, replacing m above with 2m, then we get solutions with p = 2`−2+5(2m)/2given bya2 a4 ∆E1 5ψ(2`p+ 52m) 2`−252p 22`52m+6p2E2 −2 · 5ψ(2`p+ 52m) 52m+2 2`+654m+6p• In the additional case that `−2 can be written as ˆ`/2+1 with ˆ`even, we get solutionswith p = 2ˆ`/2+1 + 5m given bya2 a4 ∆H1 5ψ(2ˆ` + 5mp) 2ˆ`−252 22ˆ`5m+6pH2 −2 · 5ψ(2ˆ` + 5mp) 5m+2p 2ˆ`+652m+6p2(7) The prime p has the form p = d2 − 2`5m with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆B1 5ψ(2`5m + p) 2`−25m+2 22`52m+6pB2 −2 · 5ψ(2`5m + p) 52p 2`+65m+6p2(8) The prime p has the form p = 2`5m − d2 with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves•a2 a4 ∆C1 5ψ(2`5m − p) 2`−25m+2 −22`52m+6pC2 −2 · 5ψ(2`5m − p) −52p 2`+65m+6p2(9) The prime p has the form p = d2 + 2`5m with ` ≥ 7, m ≥ 0 and d ∈ Z and correspondsto the following elliptic curves238•a2 a4 ∆D1 5ψ(p− 2`5m) −2`−25m+2 22`52m+6pD2 −2 · 5ψ(p− 2`5m) 52p −2`+65m+6p2(10) The prime p has the form p = 5d2 − 2` with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves•a2 a4 ∆M1 5s+1ψ(2`+p5 ) 2`−252s+1 22`56s+3pM2 −2 · 5s+1ψ(2`+p5 ) 52s+1p 2`+656s+3p2(11) The prime p has the form p = 2` − 5d2 with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves•a2 a4 ∆N1 5s+1ψ(2`−p5 ) 2`−252s+1 −22`56s+3pN2 −2 · 5s+1ψ(2`−p5 ) −52s+1p 2`+656s+3p2(12) The prime p has the form p = 5d2 + 2` with ` ≥ 7, m ≥ 0, d ∈ Z and s ∈ {0, 1} andcorresponds to the following elliptic curves•a2 a4 ∆O1 5s+1ψ(p−2`5 ) −2`−252s+1 22`56s+3pO2 −2 · 5s+1ψ(p−2`5 ) 52s+1p −2`+656s+3p2Theorem 5.0.15. Let p be a prime distinct from 2 and 5. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 200p provided that p satisfies at least one ofthe following where d > 0, `1 ∈ {2, 4, 5}, `2 ∈ {4, 5}, `3 ∈ {3, 4, 5} and m ≥ 0 are integers:1. p ∈ {3, 7, 11, 13, 17, 23, 31, 37, 41}.2. p = 2`2−2 · 5m + 1 with m ≥ 1.3. p = 2`2−2 · 5m − 1 with m ≥ 1.4. pn = 5m−14 with m ≥ 1 odd and n = 1 or Pmin(n) ≥ 7.5. pn =(5m−14)2with n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.6. pn = d2−5m4 with m odd and either n = 1 or Pmin(n) ≥ 7.7. pn = 5m−d24 with m odd and either n = 1 or Pmin(n) ≥ 7 or n even.8. pn = 5m−d22`2with m even and Pmin(n) ≥ 7.2399. pn = 2`2−2 + 5m with m ≥ 1 and either n = 1 or Pmin(n) ≥ 7.10. pn = 5m − 2`2−2 with m ≥ 1 and either n = 1 or both `2 = 4 and Pmin(n) ≥ 7.11. pn = (5m − 4)2 with m ≥ 1 and either n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.12. pn = d2 − 2`15m with n = 1 or Pmin(n) ≥ 7.13. pn = 2`15m − d2 with n = 1 or Pmin(n) ≥ 7.14. p = d2 + 2`15m.15. pn = d2+45m with m ≥ 1 odd and either n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.16. pn = d2+2`15m with m ≥ 1 and either n = 1 or Pmin(n) ≥ 7.17. pn = 5d2−14 with pn ≡ 1 (mod 4) and n = 1, n = 2, Pmin(n) ≥ 7 or both 2 ‖ n andPmin(n/2) ≥ 7.18. pn = 5d2 − 2`1 with either n = 1 or Pmin(n) ≥ 7.19. pn = 5d2 − 4 with either n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.20. pn = 5d2 + 2`2 with either n = 1 or Pmin(n) ≥ 7.21. pn = 5d2 + 4 with either n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.Theorem 5.0.16. Let p be a prime distinct from 2 and 5. Then up to the finitely manyprimes inp ∈ {3, 7, 11, 13, 17, 23, 31, 37, 41}the following gives a complete list of the elliptic curves with nontrivial rational two torsionof conductor 200p associated with the primes in the previous theorem:(1) The prime p has the form p = 2`−2 · 5m + 1 with ` ∈ {4, 5} and m > 0 and correspondsto the following elliptic curves•a2 a4 ∆A1 5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 −2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆G1 5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2G2 −2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4240(2) The prime p has the form p = 2`−2 · 5m − 1 with ` ∈ {4, 5} and m > 0 and correspondsto the following elliptic curves•a2 a4 ∆A1 5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 −2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆E1 5ψ(2`5m + pn) 2`−25m+2 22`52m+6p2E2 −2 · 5ψ(2`5m + pn) 52p2 2`+65m+6p4(3) The prime p has the form pn = 5m−14 with and m ≥ 1 odd and corresponds to the followingelliptic curves• Either n = 1 or Pmin(n) ≥ 7 and the curve is given bya2 a4 ∆A1 5ψ(2`5mpn + 1) 225m+2pn 2852m+6p2nA2 −2 · 5ψ(2`5mpn + 1) 52 2105m+6pn• n = 1 and the curve is given bya2 a4 ∆H1 −5ψ(4p2 + 5m) 52p2 245m+6p4H2 2 · 5ψ(4p2 + 5m) 5m+2 2852m+6p2• We have n = 1 and in addition, replacing m above with 2m, then we get solutionsvia p = 5(2m)/2−14 given by,a2 a4 ∆M1 5ψ(52m − 24pn) −2252p 2852m+6p2M2 −2 · 5ψ(52m − 24pn) 52m+2 −21054m+6p(4) The prime p has the form pn =(5m−14)2with and m ≥ 1 odd, n = 2 or 2 ‖ n andPmin(n/2) ≥ 7 and corresponds to the following elliptic curves•a2 a4 ∆H1 −5ψ(4pn + 5m) 52pn 245m+6p2nH2 2 · 5ψ(4pn + 5m) 5m+2 2852m+6pn(5) The prime p has the form pn = d2−5m4 with m ≥ 1 odd and d ∈ Z and corresponds to thefollowing elliptic curves241• Either n = 1 or Pmin(n) ≥ 7 and the curve is given bya2 a4 ∆H1 −5ψ(4pn + 5m) 52pn 245m+6p2nH2 2 · 5ψ(4pn + 5m) 5m+2 2852m+6pn(6) The prime p has the form pn = 5m−d24 with m ≥ 1 odd and d ∈ Z and corresponds to thefollowing elliptic curves• Either n = 1, Pmin(n) ≥ 7 or n is even, m is odd and the curve is given bya2 a4 ∆J1 −5ψ(5m − 4pn) −52pn 245m+6p2nJ2 2 · 5ψ(5m − 4pn) 5m+2 −2852m+6pn(7) The prime p has the form pn = 5m−d22` with ` ∈ {4, 5}, m even, Pmin(n) ≥ 7 and corre-sponds to the following elliptic curve•a2 a4 ∆M1 5ψ(5m − 2`pn) −2`−252pn 22`5m+6p2nM2 −2 · 5ψ(5m − 2`pn) 5m+2 −2`+652m+6pn(8) The prime p has the form pn = 2`−2+5m with and m ≥ 1 and corresponds to the followingelliptic curves• n = 1, ` ∈ {4, 5} anda2 a4 ∆G1 5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2G2 − · 2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4• n = 1 or Pmin(n) ≥ 7, ` ∈ {4, 5} with pn = 2`−2 + 5(2m)/2 anda2 a4 ∆K1 5ψ(2`pn + 5m) 2`−252pn 22`52m+6p2nK2 − · 2 · 5ψ(2`pn + 5m) 52m+2 2`+654m+6pn• ` = 4 so `− 2 = 2, n = 1 or Pmin(n) ≥ 7 anda2 a4 ∆N1 −5ψ(4 + 5mpn) 52 245m+6pnN2 2 · 5ψ(4 + 5mpn) 5m+2pn 2852m+6p2n• n = 1, ` = 5 so `− 2 = 3 = 4/2 + 1 and242a2 a4 ∆O1 5ψ(16 + 5mp) 2252 285m+6pO2 −2 · 5ψ(16 + 5mp) 5m+2p 21052m+6p2(9) The prime p has the form pn = 5m−2`−2 with and m ≥ 1 and corresponds to the followingelliptic curves• n = 1 and ` ∈ {4, 5} and the elliptic curve is given bya2 a4 ∆M1 5ψ(5m − 2`p) −2`−252p 22`5m+6p2M2 −2 · 5ψ(5m − 2`p) 5m+2 −2`+652m+6p• Either n = 1 or Pmin(n) ≥ 7 and ` = 4 so `− 2 = 2 = 2/2 + 1givinga2 a4 ∆N1 −5ψ(4 + 5mpn) 52 245m+6pnN2 2 · 5ψ(4 + 5mpn) 5m+2pn 2852m+6p2n• n = 1 with ` ∈ {4, 5} and the elliptic curve is given bya2 a4 ∆E1 5ψ(2`5m + p) 2`−25m+2 22`52m+6pE2 −2 · 