"Arts and Sciences, Irving K. Barber School of (Okanagan)"@en . "Computer Science, Mathematics, Physics and Statistics, Department of (Okanagan)"@en . "DSpace"@en . "UBCO"@en . "Johnstone, Jennifer Ann"@en . "2010-07-29T15:27:39Z"@en . "2010"@en . "Master of Science - MSc"@en . "University of British Columbia"@en . "Throughout this thesis we will be primarily concerned with the area of a rational right angle triangle, also known as a congruent number. The purpose of this thesis is to present a family of congruent number elliptic curves with rank at least three, as well as provide some insight into the distribution of congruent numbers. We provide an in depth background on congruent numbers and elliptic curves, as well as an overview of one of the key methods that will be used in determining the rank of an elliptic curve."@en . "https://circle.library.ubc.ca/rest/handle/2429/26993?expand=metadata"@en . "Congruent Numbers and Elliptic Curves by Jennifer Ann Johnstone B.Sc. Hons., The University of British Columbia, 2009 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The College of Graduate Studies (Mathematics) THE UNIVERSITY OF BRITISH COLUMBIA (Okanagan) July 2010 c\u00C2\u00A9 Jennifer Ann Johnstone 2010 Abstract Throughout this thesis we will be primarily concerned with the area of a rational right angle triangle, also known as a congruent number. The pur- pose of this thesis is to present a family of congruent number elliptic curves with rank at least three, as well as provide some insight into the distribution of congruent numbers. We provide an in depth background on congruent numbers and elliptic curves, as well as an overview of one of the key methods that will be used in determining the rank of an elliptic curve. ii Table of Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v List of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . vii Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 Congruent Numbers . . . . . . . . . . . . . . . . . . . . . . . . 7 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Families of Congruent Numbers . . . . . . . . . . . . . . . . 12 3 Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 3.2 Group Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.3 Mordell\u00E2\u0080\u0099s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 21 4 Rank Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . 24 4.1 Method of 2-descent . . . . . . . . . . . . . . . . . . . . . . . 24 4.2 Worked Example . . . . . . . . . . . . . . . . . . . . . . . . . 26 5 Congruent Number Curves of Moderate Rank . . . . . . . 29 5.1 Rank Three Results . . . . . . . . . . . . . . . . . . . . . . . 29 5.2 Proof of the Main Theorem . . . . . . . . . . . . . . . . . . . 35 iii Table of Contents 6 The Distribution of Congruent Numbers . . . . . . . . . . . 40 6.1 Rank Two Results . . . . . . . . . . . . . . . . . . . . . . . . 40 6.2 Proof of the Main Theorem . . . . . . . . . . . . . . . . . . . 43 7 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Appendices A MAPLE Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 A.1 Worked Example Calculations from Section4.2 . . . . . . . . 51 A.2 Chapter 5 Calculations . . . . . . . . . . . . . . . . . . . . . 52 A.3 Chapter 6 Calculations . . . . . . . . . . . . . . . . . . . . . 53 B MAGMA Code . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 B.1 Elliptic Curve Calculations . . . . . . . . . . . . . . . . . . . 54 iv List of Figures 1.1 Rational right angle triangle with area 6. . . . . . . . . . . . 1 3.1 Elliptic curve with three real roots, y2 = x(x+ 1)(x\u00E2\u0088\u0092 1). . . 19 3.2 Elliptic curve with one real root, y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). . 19 3.3 Cubic equation with a triple root, y2 = x3. . . . . . . . . . . . 19 3.4 Cubic equation with a double root, y2 = x2(x+ 1). . . . . . . 19 3.5 Geometric interpretation of P \u00E2\u0088\u0097Q on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). . . . . . . . . . . . . . . . . . . . . 20 3.6 Geometric interpretation of P \u00E2\u0088\u0097 P on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). . . . . . . . . . . . . . . . . . . . . 20 3.7 Geometric interpretation of P +Q on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). . . . . . . . . . . . . . . . . . . . . 21 v List of Notation Z Set of integers {. . . ,\u00E2\u0088\u00922,\u00E2\u0088\u00921, 0, 1, 2, . . .} . . . . 2. a \u00E2\u0089\u00A1 b(modm) Congruent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3. gcd(a, b) Greatest common divisor . . . . . . . . . . . . . . . . . 