5ψ(2`5m + p) 52p 2`+65m+6p2• n = 1, ` = 5 and so `− 2 = 3 = 4/2 + 1 anda2 a4 ∆O1 5ψ(16 + 5mp) 2252 285m+6pO2 −2 · 5ψ(16 + 5mp) 5m+2p 21052m+6p2(10) The prime p has the form pn = (5m − 4)2 with m ≥ 1 odd and either n = 2 or 2 ‖ n andPmin(n/2) ≥ 7 and corresponds to the following elliptic curvea2 a4 ∆E1 5ψ(245m + pn) 225m+2 2852m+6pnE2 −2 · 5ψ(245m + pn) 52pn 2105m+6p2n(11) The prime p has the form pn = d2 − 2` · 5m with m ≥ 0, d ∈ Z, ` ∈ {2, 4, 5} and eithern = 1 or Pmin(n) ≥ 7 and corresponds to the following elliptic curves• ` = 2 and243a2 a4 ∆B1 −5ψ(4 · 5m + pn) 5m+2 2452m+6pnB2 2 · 5ψ(4 · 5m + pn) 52pn 285m+6p2n• ` ∈ {4, 5} anda2 a4 ∆E1 5ψ(2`5m + pn) 2`−25m+2 22`52m+6pnE2 −2 · 5ψ(2`5m + pn) 52pn 2`+65m+6p2n(12) The prime p has the form pn = 2` · 5m − d2 with m ≥ 0, d ∈ Z, ` ∈ {2, 4, 5} and eithern = 1 or Pmin(n) ≥ 7 and corresponds to the following elliptic curves• ` = 2 anda2 a4 ∆C1 −5ψ(4 · 5m − pn) 5m+2 −2452m+6pnC2 2 · 5ψ(4 · 5m − pn) −52pn 285m+6p2n• ` ∈ {4, 5} anda2 a4 ∆F1 5ψ(2`5m − pn) 2`−25m+2 −22`52m+6pnF2 −2 · 5ψ(2`5m − pn) −52pn 2`+65m+6p2n(13) The prime p has the form p = d2 + 2` · 5m with m ≥ 0, d ∈ Z and ` ∈ {2, 4, 5} andcorresponds to the following elliptic curves• ` = 2 anda2 a4 ∆D1 −5ψ(p− 4 · 5m) −5m+2 2452m+6pD2 2 · 5ψ(p− 4 · 5m) 52p −285m+6p2• ` ∈ {4, 5} anda2 a4 ∆G1 5ψ(p− 2`5m) −2`−25m+2 22`52m+6pG2 −2 · 5ψ(p− 2`5m) 52p −2`+65m+6p2(14) The prime p has the form pn = d2+165m with m ≥ 0, d ∈ Z and either n = 1 or Pmin(n) ≥ 7and corresponds to the following elliptic curves244•a2 a4 ∆Q1 5ψ(5mpn − 16) −2252 285m+6pnQ2 −2 · 5ψ(5mpn − 16) 5m+2pn −21052m+6p2n(15) The prime p has the form pn = 5d2−14 with pn ≡ 1 (mod 4), s ∈ {0, 1}, one of n = 1,n = 2, Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥ 7 and corresponds to the following ellipticcurves•a2 a4 ∆R1 −5s+1ψ(4pn+15 ) 52s+1pn 2456s+3p2nR2 2 · 5s+1ψ(4pn+15 ) 52s+1 2856s+3pn(16) The prime p has the form pn = 5d2 − 2` with ` ∈ {2, 4, 5}, s ∈ {0, 1} and either n = 1or Pmin(n) ≥ 7 and corresponds to the following elliptic curves• ` = 2 anda2 a4 ∆V1 −5s+1ψ(4+pn5 ) 52s+1 2456s+3pnV2 2 · 5s+1ψ(4+pn5 ) 52s+1pn 2856s+3p2n• ` ∈ {4, 5} anda2 a4 ∆X1 5s+1ψ(2`+pn5 ) 2`−252s+1 22`56s+3pnX2 −2 · 5s+1ψ(2`+pn5 ) 52s+1pn 2`+656s+3p2n(17) The prime p has the form pn = 5d2 + 2` with ` ∈ {4, 5}, s ∈ {0, 1} and either n = 1 orPmin(n) ≥ 7 and corresponds to the following elliptic curves•a2 a4 ∆Z1 5s+1ψ(pn−2`5 ) −2`−252s+1 22`56s+3pnZ2 −2 · 5s+1ψ(pn−2`5 ) 52s+1pn −2`+656s+3p2nTheorem 5.0.17. Let p be a prime distinct from 2 and 5. Then there exists an elliptic curvewith nontrivial rational two torsion of conductor 400p provided that p satisfies at least oneof the following where d > 0, `1 ≥ 2, `2 ≥ 4, `3 = λ with λ = 2 or λ ≥ 4 and m ≥ 0 areintegers:1. p ∈ {3, 7, 11, 13, 17, 23, 31, 37, 41}.2. p = 2`15m + 1.2453. p = 2`15m − 1.4. p = 2`1+15m .5. pn = 5m−14 with m odd and n = 1 or Pmin(n) ≥ 7.6. pn =(5m−14)2with m ≥ 1 and n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.7. pn = d2−5m4 with m odd and either n = 1 or Pmin(n) ≥ 7.8. pn = 5m−d24 with m odd and either n = 1, Pmin(n) ≥ 7 or n even.9. pn = 5m−d22`3with `3 ∈ {4, 5}, m even and either n = 1 or Pmin(n) ≥ 7.10. pn = 2`1 + 5m with n = 1 or Pmin(n) ≥ 7.11. pn = 2`1 − 5m with n = 1 or Pmin(n) ≥ 7.12. p = 5m − 2`2 with n = 1 or both Pmin(n) ≥ 7 and `1 = 2.13. pn = (5m − 4)2 with n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.14. pn = d2 − 2`35m with either n = 1 or both Pmin(n) ≥ 7 and `3 ∈ {2, 4, 5}.15. pn = 2`35m − d2 with either n = 1 or both Pmin(n) ≥ 7 and `3 ∈ {2, 4, 5}.16. p = d2 + 2`35m.17. pn = d2+2`35m with m ≥ 1, n = 1 or Pmin(n) ≥ 7.18. pn = d2+45m with m ≥ 1 odd and either n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.19. pn = 5d2−14 with pn ≡ 1 (mod 4) and one of n = 1, n = 2, Pmin(n) ≥ 7 or 2 ‖ n andPmin(n/2) ≥ 7.20. pn = 5d2 − 2`3 with n = 1 or both Pmin(n) ≥ 7 and `3 ∈ {4, 5}.21. pn = 5d2 − 4 with n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.22. p = 2`3 − 5d2.23. pn = 5d2 + 2`3 with n = 1 or both Pmin(n) ≥ 7 and `3 ∈ {4, 5}.24. pn = 5d2 + 4 with n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7.246Theorem 5.0.18. Let p be a prime distinct from 2 and 5. Then up to the finitely manyprimes inp ∈ {3, 7, 11, 13, 17, 23, 31, 37, 41}the following gives a complete list of the elliptic curves with nontrivial rational two torsionof conductor 400p associated with the primes in the previous theorem:(1) The prime p has the form p = 2`−2 · 5m + 1 with ` ≥ 4 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 −5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆G1 −5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2G2 2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4(2) The prime p has the form p = 2`−2 · 5m− 1 with ` ≥ 4 and m ≥ 0 and corresponds to thefollowing elliptic curves•a2 a4 ∆A1 −5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p•a2 a4 ∆E1 −5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2E2 2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4(3) The prime p has the form p = 2L+15m with L,m ≥ 0 and corresponds to the followingelliptic curves• If L = `− 2 for ` ≥ 4 even, thena2 a4 ∆A1 −5ψ(2`5mp+ 1) 2`−25m+2p 22`52m+6p2A2 2 · 5ψ(2`5mp+ 1) 52 2`+65m+6p• If L = `/2 for ` ≥ 4 even and p = 2`/2+15(2m)/2givinga2 a4 ∆R1 −5ψ(52mpn − 2`) −2`−252 22`52m+6p2R2 2 · 5ψ(52mpn − 2`) 52m+2p2 −2`+654m+6p4247(4) The prime p has the form pn = 5m−14 with m > 0 odd and corresponds to the followingelliptic curves• Either n = 1 or Pmin(n) ≥ 7 anda2 a4 ∆A1 −5ψ(2`5mpn + 1) 225m+2pn 2852m+6p2nA2 2 · 5ψ(2`5mpn + 1) 52 2105m+6pn• n = 1, p ≡ 1 (mod 4) anda2 a4 ∆J1 5ψ(5m − 4p) −52p 245m+6p2J2 −2 · 5ψ(5m − 4p) 5m+2 −2852m+6p• n = 1 and we have a solution with p = 5(2m)/2−14 given bya2 a4 ∆M1 −5ψ(52m − 16p) −2252p 2852m+6p2M2 2 · 5ψ(52m − 16p) 52m+2 −21054m+6p(5) The prime p has the form pn =(5m−14)2with m ≥ 1 odd and either n = 2 or 2 ‖ n andPmin(n/2) ≥ 7 and corresponds to the following curve•a2 a4 ∆J1 5ψ(5m − 4pn) −52pn 245m+6p2nJ2 −2 · 5ψ(5m − 4pn) 5m+2 −2852m+6pn(6) The prime p has the form pn = d2−5m4 with m > 0 odd, n = 1 or Pmin(n) ≥ 7 andcorresponds to the following elliptic curves•a2 a4 ∆H1 5ψ(4pn + 5m) 52pn 245m+6p2nH2 −2 · 5ψ(4pn + 5m) 5m+2 2852m+6pn(7) The prime p has the form pn = 5m−d24 with m ≥ 0 and either n = 1 or both m is odd andPmin(n) ≥ 7 and corresponds to the following elliptic curve•a2 a4 ∆J1 5ψ(5m − 4pn) −52pn 245m+6p2nJ2 −2 · 5ψ(5m − 4pn) 5m+2 −2852m+6pn248(8) The prime p has the form pn = 5m−d22` with m ≥ 0, ` ∈ {4, 5} and Pmin(n) ≥ 7 andcorresponds to the following elliptic curve•a2 a4 ∆M1 −5ψ(5m − 2`pn) −2`−252pn 22`52m+6p2nM2 2 · 5ψ(5m − 2`pn) 52m+2 −2`+654m+6pn(9) The prime p has the form pn = 2L + 5m with L > 0, m ≥ 0 and corresponds to thefollowing elliptic curves• If L = `− 2, ` ≥ 4 and in this case pn = 2`−2 + 5(2m)/2 with n = 1 or Pmin(n) ≥ 7and ` ∈ {4, 5}, then we havea2 a4 ∆K1 −5ψ(2`pn + 52m) 2`−252pn 22`52m+6p2nK2 2 · 5ψ(2`pn + 52m) 52m+2 2`+654m+6pn• If L = `− 2, ` ≥ 4, and n = 1, thena2 a4 ∆G1 −5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2G2 2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4• If L = `/2 + 1 for ` ≥ 4 even and n = 1, thena2 a4 ∆P1 −5ψ(2` + 5mp) 2`−252 22`5m+6pP2 2 · 5ψ(2` + 5mp) 5m+2p 2`+652m+6p2• If L = 2 and n = 1 or Pmin(n) ≥ 7, then we havea2 a4 ∆N1 5ψ(4 + 5mpn) 52 245m+6pnN2 −2 · 5ψ(4 + 5mpn) 5m+2pn 2852m+6p2n(10) The prime p has the form pn = 2L − 5m with L > 0, m ≥ 0 and corresponds to thefollowing elliptic curves• If n = 1 and L = `− 2 ≥ 2 hold we havea2 a4 ∆E1 −5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2E2 2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4249• If L = `− 2, ` ≥ 4 and in this case pn = 2`−2 + 5(2m)/2 with n = 1 or Pmin(n) ≥ 7and ` ∈ {4, 5}, then we havea2 a4 ∆K1 −5ψ(2`pn + 52m) 2`−252pn 22`52m+6p2nK2 2 · 5ψ(2`pn + 52m) 52m+2 2`+654m+6pn• If n = 1 and L = `− 2 with ` ≥ 4, then we havea2 a4 ∆G1 −5ψ(p2 − 2`5m) −2`−25m+2 22`52m+6p2G2 2 · 5ψ(p2 − 2`5m) 52p2 −2`+65m+6p4• If n = 1 and L = `/2 + 1 for ` ≥ 4 even we havea2 a4 ∆Q1 −5ψ(2` − 5mp) 2`−252 −22`5m+6pQ2 2 · 5ψ(2` − 5mp) −5m+2p 2`+652m+6p2(11) The prime p has the form pn = 5m − 2L with L > 0, m ≥ 0 and corresponds to thefollowing elliptic curves• If L = `− 2, ` ≥ 4 and in this case p = 5(2m)/2 − 2`−2, then we havea2 a4 ∆M1 −5ψ(52m − 2`p) −2`−252p 22`52m+6p2M2 2 · 5ψ(52m − 2`p) 52m+2 −2`+654m+6p• If n = 1 and L = `− 2 ≥ 2 hold we havea2 a4 ∆E1 −5ψ(2`5m + p2) 2`−25m+2 22`52m+6p2E2 2 · 5ψ(2`5m + p2) 52p2 2`+65m+6p4• If L = `/2 + 1 for ` ≥ 4 even with n = 1, we havea2 a4 ∆P1 −5ψ(2` + 5mp) 2`−252 22`5m+6pP2 2 · 5ψ(2` + 5mp) 5m+2p 2`+652m+6p2• If L = 2 and n = 1 or Pmin(n) ≥ 7, then we havea2 a4 ∆N1 5ψ(4 + 5mpn) 52 245m+6pnN2 −2 · 5ψ(4 + 5mpn) 5m+2pn 2852m+6p2n250(12) The prime p has the form pn = (5m − 4)2 with m ≥ 0 odd, n = 2 or 2 ‖ n andPmin(n/2) ≥ 7 and corresponds to the following elliptic curvea2 a4 ∆E1 −5ψ(16 · 5m + pn) 225m+2 2852m+6pnE2 2 · 5ψ(16 · 5m + pn) 52pn 2105m+6p2n(13) The prime p has the form pn = d2−2`5m with ` = 2 or ` ≥ 4 and m ≥ 0 and correspondsto the following elliptic curves• Either n = 1 and ` ≥ 4 or Pmin(n) ≥ 7 and ` ∈ {4, 5} anda2 a4 ∆E1 −5ψ(2`5m + pn) 2`−25m+2 22`52m+6pnE2 2 · 5ψ(2`5m + pn) 52pn 2`+65m+6p2n• ` = 2 and either n = 1 or Pmin(n) ≥ 7 and the elliptic curve is given bya2 a4 ∆B1 5ψ(4 · 5m + pn) 5m+2 2452m+6pnB2 −2 · 5ψ(4 · 5m + pn) 52pn 285m+6p2n(14) The prime p has the form pn = 2`5m−d2 with ` = 2 or ` ≥ 4 and m ≥ 0 and correspondsto the following elliptic curves• Either n = 1 or Pmin(n) ≥ 7, ` = 2 and p ≡ 3 (mod 8)a2 a4 ∆C1 5ψ(4 · 5m − pn) 5m+2 −2452m+6pnC2 −2 · 5ψ(4 · 5m − pn) −52pn 285m+6p2n• Either n = 1 with ` ≥ 4 or Pmin(n) ≥ 7 with ` ∈ {4, 5}, p ≡ 7 (mod 8) and theelliptic curve is given bya2 a4 ∆F1 −5ψ(2`5m − pn) 2`−25m+2 −22`52m+6pnF2 2 · 5ψ(2`5m − pn) −52pn 2`+65m+6p2n(15) The prime p has the form p = d2 + 2`5m with ` = 2 or ` ≥ 4 and m ≥ 0 and correspondsto the following elliptic curves• Either m = 0 and ` = 2 or ` ≥ 4 and251a2 a4 ∆G1 −5ψ(p− 2`5m) −2`−25m+2 22`52m+6pG2 2 · 5ψ(p− 2`5m) 52p −2`+65m+6p2(16) The prime p has the form pn = d2+2`5m with ` = 2 or ` ≥ 4 and m ≥ 0 and corresponds tothe following elliptic curves• ` = 2 and either n = 1, n = 2, Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥ 7 anda2 a4 ∆O1 5ψ(5mpn − 4) −52 245m+6pnO2 −2 · 5ψ(5mpn − 4) 5m+2pn −21052m+6p2n• Either n = 1 with ` ≥ 4 or Pmin(n) ≥ 7 with ` = 4 anda2 a4 ∆R1 −5ψ(5mpn − 2`) −2`−252 22`5m+6pnR2 2 · 5ψ(5mpn − 2`) 5m+2pn −2`+652m+6p2n(17) The prime p has the form pn = 5d2−14 with pn ≡ 1 (mod 4) and either n = 1, n = 2,Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥ 7 and corresponds to the following elliptic curve•a2 a4 ∆S1 5s+1ψ(4pn+15 ) 52s+1pn 2456s+3p2nS2 −2 · 5s+1ψ(4pn+15 ) 52s+1 2856s+3pn(18) The prime p has the form pn = 5d2 − 2` with either ` = 2 or ` ≥ 4, s ∈ {0, 1} andcorresponds to the following elliptic curves• We have ` = 2 with either n = 1, n = 2, Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥ 7givinga2 a4 ∆W1 5s+1ψ(4+pn5 ) 52s+1 2456s+3pnW2 −2 · 5s+1ψ(4+pn5 ) 52s+1pn 2856s+3p2n• Either n = 1 and ` ≥ 4 or both ` ∈ {4, 5} and Pmin(n) ≥ 7.a2 a4 ∆Y1 −5s+1ψ(2`+pn5 ) 2`−252s+1 22`56s+3pnY2 2 · 5s+1ψ(2`+pn5 ) 52s+1pn 2`+656s+3p2n252(19) The prime p has the form p = 2` − 5d2 with ` ≥ 4, s ∈ {0, 1} and corresponds to thefollowing elliptic curves•a2 a4 ∆Z1 −5s+1ψ(2`−pn5 ) 2`−252s+1 −22`56s+3pZ2 2 · 5s+1ψ(2`−pn5 ) −52s+1p 2`+656s+3p2(20) The prime p has the form pn = 5d2 + 2` with either ` = 2 or ` ≥ 4, s ∈ {0, 1} andcorresponds to the following elliptic curves• We have ` = 2 with either n = 1, n = 2, Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥ 7givinga2 a4 ∆X1 −5s+1ψ(pn−45 ) −52s+1 2456s+3pnX2 2 · 5s+1ψ(pn−45 ) 52s+1pn −2856s+3p2n• Either n = 1 and ` ≥ 4 or both Pmin(n) ≥ 7 and ` ∈ {4, 5} givinga2 a4 ∆AA1 −5s+1ψ(pn−2`5 ) −2`−252s+1 22`56s+3pnAA2 2 · 5s+1ψ(pn−2`5 ) 52s+1pn −2`+656s+3p2n253Chapter 6On the Diophantine Equationx5 + y5 = pαznIn this section, we discuss the results on the equation x5 +y5 = pαzn with n, p ≥ 7 primes,α ≥ 1 an integer and (x, y, z) a nontrivial solution. Throughout let S5 be the set of primesp′ ≥ 7 such that there is an elliptic curve E/Q with conductor NE ∈ {50p′, 200p′, 400p′} andat least one nontrivial rational 2-torsion point. The set S5 contains the primes between 7and 41 with the first exception being 43. Denote by C5(p) = p13p.Let p = 3 or p ≥ 7 be a prime and suppose n ≥ C5(p) is prime. Suppose that (a, b, c) isa proper nontrivial solution to x5 + y5 = pαzn, that is,a5 + b5 = pαcnwith a, b, c pairwise coprime. Suppose further that p - c, ac is even and that b is odd (possiblesince at least one of a or b is odd). Further, suppose that b ≡ −1 (mod 4) if c is even andb ≡ 1 (mod 4) if c is odd (to keep consistent with the q = 3 case). Following [BD10] via[Kra99]1, we consider the Frey-Hellegouarch curveE5,a,b : y2 = x3 − 5(a2 + b2)x2 + 5(a5 + b5a+ b)x.Setting R := rad10p(c), we can observe through the use of Tate’s Algorithm [Bil07, p.174] or1We have changed the Frey-Hellegouarch curve slightly from this reference inserting a negative coefficientfor x2254more generally using [Mul06, p.13-16] that the conductor of the curve isN(E5,a,b) =50pR if c is even, b ≡ −1 (mod 4)200pR if c is odd, v2(a) ≥ 2 and b ≡ 1 (mod 4)400pR if c is odd, v2(a) = 1 and b ≡ 1 (mod 4)where v2 denotes the usual 2-adic valuation. The standard invariants are given by [Bil07,p.165]c4 = 24 · 5(5(a2 + b2)2 − 3a5 + b5a+ b)c6 = 25 · 52(a2 + b2)(2 · 5(a2 + b2)2 − 32a5 + b5a+ b)∆ = 24 · 53(a+ b)2(a5 + b5)2 = 24 · 53(a+ b)2p2αc2nj(E5,a,b) : =c34∆=256(5(a2 + b2)2 − 3a5+b5a+b)3(a+ b)2p2αc2n.Since n ≥ C5(p) ≥ 17 and j(E5,a,b) /∈ Z[12 ], we may apply the result from Mazur (Theorem1.4.7) to see that E5,a,b does not have any p-isogenies. Thus we can apply Ribet’s LevelLowering Theorem. In the case where n | α, we get a modular form at level NE5,a,b/(Rp)(there is an extra power of n in the discriminant given by p) and when n - α, we get amodular form of level NE5,a,b/R.For now, assume that n | α. Then level lowering gives us a newform f at level 50, 200 or400. All the newforms at these levels are rational and hence correspond to elliptic curves. Forthe newforms whose corresponding elliptic curves do not have rational two torsion (curves50a1, 50b1, 200a1, 200e1, 400b1, 400c1, 400g1, 400h1), we can use Theorem 1.4.12 and seethat the B3 values for these curves are given by{45,−45, 105,−105, 45,−45,−105, 105}respectively. Thus so long as p ≥ 11, we get a contradiction for these curves. For the other7 curves at levels 200 and 400, we can use Theorem 1.4.4 to see that n | p + 1 ± ap whichcombining with the Hasse bound gives n ≤ p+ 1 + 2√p contradicting that n ≥ C5(p).Hence we assume from now on that n - α so that we get a modular form of level NE5,a,b/R.Now, we can prove our theorem.Theorem 6.0.19. Suppose that p ≥ 7 and that p /∈ S5. Let α ≥ 1 be an integer. Then the255equationx5 + y5 = pαznhas no solutions in coprime nonzero integers x, y, z and prime n satisfying n ≥ p13pProof. Let g+0 (N) = dimC(Γ0(N)). Suppose that f is a corresponding normalized newform ofconductor {50p, 200p, 400p} to Ea,b which corresponds to a solution of x5 +y5 = pαzn. Let Kfbe the associated number field corresponding to the Fourier coefficients of the newform f . Bylemme 1 of [Kra97], there exists a prime ` satisfying in all cases ` ≤ [SL2(Z) : Γ0(24 ·52 ·p)] =120 · (p + 1) with a`(f) /∈ Z. Since f is normalized, the Fourier coefficients a2, a5 and ap alllive in {0,±1} and hence ` /∈ {2, 5, p}. Notice that a`(Ea,b) is a rational integer satisfyingthe Hasse bound, namely |a`(Ea,b)| ≤ 2√` and that for any embedding σ : Kf → R, we havethat |σ(a`(f))| ≤ 2√`. We can apply the the results from Kraus and Oesterle´ from Theorem1.4.4:n | NormKf/Q(a`(f)− a`(Ea,b))n | NormKf/Q(a`(f)− (`+ 1))holding for each prime ` distinct from n and dividing R to obtainn ≤ (`+ 1 + 2√`)[Kf :Q] ≤ (√`+ 1)2g+0 (N).Next, we apply theorem 1 of [Mar05] and see that 2g+0 (N) ≤19p2 . Hence, we obtainn ≤ (√120(p+ 1) + 1)19p/2 < p13pholding for2 p ≥ 43. This is a contradiction.Now, suppose that all the Fourier coefficients are integers. This means that our f corre-sponds to an isogeny class of elliptic curves over Q with conductor N ∈ {50p, 200p, 400p}.Applying Pproposition 2 of Appendice II of [Kra97], we see that we must have one of thetwo following claims.1. There exists a prime ` ≤ 120(p+ 1) coprime to 10p with a`(f) ≡ 1 (mod 2).2. We have a`(f) ≡ 0 (mod 2) for all primes ` coprime to 10p.In the first case, as before sincen | NormKf/Q(a`(f)− a`(Ea,b)) = a`(f)− a`(Ea,b) ≤ `+ 1 + 2√`2The value p13p can be improved at the sacrifice of the lower bound being increased. However note thatthe smallest prime not in S5 is p = 43.256we have thatn ≤ `+ 1 + 2√` ≤ 120(p+ 1) + 1 + 4√30(p+ 1) < p13pwhere the last inequality holds for all p ≥ 23. This is a contradiction.Now for a small lemma.Lemma 6.0.20. If E/Q is an elliptic curve with a`(E) ≡ 0 (mod 2) for all primes ` coprimeto the conductor, then E has rational 2 torsion.Proof. Let E : y2 = p(x) for p(x) ∈ Z[x] a degree 3 monic polynomial. Since we know thata`(E) ≡ 0 (mod 2) for all primes ` coprime to the conductor, we have that the polynomialp(x) must have a linear factor for all but finitely many primes. If p(x) were irreducible, thenthe Galois group Gal(L/Q) for the splitting field L must contain a 3 cycle. The ChebotarevDensity Theorem states that there must exist infinitely many primes that also have thesame cycle structure as elements of the Galois group. Thus, for infinitely many primes, thepolynomial must be irreducible which is a contradiction. Hence p(x) is reducible and sinceit has degree 3 it must contain a rational linear factor by the Rational Root Theorem. ThusE has a rational 2 torsion point. Applying this in our situation, this gives us that Ef is an elliptic curve over Q with rational2-torsion and conductor 50p, 200p or 400p. Thus p ∈ S5, contradicting our assumptions onp.To finish this section off, I now show that the primes not in S5 comprise most of theprimes. That is to say,Theorem 6.0.21. We havepiS5 := #{p ≤ x : p ∈ S5} √x log2(x)}Proof. We proceed in a similar fashion to [BLM11]. First, suppose that n > 3 for the primesin the prime families in Theorems 5.0.13, 5.0.15, 5.0.17 that allow for n > 3. In this case,we can use [ST86, p.180] to see that the largest prime divisor of Apn + Bym gets large as ntends to infinity (where m > 1 and gcd(p, y) = 1). In our setting, the largest prime divisoris always at worst 5 and so n must be bounded by an absolute constant. Using [DG95, p,4Theorem 2], we see that the number of solutions is finite for p, n, `,m, d.We have excluded the case when n = 3 by using elliptic curves. For n = 2, the cases inwhich p2 = (5m − 4)2 or p2 =(5m−14)2reduce to the cases where n = 1. Otherwise, up tothe constant term which is okay since it only changes the solutions by its prime factors, we257get Pell equations whose solutions grow exponentially and so there are only asymptoticallylog(x) of these solutions.Thus from now on we suppose that n = 1. For most of the prime families in Theorems5.0.13, 5.0.15, 5.0.17, the work done in [BLM11] was independent of the number 3. Thus,we reduce our list to the consideration of the prime families given by the following. Unlessotherwise stated, m and ` are arbitrary non negative integers in the following list:•p =2` + 15m(6.1)•p =5m − 14(6.2)• m is odd andp =|d2 − 5m|4(6.3)•p =d2 + 2`5m(6.4)•p =5d2 − 14. (6.5)Equations 6.1 and 6.2 are covered by 6.4 and 6.5 respectively leaving only the last 3 casesabove. Throughout suppose that p ≤ x and that d is positive (allowing d negative doublesthe total possible values of d). Beginning with the last case, notice that5d2 − 14≤5d24≤ x+ 