5. a|b Divides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5. a - b Does not divide . . . . . . . . . . . . . . . . . . . . . . . . . . 5. pa \u00E2\u0080\u0096 n Exactly divides . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.\u00E2\u0088\u008F Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6. Z+ Set of positive integers {1, 2, 3, . . .} . . . . . . . . 7. Q Set of rational numbers . . . . . . . . . . . . . . . . . . . 7. E(Q) Set of rational points on an elliptic curve . . 19. O Point at infinity on an elliptic curve . . . . . . . 20. Q\u00E2\u0088\u0097 Set of non-zero rational numbers . . . . . . . . . . 24. Q\u00E2\u0088\u00972 Set of squares of non-zero rational numbers 24. Q\u00E2\u0088\u0097 Q\u00E2\u0088\u00972 Quotient group . . . . . . . . . . . . . . . . . . . . . . . . . . . 24. vi Acknowledgements To start, the author would like to express her deepest gratitude to her super- visor Dr. Blair Spearman for his constant support and guidance throughout this process. The ideas presented in this thesis are the result of two summers of research with Dr. Spearman, which would not have been possible with- out the support of the Natural Sciences and Engineering Research Council of Canada. The author would also like to thank her committee Dr. Qiduan Yang and Dr. Yves Lucet for their helpful suggestions and all of the time they devoted to this success. Finally, the author would also like to acknowledge and thank all of the people who encouraged her to complete her Masters and supported her throughout. vii Dedication To my mother. viii Chapter 1 Introduction Although there are several equivalent definitions, the best way to define a congruent number is to first recall a similar, well known, property involving pythagorean triples. Pythagorean triples are defined as triples of positive integers satisfying the equation x2 + y2 = z2, such that each triple forms a right angle triangle with integer sides. Specifically, x, y and z can be defined to generate all possible pythagorean triples [Ros05]. This in turn allows us to find all possible integers m that will produce a right angle triangle with integer sides [Ros05]. To turn this to the definition of a congruent number we need to consider the possible integers n that will result in a right angle triangle with rational sides. That is to say, congruent numbers are defined as follows. Definition 1.1. A positive integer n is a congruent number if there exists a rational right angle triangle with area n. For example, 6 is a congruent number since there exists a rational right angle triangle with area 6, namely the right angle triangle seen in Figure 1.1. Figure 1.1: Rational right angle triangle with area 6. Notably, congruent numbers were first documented in a 10th century Arab manuscript and have since been studied by numerous scholars, in- 1 Chapter 1. Introduction cluding Euler, Fibonacci, and Fermat [AC74, Alt80, Kra86, Coa05, Cip09, Cha06, Hem06, Tun83]. In 1225 Fibonacci was the first to show that 5 and 7 are congruent numbers [Cha06]. Fibonacci also stated, without proof, that no congruent number n is exactly a square [Cha98]. It was not until four centuries later that Fermat proved Fibonacci\u00E2\u0080\u0099s statement, using the method of infinite descent. Specifically, Fermat proved that 1 is not a congruent number [Cha98, DJS09, Coa05, Cip09, Cha06]. When discussing congruent numbers we are often interested in the con- gruent number problem, which is as follows. Given a positive integer n can we determine whether or not n is a congruent number in a finite number of steps? [Cha98, Coa05, Hem06] Since the 10th century many different definitions of congruent numbers have been derived in order to try and solve the congruent number problem. How- ever, unlike right angle triangles with integer sides the congruent number problem remains unsolved [Cha98]. Even though congruent numbers in general have been around for a very long time, and lots of work has gone into finding them, it took until 1915 to list all congruent numbers less than 100. This in turn led to another goal \u00E2\u0080\u009CFinding all square-free congruent numbers less than 1000\u00E2\u0080\u009D [Alt80, AC74, Cip09]. It was not until 1983 when Tunnell determined an equivalent definition for a congruent number that ultimately allowed for all congruent numbers less than 1000 to be determined [Cip09, Hem06]. Tunnell also provided a simple criterion to determine whether or not a positive integer n is a congruent number [Hem06, Ros05]. Theorem 1.2. [Hem06, Ros05, Kob92, Tunnell\u00E2\u0080\u0099s Theorem] Define An = # { x, y, z \u00E2\u0088\u0088 Z|n = 2x2 + y2 + 32z2} Bn = # { x, y, z \u00E2\u0088\u0088 Z|n = 2x2 + y2 + 8z2} Cn = # { x, y, z \u00E2\u0088\u0088 Z|n = 4x2 + 2y2 + 64z2} Dn = # { x, y, z \u00E2\u0088\u0088 Z|n = 8x2 + 2y2 + 16z2} . Suppose n is congruent, if n is even then An = Bn and if n is odd, then 2Cn = Dn. If the Birch and Swinnerton-Dyer Conjecture (BSD Conjecture) holds for curves of the form y2 = x3 \u00E2\u0088\u0092 n2x then, conversely, these equalities imply that n is a congruent number. 2 Chapter 1. Introduction Proof. See [Kob92] for the logical structure of the argument. As we can see the converse of Tunnell\u00E2\u0080\u0099s Theorem hinges on the BSD Conjecture, which is still an open problem today. Details about the BSD Conjecture are, however, beyond the scope of this thesis. Fortunately, the results found in this thesis do not assume the BSD Conjecture, but it is worth mentioning that the BSD Conjecture is one of the Clay Mathematics Institutes Millennium Prize Problems and it is widely assumed to be true [Hem06, Cip09]. Since 1983 a lot more work has been done to determine congruent num- bers, including some results that do not use Tunnell\u00E2\u0080\u0099s Theorem. In 1986 Kramarz verified the converse of Tunnell\u00E2\u0080\u0099s Theorem for all square-free inte- gers less than 2000 [Kra86]. By 1993, with the help of computers and the use of Tunnell\u00E2\u0080\u0099s Theorem, all congruent numbers less than 10000 had been determined [NW93]. Today\u00E2\u0080\u0099s results include computations for congruent numbers up to 1 trillion, once again assuming Tunnell\u00E2\u0080\u0099s Theorem [Cip09]. Currently, Rubinstein and others have predicted that the number of congruent numbers less than x, arising from even rank elliptic curves, is asymptotically cx3/4 log(x)11/8, (1.1) where c is a constant. It is their hope to provide a better determination of c, using the data produced from the computations for congruent numbers up to 1 trillion [Cip09]. We now provide some of the other known results that do not assume the BSD Conjecture. For instance, given distinct primes pi and qi where pi \u00E2\u0089\u00A1 qi \u00E2\u0089\u00A1 i(mod 8) , i.e. pi = 8k + i and qi = 8l + i for some k, l \u00E2\u0088\u0088 Z, we have the following results, as stated in [Hem06, Tun83]. 1. p3 is not a congruent number. 2. p3q3, 2p5, 2p5q5 are not congruent numbers. 3. p5, p7 are congruent numbers. 4. 2p7, 2p3, p3q7, 2p3q5, 2p5q7 are congruent numbers. One of the main results, to be presented in Chapter 5, is that we were able to find a family of congruent numbers, for which the associated elliptic curve satisfies a rank condition. These results have been accepted for publication in the Canadian Mathematical Bulletin under the title \u00E2\u0080\u009CCongruent Number 3 1.1. Preliminaries Elliptic Curves with rank at least Three\u00E2\u0080\u009D. Another main result, to be pre- sented in Chapter 6, involves the distribution of congruent numbers and has been recently published in the Proceedings of the Japan Academy, Series A, under the title \u00E2\u0080\u009COn the Distribution of Congruent Numbers\u00E2\u0080\u009D. Before we present these results we will provide the necessary background information in the next section followed by an in depth chapter on congruent numbers. In Chapter 3 we provide the necessary definitions and theorems for elliptic curves followed by a discussion on how to calculate the rank of an elliptic curve in Chapter 4. 1.1 Preliminaries To start, we recall some basics from Abstract Algebra that can be found in most first year Abstract Algebra textbooks. We begin with some definitions involving binary algebraic structures, denoted by < G, \u00E2\u0088\u0097 >, where G is a set and \u00E2\u0088\u0097 is a binary operator. Definition 1.3. A group < G, \u00E2\u0088\u0097 > is a set G, closed under a binary opera- tion \u00E2\u0088\u0097, such that 1. (a\u00E2\u0088\u0097b)\u00E2\u0088\u0097c = a\u00E2\u0088\u0097 (b\u00E2\u0088\u0097c) for all a, b, c \u00E2\u0088\u0088 G (i.e. \u00E2\u0088\u0097 is associative with respect to G). 2. e \u00E2\u0088\u0097 a = a \u00E2\u0088\u0097 e = a for all a \u00E2\u0088\u0088 G and some element e \u00E2\u0088\u0088 G (i.e. there exists an identity element e in G for \u00E2\u0088\u0097). 