1.Thus the total number possible values of d is O(√x), let alone primes.For equation 6.3, write m = 2m0 + 1. Factoring gives us that|d− 5m0√5||d+ 5m0√5| = 4p ≤ 4x.Hence as d > 0, ∣∣∣∣√5−d5m0∣∣∣∣ =4p5m0(d+√55m0)≤4x52m0.Next, we use a theorem of Ridout [Rid57, p.125]3 which can be thought of as a p-adic analogue3In the paper, we set µ = 1, ν = 0, κ = 1 + , Q1 = 5258to Roth’s theorem, to see that there exists a constant depending only on such that∣∣∣∣√5−d5m0∣∣∣∣ >C()5m0(1+).We set = 1/2 and combine the previous two displayed equations to see5m0/2 x.Hence m log(x). For each fixed value of m, notice that there are O(√x) values for d.Thus the number of primes up to x is O(√x log x).For the last case, the case of p = d2+2`5m , we argue similarly to [BLM11] and the caset2+2a3b .To do this, we begin by noting that modulo 5 considerations give us that d2 ≡ −2` and so `must be even otherwise the right hand side is not a quadratic residue modulo 5.Next, since d2 + 2` ≡ 0 (mod 5) has two simple roots, we lift using Hensel’s lemma to seethat d2 + 2` ≡ 0 (mod 5m) has only two solutions. Let d1 and d2 be the least positive suchsolutions. Then d = dj + 5mM for some integer M ≥ 0 and j ∈ {1, 2}. Thend2 + 2`5m=d2j + 2`5m+ 2djM + 5mM2 ≤ xHence 5mM2 < x hence M < 5−m/2√x. Thus, the number of positive integer values possiblefor d is2( √x5m/2+ 1)where we double the above to account for d1 and d2.Also notice that2` ≤ d2 + 2` = 5mp ≤ 5mx.Hence ` m + log x. Thus with the above it suffices to show that m log x for thencombined with the estimates for the number of admissible values for d gives us an upperbound for O(√x log x).Assume towards a contradiction that m > κ log x where κ is some large positive constant.Recalling that ` is even and that Z[i] has class number 1, we have5mp = d2 + 2` = (d+ 2`/2i)(d− 2`/2i).259Thus, we haved+ 2`/2i = αmp1d− 2`/2i = αmp1where α is one of 1± 2i and p1 = u+ vi is such that u2 + v2 = p. Eliminating for d yields2`/2+1i = αmp1 − αmp1.Now the 2-adic valuation of the left hand side above is `/2 + 1 and the right hand side has2-adic valuation ofv2 (αmp1 − αmp1) = v2((αα)m p1p1− 1)valid since α and p1 have odd norms. Using [BL96, The´ore´m 4] a 2-adic linear forms inlogarithms estimate, we see that there exists a constant C such thatv2((αα)m p1p1− 1)< C log2(mlog x)log x.Thus, we may conclude that` < `+12< 2C log2(mlog x)log x.Assume towards a contradiction that ` > (log κ)−1m. It follows thatmlog xlog2(mlog x) < 2C log κ.and recalling that we assumed that m > κ log x and that x/ log2 x is an increasing function,we haveκ < 2C log3 κ.which is a contradiction if κ is large enough. Hence, we have that ` ≤ (log κ)−1m. This givesus that2`/2 ≤ 2m/(2 log κ) ≤ 5m/4 (6.6)following since 2 log 2log 5 < log κ for sufficiently large κ. We now apply a generalization of theSchmidt Subspace Theorem due to Schlickewei [Sch77], [Sch91, p.177 Theorem 1D] to theequation2`/2+1i = αmp1 − αmp1.Let K = Q[i] and take a set of valuations to be S = {α, α,∞}. Let x = (x1, x2). For j = 1, 2260and ν ∈ S, takeLj,ν(x) =x1 − x2 if (j, ν) = (2,∞)xj otherwise.Next, for x = (αmp1, αmp1), we show thatA :=∏(j,ν)∈{1,2}×S}|Lj,ν(x)|ν 1(max{|x1, |x2|})1/8. (6.7)First, compute that∏ν∈S|L1,ν(x)|ν = |αmp1|α|αmp1|α|αmp1|= α−mα−m|αmp1| = |p1|.Further,∏ν∈S\{∞}|L2,ν(x)|ν = |αmp1|α|αmp1|α = |α−m| = 5−m/2.Lastly,|L2,∞(x)| = |x1 − x2| = |αmp1 − αmp1| = 2`/2+1.Hence from equations 6.6, the product in equation 6.7 becomesA ≤|p1|2`/2+15m/2≤2√x2`/25m/2≤2√x5m/4≤ 5−m/8.where the last inequality holds since m > κ log x and so x < em/κ giving 2√x < 2em/(2κ) ≤5m/8 provided κ is large enough. This also gives that x ≤ 5m/2 and hence,|αmp1| = |αmp1| ≤ 5m/2x ≤ 5m.Combining the above shows that equation 6.7 holds. The Generalized Schmidt SubspaceTheorem [Sch91, p.177 Theorem 1D] asserts that there exists finitely many pairs (ck, dk) ∈K2\{(0, 0)} for 1 ≤ k ≤ s such that equation 6.6 satisfies ckx1 = dkx2. Without loss ofgenerality, we may assume that ck and dk are coprime. For a fixed k, this implies thatckαmp1 = ckx1 = dkx2 = dkαmp1which shows that αm | dkαmp1. Since α and α are coprime, we get that αm | dkp1. Since261|p1| ≤√p ≤√x, choosing κ ≥ max1≤k≤s |dk| shows that αm ≤ κ√x. This will contradictthat m > κ log x. Hence m < κ log x262Chapter 7Strengthening Results on theDiophantine Equation xq + yq = pαznIn this section, we will discuss solutions to xq+yq = pαzn for primes not considered aboveor in [BLM11]. LetS3,p = {18p, 36p, 72p} C3(p) = p2pS5,p = {50p, 200p, 400p} C5(p) = p13pand let Sq be the set of primes p ≥ 5 (or p ≥ 7 if q = 5) for which there exists an ellipticcurve E with conductor NE ∈ Sq,p with at least one non-trivial rational two torsion point.The trick to strengthening the theorem above and the results from [BLM11] is to exploitmore information contained in our Frey-Hellegouarch curve other than the fact that it hasa non-trivial rational two torsion point. Recall that for the curves xq + yq = pαzn withq ∈ {3, 5}, we have the following associated Frey-Hellegouarch curves given byE3,a,b : y2 = (x+ b− a)(x2 + (a− b)x+ (a2 + ab+ b2)) =: f3(x).E5,a,b : y2 = x3 − 5(a2 + b2)x2 + 5(a5 + b5a+ b)x= x(x2 − 5(a2 + b2)x+ 5(a5 + b5a+ b))=: f5(x).where (a, b, c) is a solution to xq +yq = pαzn. As notation, we denote by ∆E the discriminantof the elliptic curve, ∆E,3 the elliptic curve of E3,a,b and ∆E,5 the elliptic curve of E5,a,b andsimilarly for ∆E,min to denote the minimal discriminant and NE to denote the conductor forthe elliptic curve E. We make the assumption that if one of a or b is even, we assume withoutloss of generality that a is even. This only occurs when c is odd. For the above, we have by263[BLM11] and [Bil07] that the invariants are∆E,3,min = −2433p2αc2n,NE,3 =18p · rad6p(c) if c is even, b ≡ −1 (mod 4)36p · rad6p(c) if c is odd, v2(a) ≥ 2 and b ≡ 1 (mod 4)72p · rad6p(c) if c is odd, v2(a) = 1 and b ≡ 1 (mod 4),∆E,5,min = 2453(a+ b)2p2αc2nNE,5, =50p · rad10p(c) if c is even, b ≡ −1 (mod 4)200p · rad10p(c) if c is odd, v2(a) ≥ 2 and b ≡ 1 (mod 4)400p · rad10p(c) if c is odd, v2(a) = 1 and b ≡ 1 (mod 4).For curves with a rational two torsion point say y2 = x3 + a2x2 + a4x, let ∆q de-note the discriminant of the quadratic polynomial x2 + a2x + a4. In the above, let ∆Q,3be the discriminant of x2 + (a − b)x + (a2 + ab + b2) and let ∆Q,5 be the discriminant of(x2 − 5(a2 + b2)x+ 5(a5+b5a+b)).Notice that ∆Q,3 = −3(a + b)2 and that ∆Q,5 = 5(a + b)2. Thus, f3(x) splits completelyover F`′ , where `′ is prime, whenever(−3`′)= 1 and f5(x) splits completely over F`′ whenever(5`′)= 1. The first case occurs whenever `′ ≡ 1 (mod 6) and the second occurs whenever`′ ≡ ±1 (mod 5). Modulo these primes `′, we see that our elliptic curve does not havefull rational two torsion but modulo the primes `′ is pretending to have full rational twotorsion. Mathematically, this means that while #Etor(Q) 6≡ 0 (mod 4) we do have that#Etor(F`′) ≡ 0 (mod 4) for E being one of E3,a,b or E5,a,b as appropriate. Via Ribet’s LevelLowering Theorem as in Theorem 1.4.6 (see [BLM11] and [Bil07] for the details) we see thatthere is a modular form f of level N whereN =18p if q = 3, c is even, b ≡ −1 (mod 4)36p if q = 3, c is odd, v2(a) ≥ 2 and b ≡ 1 (mod 4)72p if q = 3, c is odd, v2(a) = 1 and b ≡ 1 (mod 4)50p if q = 5, c is even, b ≡ −1 (mod 4)200p if q = 5, c is odd, v2(a) ≥ 2 and b ≡ 1 (mod 4)400p if q = 5, c is odd, v2(a) = 1 and b ≡ 1 (mod 4)such that a`′(f) ≡ a`′(E) (mod p). Let’s further assume that this f is a rational newform withnontrivial 2 torsion and a`′(f) 6≡ `+ 1 (mod 4). Then we have that a`′(E) 6≡ a`′(f) (mod 4).Since at some prime `′ as mentioned above we get that a`′(f) 6= a`′(E), we see via appendice264II of [Kra97] that the prime where this difference occurs is small when compared to p2.This tells us that for such prime p, the equation xq + yq = pαzn has no nontrivial coprimesolutions in integers (x, y, z). After a few more definitions, we can summarize the above workin a theorem.Definition 7.0.22. For `′ prime and q ∈ {3, 5}, let R(`′, q) be the condition((−1q )q`′)= 1.This is equivalent to(−3`′)= 1 and(5`′)= 1 for q = 3 and q = 5 respectively.Definition 7.0.23. Let Pg,q denote the set of primes p such that there are no elliptic curves Ewith conductor in the set Sq.p such that 4|#ETor(Q) but at least one curve E ′ with conductorin the same set such that 2|#E ′Tor(Q).Primes in Pg,3 less than 200 include53, 79, 83, 103, 149, 151, 157, 163, 167, 173, 181, 199.and primes in Pg,5 less that 200 include23, 47, 53, 71, 83, 97, 107, 137, 139, 149, 151, 173, 179, 181, 191, 193, 197.See the appendix for a longer list.Note: According to the definition, there needs to be only one curve with conductor in theset Sq. It could be, for example when q = 3, that you have a curve of conductor 18p withthe required property and no curves with conductor 36p or 72p with 2 | ETor(Q).Definition 7.0.24. Let Pb,q ⊆ Pg,q be the set of primes p such that p ∈ Pg,q and for everyelliptic curves E with conductor in the set Sq,p such that 2|#ETor(Q), there exists a prime`′ satisfying both R(`′, q) and a`′(E) 6≡ `′ + 1 (mod 4). Equivalently, there exists a prime `′satisfying both R(`′, q) and #ETor(F`′) 6≡ 0 (mod 4).Primes in Pb,3 less than 200 include153, 83, 149, 167, 173, 199.Primes in Pb,5 less that 200 include23, 53, 71, 73, 83, 97, 107, 137, 151, 173, 181, 191, 193, 197.Again for longer lists, see the appendix. We now have enough notation to state the mainresult of this section.1Note that in [BLM11] paper, the prime 53 listed above was omitted for Pb,3.265Theorem 7.0.25. Suppose that q ∈ {3, 5}. Suppose that either p ∈ Pb,q ⊆ Sq,p or thatp /∈ Sq,p. Then the equationxq + yq = pαznhas no solutions in nonzero, coprime integers x, y and z, integer α ≥ 1 and prime n ≥ Cq(p).The theorem above tells us that if we can give conditions for primes belonging to Pb,q,then we can show which primes satisfy the above theorem. First, we show which primes donot belong to Pg,q.Theorem 7.0.26. Let p be a prime, q ∈ {3.5} and let E : y2 = x3 + a2x2 + a4 be anelliptic curve with integer coefficients and nontrivial two torsion and conductor in the setSq,p. Suppose further that ∆E is a positive square. Then p /∈ Pg,q.Proof. Notice that the discriminant of x2+a2x+a4 is ∆Q = a22−4a4. Further, the discriminantof E is∆E = 16a24(a22 − 4a4) = (4a4)2∆QHence, if ∆E is a square, then ∆Q is also a square. Thus, the polynomial x2 + a2x+ a4 splitsover Z. This means thaty2 = x3 + a2x2 + a4x = x(x− α)(x− β)for some α, β ∈ Z. Hence the curve has full rational two torsion and so p /∈ Pg,q as required.Now, we display the types of primes that are not in Pg,3 and Pg,5. To form these lists, weuse Theorems 4.0.8, 4.0.10, 4.0.12, 5.0.14, 5.0.16 and 5.0.18 to figure out which cases have thepossibility of having an elliptic curve with square conductor. If there are any present, thenthese primes are not members of Pg,q and thus are included below. In all of the tables below,m ≥ 0 and d ∈ Z unless stated otherwise. The third column throughout we will abbreviateusing the letters TC to mean “Theorem Case”, the entry in the list in the associated theorem.266pn Conductor TC Extra Information2`−2 · 3m + 1 18p 1 n = 1, ` ≥ 72`−2 · 3m − 1 18p 2 n = 1, ` ≥ 73m + 2`−2 18p 3 n = 1, ` ≥ 73m − 2`−2 18p 4 n = 1, ` ≥ 72`−2 − 3m 18p 5 n = 1, ` ≥ 7d2 − 2`3m 18p 7 n = 1, ` ≥ 7 even, m even2`3m − d2 18p 8 n = 1, ` ≥ 7 even, m evend2+2L3m 18p 9 n = 1, L ≥ 52`−2 · 3m + 1 72p 1 n = 1, ` ∈ {4, 5}2`−2 · 3m − 1 72p 2 n = 1, ` ∈ {4, 5}3m + 2`−2 72p 3 n = 1, ` ∈ {4, 5}3m − 2`−2 72p 4 n = 1, ` ∈ {4, 5}d2 − 2`3m 72p 6 n = 1 or Pmin(n) ≥ 7, ` = 4, m even2`3m − d2 72p 7 n = 1 or Pmin(n) ≥ 7, ` = 4, m even3m+14 72p 8 n = 1Table 7.1: Primes not in Pg,3.267pn Conductor TC Extra Information2`−2 · 5m + 1 50p 1 n = 1, ` ≥ 72`−2 · 5m − 1 50p 2 n = 1, ` ≥ 72`−2+15m 50p 3 n = 1, ` ≥ 72`−2 − 5m 50p 4 n = 1, ` ≥ 75m − 2`−2 50p 5 n = 1, ` ≥ 72`−2 + 5m 50p 6 n = 1, ` ≥ 7d2 − 2`5m 50p 7 n = 1, ` ≥ 7, `,m even2`5m − d2 50p 8 n = 1, ` ≥ 7, `,m even2`−2 · 5m + 1 200p 1 n = 1, ` ∈ {4, 5}2`−2 · 5m − 1 200p 2 n = 1, ` ∈ {4, 5}5m−14 200p 3 n = 1 or Pmin(n) ≥ 7(5m−14)2200p 4 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m odd5m−d22` 200p 7 Pmin(n) ≥ 7, ` ∈ {2, 4, 5}, m even2`−2 + 5m 200p 8 n = 1 or Pmin(n) ≥ 7, ` ∈ {4, 5}5m − 2`−2 200p 9 n = 1 and ` ∈ {4, 5} or Pmin(n) ≥ 7 when ` = 4(5m − 4)2 200p 10 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m oddd2 − 2`5m 200p 11 n = 1 or Pmin(n) ≥ 7, ` ∈ {2, 4}, m even2`5m − d2 200p 12 n = 1 or Pmin(n) ≥ 7, ` ∈ {2, 4}, m even2`−2 · 5m + 1 400p 1 n = 1, ` ≥ 42`−2 · 5m − 1 400p 2 n = 1, ` ≥ 42L+15m 400p 3 n = 1, L ≥ 25m−14 400p 4 n = 1 or Pmin(n) ≥ 7, m odd5m−d24 400p 7 n = 1 and m even5m−d24 400p 8 Pmin(n) ≥ 7, ` ∈ {4, 5}2L + 5m 400p 9 Pmin(n) ≥ 7 and (L+ 2) ∈ {4, 5}, n = 1 and L ≥ 22L − 5m 400p 10 Pmin(n) ≥ 7 and (L+ 2) ∈ {4, 5}, n = 1 and L ≥ 25m − 2L 400p 11 Pmin(n) ≥ 7 and L = 2, n = 1 and L ≥ 2(5m − 4)2 400p 12 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m oddd2 − 2`5m 400p 13 Either n = 1 and ` ≥ 4 even or both Pmin(n) ≥ 7 and` ∈ {2, 4}, m even2`5m − d2 400p 14 Either n = 1 and ` ≥ 4 even or both Pmin(n) ≥ 7 and` ∈ {2, 4}, m evend2+45m 400p 16 