3. Corresponding to each a \u00E2\u0088\u0088 G, there is an element a\u00E2\u0080\u00B2 \u00E2\u0088\u0088 G such that a \u00E2\u0088\u0097 a\u00E2\u0080\u00B2 = a\u00E2\u0080\u00B2 \u00E2\u0088\u0097 a = e (i.e. G contains inverses with respect to \u00E2\u0088\u0097). Definition 1.4. An abelian group < G, \u00E2\u0088\u0097 > is a group G where a \u00E2\u0088\u0097 b = b \u00E2\u0088\u0097 a for all a, b \u00E2\u0088\u0088 G (i.e. \u00E2\u0088\u0097 is commutative). Definition 1.5. Let < G, \u00E2\u0088\u0097 > and < G\u00E2\u0080\u00B2, \u00E2\u0088\u0097\u00E2\u0080\u00B2 > be binary algebraic struc- tures, where G and G\u00E2\u0080\u00B2 are both groups. Then a map \u00CF\u0086 of G into G\u00E2\u0080\u00B2 is a homomorphism if \u00CF\u0086(a \u00E2\u0088\u0097 b) = \u00CF\u0086(a) \u00E2\u0088\u0097\u00E2\u0080\u00B2 \u00CF\u0086(b) for all a, b \u00E2\u0088\u0088 G. Definition 1.6. Let < G, \u00E2\u0088\u0097 > and < G\u00E2\u0080\u00B2, \u00E2\u0088\u0097\u00E2\u0080\u00B2 > be binary algebraic structures. An isomorphism of G with G\u00E2\u0080\u00B2, denoted by G ' G\u00E2\u0080\u00B2, is a one-to-one function \u00CF\u0086 mapping G onto G\u00E2\u0080\u00B2 such that \u00CF\u0086(a \u00E2\u0088\u0097 b) = \u00CF\u0086(a) \u00E2\u0088\u0097\u00E2\u0080\u00B2 \u00CF\u0086(b) 4 1.1. Preliminaries for all a, b \u00E2\u0088\u0088 G. We also define a generator for a group < G, \u00E2\u0088\u0097 > as an element a \u00E2\u0088\u0088 G that generates G under the assumed binary operation \u00E2\u0088\u0097 (i.e. if G = {an|n \u00E2\u0088\u0088 Z} for some a in G then we say that a is a generator for G). Lastly, we define finitely generated abelian groups and recall the Fundamental Theorem of Finitely Generated Abelian Groups, as follows. Definition 1.7. A finitely generated abelian group < G, \u00E2\u0088\u0097 > is an abelian group that contains a finite set of generators of G. Theorem 1.8. [Fra03, Fundamental Theorem of Finitely Generated Abelian Groups] Every finitely generated abelian group G is isomorphic to a direct sum of cyclic groups in the form Z \u00E2\u008A\u0095 \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 \u00E2\u008A\u0095 Z \u00E2\u008A\u0095 Zpv11 \u00E2\u008A\u0095 \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 \u00E2\u008A\u0095 Zpvss , where Z is a cyclic group with infinite order and Zpvii is a finite cyclic group of prime power order, for some prime pi and some positive integer vi. Next we have the Number Theory portion of the preliminaries where we recall some basic definitions from this area. Throughout this thesis we will be dealing with several different of equations, some of which we describe as diophantine equations. Specifically, a diophantine equation is a type of equation that requires that the solutions come from the set of integers. Additionally, we need to define the following notation and terms. Definition 1.9. The greatest common divisor of two integers a and b, which are not both 0, is the largest integer that divides both a and b. The notation that we use for the greatest common divisor of two integers a and b is gcd(a, b) . For integers a and b we also use the notation a|b to mean that a divides b and the notation a - b to mean that a does not divide b . Definition 1.10. Let p be a prime number and n be a positive integer. If pa|n but pa+1 - n, we say that pa exactly divides n, and we write pa \u00E2\u0080\u0096 n. Another symbol that we should be aware of is +, which we use to sym- bolize defined equality. Another symbol that we have already seen is \u00E2\u0089\u00A1, which defines a congruence. Along these lines we also need to define the term congruence class. Definition 1.11. A congruence class modulo m contains integers that are mutually congruent modulo m. 5 1.1. Preliminaries For example the set of integers modulo 2 can be put into one of 2 con- gruence classes, namely the class of integers that are congruent to 0 modulo 2 or the class of integers that are congruent to 1 modulo 2 (a.k.a the class of even integers or the class of odd integers). We conclude this section with the following two terms. Definition 1.12. The discriminant of x3 +ax2 + bx+ c is defined to be the quantity D = \u00E2\u0088\u00924a3c+ a2b2 + 18abc\u00E2\u0088\u0092 4b3 \u00E2\u0088\u0092 27c2. Definition 1.13. Given a polynomial P (x) = xn + an\u00E2\u0088\u00921x+ \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7+ a1x+ a0 of degree n with roots \u00CE\u00B1i, i = 1, . . . , n and a polynomial Q(x) = xm + bm\u00E2\u0088\u00921x+ \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7+ b1x+ b0 of degree m with roots \u00CE\u00B2j , j = 1, . . . ,m the resultant \u00CF\u0081(P,Q) is defined by \u00CF\u0081(P,Q) = n\u00E2\u0088\u008F i=1 m\u00E2\u0088\u008F j=1 (\u00CE\u00B1i \u00E2\u0088\u0092 \u00CE\u00B2j). 6 Chapter 2 Congruent Numbers 2.1 Introduction Before we dive into the topic of congruent numbers we recall, from Chapter 1, the most common definition of a congruent number. Definition 2.1. A positive integer n is a congruent number if there exists a rational right angle triangle with area n. We assume that the congruent numbers we are working with or searching for are square-free (except in Chapter 6 when we discuss the distribution of congruent numbers), since it is directly apparent from the definition of a congruent number that if n is a congruent number then so too is nk2, for some integer k. Additionally, there are several equivalent definitions for a congruent number, which can be summarized by the following theorem. Theorem 2.2. The following 5 statements are equivalent definitions for a positive integer n to be a congruent number. (i) There exist x, y, z, t \u00E2\u0088\u0088 Z+ satisfying the rationalized Diophantine equa- tions x2 + ny2 = z2 and x2 \u00E2\u0088\u0092 ny2 = t2, as seen in [AC74, Alt80, Cha06, Kra86, God78, Tun83]. (ii) There exist x, y, z \u00E2\u0088\u0088 Z+ satisfying the rationalized Diophantine equation x4 \u00E2\u0088\u0092 n2y4 = z2, as seen in [AC74]. (iii) The elliptic curve Y 2 = X3 \u00E2\u0088\u0092 n2X has non-trivial solutions \u00E2\u0088\u0088 Q [Hem06, Cha06, Cip09, Ben02, Hem06, Kob92, Tun83]. 7 2.1. Introduction (iv) There exists a rational right angle triangle with area n [Cha06, NW93, Cip09, Coa05, Kra86, Nem98, DJS09, Ben02, Hem06, Ros05, Kob92, Tun83]. (v) There exist u, v, w \u00E2\u0088\u0088 Z+ with nw2 = uv(u2 \u00E2\u0088\u0092 v2), as seen in [AC74, Alt80, God78, Kra86, DJS09]. Proof. ((i)\u00E2\u0087\u0092 (ii)) Assume that there exist x, y, z, t, n \u00E2\u0088\u0088 Z+ such that x2 + ny2 = z2 (2.1) and x2 \u00E2\u0088\u0092 ny2 = t2. (2.2) Then multiplying Equations (2.1) and (2.2) gives us that (x2 + ny2)(x2 \u00E2\u0088\u0092 ny2) = z2t2 \u00E2\u0087\u0092 x4 \u00E2\u0088\u0092 n2y4 = (zt)2. Now, let w = zt such that x4 \u00E2\u0088\u0092 n2y4 = w2 is solvable, which is exactly what we needed to show. ((ii)\u00E2\u0087\u0092 (iii)) Assume that there exist x, y, w, n \u00E2\u0088\u0088 Z+ such that x4 \u00E2\u0088\u0092 n2y4 = w2. Then x4 \u00E2\u0088\u0092 n2y4 = w2 \u00E2\u0087\u0092 x4 y4 \u00E2\u0088\u0092 n2 y4 y4 = w2 y4 \u00E2\u0087\u0092 ( x y )4 \u00E2\u0088\u0092 n2 = ( w y2 )2 \u00E2\u0087\u0092 x2 y2 [( x y )4 \u00E2\u0088\u0092 n2] = x2 y2 [( w y2 )2] \u00E2\u0087\u0092 ( x y )6 \u00E2\u0088\u0092 n2 x2 y2 = ( wx y3 )2 8 2.1. Introduction Now, let X = x 2 y2 and Y = wx y3 such that( x y )6 \u00E2\u0088\u0092 n2x 2 y2 = ( wx y3 )2 \u00E2\u0087\u0092 X3 \u00E2\u0088\u0092 n2X = Y 2. This is exactly our elliptic curve definition, since X and Y are nonzero and thus nontrivial. ((iii)\u00E2\u0087\u0092 (iv)) Assume that n is a positive integer and that there exist X,Y \u00E2\u0088\u0088 Q such that Y 2 = X3 \u00E2\u0088\u0092 n2X. Now, let a = X 2\u00E2\u0088\u0092n2 Y , b = 2nX Y and c = X2+n2 Y so that a2 + b2 = c2. Thus, we have a rational right angle triangle with sides a, b, c and area equal to n, since ab 2 = (X2 \u00E2\u0088\u0092 n2)(2nX) 2Y 2 = n(X3 \u00E2\u0088\u0092 n2X) Y 2 = nY 2 Y 2 = n. ((iv) \u00E2\u0087\u0092 (v)) Assume that n is a positive square-free integer and that there exists a right angle triangle with sides a, b, c \u00E2\u0088\u0088 Q such that a2 + b2 = c2 and n = ab 2 . First, we can scale a, b, c to conclude that there exist f, g, h \u00E2\u0088\u0088 Z+ with f2 + g2 = h2, where f, g, h are relatively prime. Then, the area of this triangle would be n = fg 2 . 9 2.1. Introduction Now, using properties of pythagorean triples we know that there exist inte- gers u and v such that f = u2\u00E2\u0088\u0092v2, g = 2uv and h = u2 +v2 [Dic20, Ros05]. Then n = 2uv(u2 \u00E2\u0088\u0092 v2) 2 = uv(u2 \u00E2\u0088\u0092 v2), which is exactly what we needed to show for the case when w2 = 1. We note that a similar result occurs to include w2 when n is not square-free[Cha98]. ((v)\u00E2\u0087\u0092 (i)) Assume that there exist u, v, n, w \u00E2\u0088\u0088 Z+ such that n is square-free, nw2 = uv(u2 \u00E2\u0088\u0092 v2) and gcd(u, v) = 1 (or else we could divide out the gcd into w2). Then clearly one of u, v, u2 \u00E2\u0088\u0092 v2 is even and the other two are odd. Case #1: Assume that u and v are odd then u2 \u00E2\u0088\u0092 v2 is even. Next, let \u00CE\u00B2 = 2uv \u00E2\u0087\u0092 2|\u00CE\u00B2 and \u00CE\u00B1 = u2 \u00E2\u0088\u0092 v2 \u00E2\u0087\u0092 2|\u00CE\u00B1, by assumption. Note that \u00CE\u00B2, \u00CE\u00B1 \u00E2\u0088\u0088 Z, since u, v \u00E2\u0088\u0088 Z+. Then nw2 = \u00CE\u00B1\u00CE\u00B2 2 . Now, let y = w so that ny2 = \u00CE\u00B1\u00CE\u00B2 2 and note that x2 = ( u2 + v2 2 )2 = (\u00CE\u00B1 2 )2 + ( \u00CE\u00B2 2 )2 , where x is in Z by definition of \u00CE\u00B1 and \u00CE\u00B2. Then ny2 = \u00CE\u00B1\u00CE\u00B2 2 = \u00CE\u00B1\u00CE\u00B2 2 + x2 \u00E2\u0088\u0092 x2 = \u00CE\u00B1\u00CE\u00B2 2 + (\u00CE\u00B1 2 )2 + ( \u00CE\u00B2 2 )2 \u00E2\u0088\u0092 x2 = ( \u00CE\u00B1 2 + \u00CE\u00B2 2 )2 \u00E2\u0088\u0092 x2. 