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m evenTable 7.2: Primes not in Pg,5.At first, the goal to classify primes in Pg,q but not in Pb,q is much easier. In fact, we have268the following.Theorem 7.0.27. Let q ∈ {3, 5}. Let p ∈ Pg,q and let E/Q be an elliptic curve withconductor in Sq,p with rational two torsion. Suppose further that the discriminant of theelliptic curve is of the form∆E =(−3)m2 if q = 35m2 if q = 5for some integer m. Then p /∈ Pb,q.Proof. By shifting the rational two torsion point to (0, 0), the curve E can be seen to beisomorphic to a curve of the formy2 = x3 + a2x2 + a4x = x(x2 + a2x+ a4)so without loss of generality, suppose E is of this form. According to [Sil09, p.42], we havethat∆E = 16a24(a22 − 4a4) = (4a4)2∆Qwhere ∆Q is the discriminant of the quadratic polynomial x2 + a2x + a4. By assumption∆E = ∆Q,q and so in fact, we have that∆Q =∆E(4a4)2=(−3)n2 if q = 35n2 if q = 5where (4a4n)2 = m2. Let ∆Q,q,n represent the right most side of the above equality.Assume towards a contradiction that p ∈ Pb,q. Then there exists a prime2 `′ 6= p satisfyingR(`′, q) such that 4 - #ETor(F`′). However, in this case, we know that ∆Q = ∆Q,q,n. But forsuch `′, we have that the Legendre symbol evaluates to(∆Q,q,n`′)= 1 by definition of R(`′, q).Thus, ∆Q is a square in F`′ . Hence, the polynomial x2 + a2x+ a4 splits and thusy2 = x3 + a2x2 + a4x = x(x2 + a2x+ a4) = x(x− α)(x− β)where α, β ∈ F`′ . Hence, we have that the curve has full rational two torsion over F`′ and sowe have that 4|#ETor(F`′) which is a contradiction. Thus p /∈ Pb,q as was required. Now, we can list the families with discriminant of the form −3m2. Hence a prime p thatlives in these families and belongs to Pg,q does not belong to Pb,q. I summarize these familiesin the following table. As before, m ≥ 0 and d ∈ Z unless otherwise indicated.2we need to avoid the conductor for reduction purposes so ensure that `′ 6= p.269pn Conductor TC Extra Informationd2 + 2`3m 18p 6 n = 1, ` ≥ 7 even, m odd3d2 + 2` 18p 10 n = 1, ` ≥ 7 evend2+3m4 36p 4 n = 1, p ≡ −1 (mod 4), m odd3d2+14 36p 5 n ∈ {1, 2}, p ≡ 1 (mod 4)d2 + 2`3m 72p 5 n = 1, ` ∈ {2, 4}, m odd3m+d24 72p 9 n = 1, p ≡ 1 (mod 4), m odd3d2+14 72p 10 n ∈ {1, 2}3d2 + 2` 72p 13 n = 1, ` ∈ {2, 4}Table 7.3: Primes not in Pb,3 but in Pg,3.270pn Conductor TC Extra Informationd2 − 2`5m 50p 7 n = 1, ` ≥ 7 even, m odd2`5m − d2 50p 8 n = 1, ` ≥ 7 even, m odd5d2 − 2` 50p 10 n = 1, ` ≥ 7 even2` − 5d2 50p 11 n = 1, ` ≥ 7 evend2−5m4 200p 5 n = 1 or Pmin(n) ≥ 7, m odd5m−d24 200p 6 n = 1 or Pmin(n) ≥ 7, m odd5m−d22` 200p 7 Pmin(n) ≥ 7, ` ∈ {2, 4, 5} m oddd2 − 2`5m 200p 11 n = 1 or Pmin(n) ≥ 7, ` ∈ {2, 4}, m odd2`5m − d2 200p 12 n = 1 or Pmin(n) ≥ 7, ` ∈ {2, 4}, m odd5d2−14 200p 15 n = 1 or n = 2 or Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥7, pn ≡ 1 (mod 4)5d2 − 2` 200p 16 n = 1 or Pmin(n) ≥ 7 and ` ∈ {2, 4}(5m−14)2400p 5 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m oddd2−5m4 400p 6 n = 1 or Pmin(n) ≥ 7, m odd5m−d24 400p 7 n = 1 or Pmin(n) ≥ 7, m oddd2 − 2`5m 400p 13 Either n = 1 and ` ≥ 4 even or Pmin(n) ≥ 7 and ` ∈{2, 4}, m odd2`5m − d2 400p 14 Either n = 1 and ` ≥ 2 even or Pmin(n) ≥ 7 and ` ∈{2, 4}, m oddd2+45m 400p 16 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7, m odd5d2−14 400p 17 n = 1 or n = 2 or Pmin(n) ≥ 7 or 2 ‖ n and Pmin(n/2) ≥7, pn ≡ 1 (mod 4)5d2 − 2` 400p 18 One of n = 1 and ` ≥ 2 even, Pmin(n) ≥ 7 and ` ∈ {2, 4},n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7 and ` = 22` − 5d2 400p 19 n = 1, ` ≥ 4 even5d2 + 4 400p 20 n = 2 or 2 ‖ n and Pmin(n/2) ≥ 7Table 7.4: Primes not in Pb,5 but in Pg,5.This leaves the following primes. Below we let m ≥ 0, d > 0 be integers and s ∈ {0, 1}unless otherwise specified.271pn Conductor TC Extra Information a4 Legendre for a4 ∆ Legendre for ∆d2 + 2`3m 18p 6 n = 1,` ≥ 7, either ` odd orboth `,m even−2`−23m+2(−2·3`′)or(−2`′)or(−1`′)22`32m+6p(p`′)d2 − 2`3m 18p 7 n = 1,` ≥ 7, either ` odd orm odd2`−23m+2(2·3`′)or(2`′)or(3`′)22`32m+6p(p`′)2`3m − d2 18p 8 n = 1,` ≥ 7, either ` odd orm odd2`−23m+2(2·3`′)or(2`′)or(3`′)−22`32m+6p(−p`′)3d2 + 2` 18p 10 n = 1,` ≥ 7 odd −2`−232s+1(−2·3`′)22`36s+3p(3p`′)3d2 − 2` 18p 11 n = 1,` ≥ 7 2`−232s+1(3`′)or(2·3`′)22`36s+3p(3p`′)2` − 3d2 18p 12 n = 1,` ≥ 7 2`−232s+1(3`′)or(2·3`′)−22`36s+3p(−3p`′)d2 + 4 · 3m 36p 1 n = 1,m even −3m+2(−1`′)2432m+6p(p`′)d2 − 4 · 3m 36p 2 n = 1 or Pmin(n) ≥ 7,m odd −3m+2(3`′)2432m+6pn(p`′)4 · 3m − d2 36p 3 n = 1 or Pmin(n) ≥ 7,m odd −3m+2(3`′)−2432m+6pn(−p`′)3d2 − 4 36p 6 n = 1 32s+1(3`′)2436s+3p(3p`′)d2 + 2`3m 72p 5 n = 1, either both ` ∈ {2, 4}and m even or just ` = 5−2`−23m+2(−1`′)or(−2`′)or(−2·3`′)22`32m+6p(p`′)d2 − 2`3m 72p 6 n = 1 or Pmin(n) ≥ 7, eitherboth ` = 4 and m even orjust ` = 52`−23m+2(2`′)or(2·3`′)or(3`′)22`32m+6pn(p`′)2`3m − d2 72p 7 n = 1 or Pmin(n) ≥ 7, eitherboth ` = 4 and m even orjust ` = 52`−23m+2(2`′)or(2·3`′)or(3`′)−22`32m+6pn(−p`′)d2+323m 72p 11 n = 1 or Pmin(n) ≥ 7 −233m+2(−2`′)2103m+6pn(p`′)or(3p`′)3d2 − 2` 72p 12 n = 1, ` ∈ {4, 5} 2`−232s+1(2·3`′)or(3`′)22`36s+3p(3p`′)3d2 + 32 72p 13 n = 1 −2332s+1(−2·3`′)21036s+3p(3p`′)Table 7.5: Remaining primes for q = 3.272pn Conductor TC Extra Information a4 Legendre for a4 ∆ Legendre for ∆d2 − 2`5m 50p 7 n = 1,` ≥ 7 odd 2`−25m+2(2·5`′)or(2`′)22`52m+6p(p`′)2`5m − d2 50p 8 n = 1,` ≥ 7 odd 2`−25m+2(2·5`′)or(2`′)−22`52m+6p(−p`′)d2 + 2`5m 50p 9 n = 1,` ≥ 7 −2`−25m+2(−2·5`′)or(−2`′)or(−5`′)22`52m+6p(p`′)5d2 − 2` 50p 10 n = 1,` ≥ 7 odd 2`−252s+1(2·5`′)22`56s+3p(5p`′)2` − 5d2 50p 11 n = 1,` ≥ 7 odd 2`−252s+1(2·5`′)−22`56s+3p(−5p`′)5d2 + 2` 50p 12 n = 1,` ≥ 7 −2`−252s+1(−2·5`′)or(−5`′)22`56s+3p(5p`′)d2 − 32 · 5m 200p 11 n = 1 or Pmin(n) ≥ 7 235m+2(2`′)or(2·5`′)22`52m+6pn(p`′)32 · 5m − d2 200p 12 n = 1 or Pmin(n) ≥ 7 235m+2(2`′)or(2·5`′)−22`52m+6pn(−p`′)d2 + 2`5m 200p 13 n = 1, ` ∈ {2, 4, 5} −2`−25m+2(−5`′)or(−2·5`′)22`52m+6p(p`′)d2+165m 200p 14 n = 1 or Pmin(n) ≥ 7 −223m+2(−1`′)285m+6pn(p`′)or(5p`′)5d2 − 32 200p 16 n = 1 or Pmin(n) ≥ 7 2352s+1(2·5`′)21056s+3pn(5p`′)5d2 + 2` 200p 17 n = 1 or Pmin(n) ≥ 7 −2`−252s+1(−2·5`′)or(−5`′)22`56s+3p(5p`′)d2 − 2`5m 400p 13 n = 1 and ` ≥ 5 odd or bothPmin(n) ≥ 7 and ` = 52`−25m+2(2·5`′)or(2`′)22`52m+6pn(p`′)2`5m − d2 400p 14 n = 1 and ` ≥ 5 odd or bothPmin(n) ≥ 7 and ` = 52`−25m+2(2·5`′)or(2`′)−22`52m+6p(−p`′)d2 + 2`5m 400p 15 n = 1, either both m = 0 −2`−25m+2(−2·5`′)or(−2`′)22`52m+6p(p`′)and ` = 2 or ` ≥ 4 or(−5`′)or(−1`′)d2+2`5m 400p 16 n = 1 or Pmin(n) ≥ 7 and` ∈ {2, 4}, or n = 1 and ` ≥5−2`−252(2·5`′)or(−1`′)22`5m+6pn(5p`′)or(p`′)5d2 − 2` 400p 18 n = 1 and ` ≥ 5 odd, orPmin(n) ≥ 7 and ` = 52`−252s+1(2·5`′)22`56s+3pn(5p`′)2` − 5d2 400p 19 n = 1,` ≥ 5 odd 2`−252s+1(2·5`′)−22`56s+3p(−5p`′)5d2 + 2` 400p 20 n = 1 or Pmin(n) ≥ 7 and` ∈ {2, 4, 5}, or n = 1 and` ≥ 6−2`−252s+1(−2·5`′)or(−5`′)22`56s+3p(5p`′)Table 7.6: Remaining primes for q = 5.273In fact, the data suggest that the converse is also true, that is, primes that are not intables 7.3 and 7.1 (thus, ones that are in 7.5) are in Pb,3 and similarly, primes that are not intables 7.4 and 7.2 (thus, ones that are in 7.6) are in Pb,5 . One has to be a bit careful thoughwith the argument as the following example illustrates.Example 7.0.28. Suppose q = 3 and consider the curves in the isogeny class 25902a. As25902 = (18)(1439) and 1439 = 215 − (177)2, we have that this curve belongs to Theorem4.0.8 case 8 with t = 177, a = 15 and b = 0. Thus, we have that our curve is isogenous tosayy2 = x3 − 3(177)x2 + 21332xwith discriminant ∆ = −23036(1439). This discriminant is not of the form (−3)m2 and sothere should be a prime `′ so that 4 - #ETor(Q). Notice that ` = 7 is a prime such that(−14397)= −1 and so the polynomial does not split in F7. However, the above elliptic curveover F7 has torsion subgroup of order 4. What has happened here is that over F7 even thoughwe do not get splitting, we have that the torsion subgroup has gone from a group isomorphicto Z/2Z to a group isomorphic to Z/4Z. As it turns out, the smallest prime that shows that1439 ∈ Pb is actually ` = 139 where the torsion subgroup of the curve over F139 is isomorphicto Z/162Z.Lemma 7.0.29. Let q ∈ {3, 5}. Let a ∈ {2, 3}, m = 2aq and b ∈ (Z/mZ)×. For everyprime p ≥ 3 distinct from q and δ, ∈ {±1}, there exists a prime `′ such that(δp`′)= and`′ ≡ b (mod m).Proof. Notice that =(δp`′)= (δ)`′−12(p`′)= (δ)`′−12 (−1)p−12`′−12(`′p).By multiplying both sides by (δ)`′−12 (−1)p−12`′−12 , we see that we are then reduced to findingan `′ so that (`′p)= (δ)`′−12 (−1)p−12`′−12If `′ ≡ 1 (mod 4) is consistent with `′ ≡ b (mod m), then notice that the right hand sideis just . Alternatively, if `′ ≡ 3 (mod 4), then the right hand side becomes ±δ where the± sign is determined by p ≡ ±1 (mod 4). In either case, we need to find a value of `′ suchthat(`′p)= ±1. If we want(`′p)= −1, choose an integer d such that(dp)= −1. Such an dmust exist since the Legendre symbol is a nontrivial real character when p ≥ 3. If we want(`′p)= 1, then we can choose d = 1 and notice that(dp)= 1. Now via the Chinese RemainderTheorem, find a simultaneous solution, say `0 such that`0 ≡ d 6≡ 0 (mod p) and `0 ≡ b 6≡ 0 (mod m).274The Chinese Remainder Theorem also gives that all solutions are given by `′ ≡ `0 (mod pm).As gcd(`0, pm) = 1, by Dirichlet’s Theorem for Primes in an Arithmetic Progression, we havethat there exists a prime `′ such that `′ ≡ `0 (mod pm). This completes the proof. Now we prove the main theorem.Theorem 7.0.30. Let q ∈ {3, 5} and let p ∈ Pg,q. Then p ∈ Pb,q if and only if every ellipticcurve with conductor in Sq,p with non-trivial rational two torsion has discriminant not of theform ∆Q,q,m :=(−1q)qm2 for all integers m.Proof. The reverse direction was Theorem 7.0.27 so it suffices to show the forward direction.Notice that there are only two cases where p /∈ Pb,q.1. The polynomial x3 + a2x2 + a4x associated to our elliptic curve splits modulo everyprime `′ ≡ 1 (mod 6).2. For every prime `′ ≡ 1 (mod 6), there exists a point P on our elliptic curve such thatP 6= (0, 0) but 2P = (0, 0).Hence, we need to show that for each curve avoiding discriminants of the form(−1q)qm2, wecan find a prime ` satisfying R(`′, q) where the curve does not split and that there is no pointP as specified above. To show this, notice that the duplication formula [ST92, p. 31] forcurves of the form E : y2 = x3 + a2x2 + a4x says that if P = (x, y) is a point on E, thenx(2P ) =x4 − 2a4x2 + a244y2=(x2 − a4)24y2.According to the duplication formula, the second condition above can occur for a prime `′only if (x2 − a4)2 ≡ 0 (mod `′) and so we must have that a4 is a quadratic residue modulo`′ for every prime `′ ≡ 1 (mod 6) if our theorem is to be false. Thus, for every curve in ourfamilies from Theorems 4.0.8, 4.0.10, 4.0.12, 5.0.14, 5.0.16 and 5.0.18, we must show that wecan find a prime `′ so that1.(∆`′)= −12.(a4`′)= −13.((−1q )qm2`′)=((−1q )q`′)= 1 or equivalently R(`′, q) holds.All that’s left to do is show that this is possible for each curve in the list with the exceptionof the two forbidden families we have already discarded. We look at Tables 7.5 and 7.6. Inthose tables, the a4 and ∆ values correspond to the first curve in each case. Choosing othercurves in the family will flip the roles of ∆ and a4 so we reduce our argument to just lookingat the table given.275Notice that in the tables, the Legendre symbol for ∆ never contains a 2. Thus the value(∆`′)is always of the form(
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Twisted extensions of Fermat's Last Theorem Bruni, Carmen Anthony 2015
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Title | Twisted extensions of Fermat's Last Theorem |
Creator |
Bruni, Carmen Anthony |
Publisher | University of British Columbia |
Date Issued | 2015 |
Description | Let x, y, z, n, α ∈ ℤ with α ≥ 1, p and n ≥ 5 primes. In 2011, Michael Bennett, Florian Luca and Jamie Mulholland showed that the equation involving a twisted sum of cubes [equation omitted] has no pairwise coprime nonzero integer solutions p ≥ 5,n ≥ p²p and p ∉ S where S is the set of primes q for which there exists an elliptic curve of conductor NE ∈ {18q,36q,72q} with at least one nontrivial rational 2-torsion point. In this dissertation, I present a solution that extends the result to include a subset of the primes in S; those q ∈ S for which all curves with conductor NE ∈ {18q,36q,72q} with nontrivial rational 2-torsion have discriminants not of the form ℓ² or -3m² with ℓ,m ∈ ℤ. Using a similar approach, I will classify certain integer solutions to the equation of a twisted sum of fifth powers [equation omitted] which in part generalizes work done from Billerey and Dieulefait in 2009. I will also discuss limitations of the methods for these equations and as they extend to further prime exponents. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2015-04-21 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivs 2.5 Canada |
DOI | 10.14288/1.0166239 |
URI | http://hdl.handle.net/2429/52912 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Graduation Date | 2015-05 |
Campus |
UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/2.5/ca/ |
Aggregated Source Repository | DSpace |
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