10 2.1. Introduction Therefore, let z = \u00CE\u00B12 + \u00CE\u00B2 2 (which is in Z by definition of \u00CE\u00B1 and \u00CE\u00B2) such that ny2 = ( \u00CE\u00B1 2 + \u00CE\u00B2 2 )2 \u00E2\u0088\u0092 x2 \u00E2\u0087\u0092 ny2 = z2 \u00E2\u0088\u0092 x2 \u00E2\u0087\u0092 x2 + ny2 = z2. Similarly, x2 = (\u00CE\u00B1 2 )2 + ( \u00CE\u00B2 2 )2 = (\u00CE\u00B1 2 )2 + ( \u00CE\u00B2 2 )2 + ny2 \u00E2\u0088\u0092 ny2 = (\u00CE\u00B1 2 )2 + ( \u00CE\u00B2 2 )2 + ny2 \u00E2\u0088\u0092 \u00CE\u00B1\u00CE\u00B2 2 = ny2 + ( \u00CE\u00B1 2 \u00E2\u0088\u0092 \u00CE\u00B2 2 )2 . So, let t = \u00CE\u00B12 \u00E2\u0088\u0092 \u00CE\u00B22 (which is in Z by definition of \u00CE\u00B1 and \u00CE\u00B2) such that x2 = ny2 + ( \u00CE\u00B1 2 \u00E2\u0088\u0092 \u00CE\u00B2 2 )2 \u00E2\u0087\u0092 x2 = ny2 + t2 \u00E2\u0087\u0092 x2 \u00E2\u0088\u0092 ny2 = t2. Hence, we have obtained the rationalized Diophantine equations in (i) with x, y, z, t \u00E2\u0088\u0088 Z, which can be restricted to x, y, z, t \u00E2\u0088\u0088 Z+. Case #2: Assume that u is even and v is odd then u2\u00E2\u0088\u0092 v2 is odd. Next, we can make similar substitutions to obtain the same result. Case #3: Similarly, when v is even and u is odd then u2\u00E2\u0088\u0092 v2 is odd and we obtain the same result. This completes the proof. Historically, the term \u00E2\u0080\u009CCongruent number\u00E2\u0080\u009D is a result of the equivalent definition stated in Theorem 2.2(i), since x2 + ny2, x2 \u00E2\u0088\u0092 ny2, and x2 11 2.2. Families of Congruent Numbers are all congruent numbers modulo n [Cip09, Hem06, Kob92]. The most current definition of a congruent number is the one involving elliptic curves, as depicted in Theorem 2.2(iii) [Cip09]. This definition also gives us the term congruent number elliptic curve. 2.2 Families of Congruent Numbers Since the 10th century several families of congruent numbers have been discovered. The purpose of this section is to present five families of congruent numbers, given in Alter and Curtz\u00E2\u0080\u0099s paper [AC74], as well as proofs and examples for each. Lemma 2.3. [AC74] Given a, b \u00E2\u0088\u0088 Z+ if n = a4 + 4b4 then n is a congruent number. Proof. Given n = a4 + 4b4, for some a, b \u00E2\u0088\u0088 Z+, we need to show that there exist x, y \u00E2\u0088\u0088 Z+ such that x2 + ny2 and x2 \u00E2\u0088\u0092 ny2 are both squares. Now, for n = a4 + 4b4 we have x2 + ny2 = x2 + (a4 + 4b4)y2 (2.3) and x2 \u00E2\u0088\u0092 ny2 = x2 \u00E2\u0088\u0092 (a4 + 4b4)y2. (2.4) So, let x = a8 + 24a4b4 + 16b8 and y = 4ab(4b4 \u00E2\u0088\u0092 a4) in Equation (2.3) then x2+(a4+4b4)y2 = (a8+24a4b4+16b8)2+16(a4+4b4)a2b2(4b4\u00E2\u0088\u0092a4)2 (2.5) and upon factoring (2.5) becomes (a8 + 32a2b6 \u00E2\u0088\u0092 8a4b4 + 8a6b2 + 16b8)2. Similarly, when we let x = a8 + 24a4b4 + 16b8 and y = 4ab(4b4 \u00E2\u0088\u0092 a4) in Equation (2.4) we get that x2 \u00E2\u0088\u0092 (a4 + 4b4)y2 = (a8 \u00E2\u0088\u0092 32a2b6 \u00E2\u0088\u0092 8a4b4 \u00E2\u0088\u0092 8a6b2 + 16b8)2. Therefore, let z = (a8 + 32a2b6 \u00E2\u0088\u0092 8a4b4 + 8a6b2 + 16b8) 12 2.2. Families of Congruent Numbers and t = (a8 \u00E2\u0088\u0092 32a2b6 \u00E2\u0088\u0092 8a4b4 \u00E2\u0088\u0092 8a6b2 + 16b8). Then we have shown that there exist x, y, z, t \u00E2\u0088\u0088 Z+ such that n = a4 + 4b4 is a congruent number by Theorem 2.2(i). Example 2.4. If we let a = 1 and b = 2 in Lemma 2.3 then n = 14+4(24) = 65 is a congruent number. Using Theorem 2.2 (i) we see that for x = 97 and y = 12 x2 + 65(y2) = 18769 = 1372 and x2 \u00E2\u0088\u0092 65(y2) = 49 = 72. Hence, the rationalized Diophantine equations x2 + 65y2 = z2 and x2 \u00E2\u0088\u0092 65y2 = t2 are solvable with x = 97, y = 12, z = 137 and t = 7, which verifies that n = 65 is a congruent number. Lemma 2.5. [AC74] Given a, b \u00E2\u0088\u0088 Z+ if n = 2a4+2b4 then n is a congruent number. Proof. Given n = 2a4 + 2b4, for some a, b \u00E2\u0088\u0088 Z+, we need to show that there exist x, y \u00E2\u0088\u0088 Z+ such that x2 + ny2 and x2 \u00E2\u0088\u0092 ny2 are both squares. Now, for n = 2a4 + 2b4 we have x2 + ny2 = x2 + (2a4 + 2b4)y2 (2.6) and x2 \u00E2\u0088\u0092 ny2 = x2 \u00E2\u0088\u0092 (2a4 + 2b4)y2. (2.7) So, let x = a8 + 6a4b4 + b8 and y = 2ab(a4 \u00E2\u0088\u0092 b4) in Equation (2.6) then x2 + (2a4 + 2b4)y2 = (a8 + 6a4b4 + b8)2 + 4(2a4 + 2b4)a2b2(a4 \u00E2\u0088\u0092 b4)2 (2.8) and upon factoring (2.8) becomes (a8 + 4a2b6 \u00E2\u0088\u0092 2a4b4 + 4a6b2 + b8)2. 13 2.2. Families of Congruent Numbers Similarly, when we let x = a8 + 6a4b4 + b8 and y = 2ab(a4\u00E2\u0088\u0092 b4) in Equation (2.7) we get that x2 \u00E2\u0088\u0092 (2a4 + 2b4)y2 = (a8 \u00E2\u0088\u0092 4a2b6 \u00E2\u0088\u0092 2a4b4 \u00E2\u0088\u0092 4a6b2 + b8)2. Therefore, let z = (a8 + 4a2b6 \u00E2\u0088\u0092 2a4b4 + 4a6b2 + b8) and t = (a8 \u00E2\u0088\u0092 4a2b6 \u00E2\u0088\u0092 2a4b4 \u00E2\u0088\u0092 4a6b2 + b8). Then we have shown that there exist x, y, z, t \u00E2\u0088\u0088 Z+ such that n = 2a4 + 2b4 is a congruent number by Theorem 2.2(i). Example 2.6. If we let a = 1 and b = 2 in Lemma 2.5 then n = 2(14) + 2(24) = 34 is a congruent number. Using Theorem 2.2(i) we see that for x = 145 and y = 12 x2 + 34(y2) = 25921 = 1612 and x2 \u00E2\u0088\u0092 34(y2) = 16129 = 1272. Hence, the rationalized Diophantine equations x2 + 34y2 = z2 and x2 \u00E2\u0088\u0092 34y2 = t2 are solvable with x = 145, y = 12, z = 161 and t = 127, which verifies that n = 34 is a congruent number. Lemma 2.7. [AC74] Given a, b \u00E2\u0088\u0088 Z+ if n = a4 \u00E2\u0088\u0092 b4 then n is a congruent number. Proof. Given n = a4 \u00E2\u0088\u0092 b4, for some a, b \u00E2\u0088\u0088 Z+, we need to show that there exist x, y \u00E2\u0088\u0088 Z+ such that x2 + ny2 and x2 \u00E2\u0088\u0092 ny2 are both squares. Now, for n = a4 \u00E2\u0088\u0092 b4 we have x2 + ny2 = x2 + (a4 \u00E2\u0088\u0092 b4)y2 (2.9) and x2 \u00E2\u0088\u0092 ny2 = x2 \u00E2\u0088\u0092 (a4 \u00E2\u0088\u0092 b4)y2. (2.10) 14 2.2. Families of Congruent Numbers So, let x = b2(a4 + b4) and y = 2ab3 in Equation (2.9) then x2 + (a4 \u00E2\u0088\u0092 b4)y2 = b4(a4 + b4)2 + 4(a4 \u00E2\u0088\u0092 b4)a2b6 (2.11) and upon factoring (2.11) becomes b4(a4 + 2a2b2 \u00E2\u0088\u0092 b4)2. Similarly, when we let x = b2(a4 + b4) and y = 2ab3 in Equation (2.10) we get that x2 \u00E2\u0088\u0092 (a4 \u00E2\u0088\u0092 b4)y2 = b4(a4 \u00E2\u0088\u0092 2a2b2 \u00E2\u0088\u0092 b4)2. Therefore, let z = b4(a4 + 2a2b2 \u00E2\u0088\u0092 b4)2 and t = b4(a4 \u00E2\u0088\u0092 2a2b2 \u00E2\u0088\u0092 b4) then we have shown that there exist x, y, z, t \u00E2\u0088\u0088 Z+ such that n = a4 \u00E2\u0088\u0092 b4 is a congruent number by Theorem 2.2(i). Example 2.8. If we let a = 2 and b = 1 in Lemma 2.7 then n = 24\u00E2\u0088\u009214 = 15 is a congruent number. Using Theorem 2.2(i) we see that for x = 17 and y = 4 x2 + 15y2 = 529 = 232 and x2 \u00E2\u0088\u0092 15y2 = 49 = 72. Hence, the rationalized Diophantine equations x2 + 15y2 = z2 and x2 \u00E2\u0088\u0092 15y2 = t2 are solvable with x = 17, y = 4, z = 23 and t = 7, which verifies that n = 15 is a congruent number. Lemma 2.9. [AC74] For integers a and b with opposite parity if nk2 = a4 + 6a2b2 + b4, for some integer k, then n is a congruent number. Proof. Recall by Theorem 2.2(v) that every congruent number n is of the form nk2 = uv(u2 \u00E2\u0088\u0092 v2), with k, u, v \u00E2\u0088\u0088 Z+. So, let u = f2, v = g2, u\u00E2\u0088\u0092 v = h2 and u+ v = nk2 where f, g, h, k, n \u00E2\u0088\u0088 Z+. Then n is a congruent number since uv(u2 \u00E2\u0088\u0092 v2) = n(fghk)2. Now, h2 + g2 = f2 (since u - v + v = u) 15 2.2. Families of Congruent Numbers is a pythagorean triple such that there exist a, b \u00E2\u0088\u0088 Z+ with h = a2 \u00E2\u0088\u0092 b2, g = 2ab, and f = a2 + b2, (2.12) where exactly one of a, b is even and the other is odd [Dic20, Ros05]. So, assume that h, g, f are defined as in (2.12) then f2 + g2 = nk2 \u00E2\u0087\u0092 (a2 + b2)2 + (2ab)2 = nk2 \u00E2\u0087\u0092 a4 + 6a2b2 + b4 = nk2. Hence, nk2 = a4 + 6a2b2 + b4 is a family of congruent numbers when a and b have opposite parity. Example 2.10. If we let a = 1 and b = 2 in Lemma 2.9 then n = 14 + 6(12)(22) + 24 = 41 is a congruent number. Using Theorem 2.2(i) we see that for x = 881 and y = 120 x2 + 41y2 = 1366561 = 11692 and x2 \u00E2\u0088\u0092 41y2 = 185761 = 4312. Hence, the rationalized Diophantine equations x2 + 41y2 = z2 and x2 \u00E2\u0088\u0092 41y2 = t2 are solvable with x = 881, y = 120, z = 1169 and t = 431, which verifies that n = 41 is a congruent number. Lemma 2.11. [AC74] For integers a, b with opposite parity if nk2 = a4 \u00E2\u0088\u0092 6a2b2 + b4 then n is a congruent number. Proof. Once again, recall by Theorem 2.2(v) that every congruent number n is of the form nk2 = uv(u2 \u00E2\u0088\u0092 v2), with k, u, v \u00E2\u0088\u0088 Z+. So, let u = f2, v = g2, u+ v = h2 and u\u00E2\u0088\u0092 v = nk2 where f, g, h, k, n \u00E2\u0088\u0088 Z+. Then n is a congruent number since uv(u2 \u00E2\u0088\u0092 v2) = n(fghk)2. Now, f2 + g2 = h2 16 2.2. Families of Congruent Numbers is a pythagorean triple such that there exist a, b \u00E2\u0088\u0088 Z+ with f = a2 \u00E2\u0088\u0092 b2, g = 2ab, and h = a2 + b2, (2.13) where exactly one of a, b is even and the other is odd [Dic20, Ros05]. So, assume that f, g, h are defined as in (2.13) then f2 \u00E2\u0088\u0092 g2 = nk2 \u00E2\u0087\u0092 (a2 \u00E2\u0088\u0092 b2)2 + (2ab)2 = nk2 \u00E2\u0087\u0092 a4 \u00E2\u0088\u0092 6a2b2 + b4 = nk2. Hence, nk2 = a4 \u00E2\u0088\u0092 6a2b2 + b4 is a family of congruent numbers when a and b have opposite parity. Example 2.12. If we let a = 4 and b = 1 in Lemma 2.11 then n = 44 \u00E2\u0088\u0092 6(42)(12) + 14 = 161 is a congruent number. Using Theorem 2.2(i) we see that for x = 305 and y = 24 x2 + 161y2 = 185761 = 4312 and x2 \u00E2\u0088\u0092 161y2 = 289 = 172. Hence, the rationalized Diophantine equations x2 + 161y2 = z2 and x2 \u00E2\u0088\u0092 161y2 = t2 are solvable with x = 305, y = 24, z = 431 and t = 17, verifying that n = 161 is a congruent number. 17 Chapter 3 Elliptic Curves 3.1 Introduction Although elliptic curves do not resemble ellipses it is of some interest to note that the curves originated from the study of computing the arc length of an ellipse [ST93]. In general, an equation of the form y2 + axy + by = x3 + cx2 + dx+ e, with coefficients a, b, c, d and e in Q, is an elliptic curve under certain restric- tions. However, for our purposes we are primarily interested in a shorter form and as such define the following. Definition 3.1 (Elliptic Curve in Weierstrass Normal Form). An equation of the form E : y2 = x3 + ax2 + bx+ c where a, b, and c are integers is an elliptic curve if the discriminant of x3 + ax2 + bx+ c is not 0 (i.e. the discriminant D = \u00E2\u0088\u00924a3c+a2b2 + 18abc\u00E2\u0088\u00924b3\u00E2\u0088\u0092 27c2 6= 0). We note that the non-zero restriction on the discriminant of the cubic is nec- essary, since certain properties about elliptic curves do not hold otherwise. The discriminant requirement directly corresponds to a cubic with three dis- tinct roots (real and/or complex). That is to say if the cubic, x3+ax2+bx+c, has a double root or a triple root then the equation y2 = x3 + ax2 + bx+ c is not an elliptic curve. Examples of elliptic curves can be seen in Figure 3.1 and Figure 3.2 and examples of cubic equations with double and triple roots (and hence not elliptic curves) can be seen in Figure 3.3 and Figure 3.4, respectively. 18 3.2. Group Law Figure 3.1: Elliptic curve with three real roots, y2 = x(x+1)(x\u00E2\u0088\u0092 1). Figure 3.2: Elliptic curve with one real root, y2 = (x+1)(x2\u00E2\u0088\u00924x+5). Figure 3.3: Cubic equation with a triple root, y2 = x3. Figure 3.4: Cubic equation with a double root, y2 = x2(x+ 1). Given an elliptic curve E we are interested in the set of rational points on E, denoted by E(Q) . Before we can discuss the structure of E(Q) we must define the group law associated with rational points on elliptic curves. 3.2 Group Law Given two rational points on elliptic curve E can we find another rational point? The answer is yes. Before we can determine the other rational point, which is defined by the group law, we must first define the composition law. So, let E(Q) be the set of rational points on an elliptic curve E and let \u00E2\u0088\u0097 be the binary composition law operator that maps E(Q)\u00C3\u0097E(Q) into E(Q). Then for each (P,Q) \u00E2\u0088\u0088 E(Q) \u00C3\u0097 E(Q) we denote the element \u00E2\u0088\u0097((P,Q)) by 19 3.2. Group Law P \u00E2\u0088\u0097Q. We define P \u00E2\u0088\u0097Q to be the third intersection point of the line PQ with the elliptic curve E [ST93, SZ03, Hus04, Sil86]. Similarly, if there is only one rational point P on the elliptic curve then we consider P \u00E2\u0088\u0097P to be the third intersection point of the tangent line created at P with the elliptic curve E (where the tangent line is assumed to pass through the point P twice) [ST93, SZ03, Hus04, Sil86]. The composition law, also known as the chord and tangent method, is illustrated in Figure 3.5 and Figure 3.6 [Hus04]. Figure 3.5: Geometric interpreta- tion of P \u00E2\u0088\u0097Q on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). Figure 3.6: Geometric interpreta- tion of P \u00E2\u0088\u0097 P on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). Now, this might be all well and good but what if we do not even have one rational point on the elliptic curve? To solve this problem we assume that the elliptic curve has a rational point, known as the rational point at infinity, denoted by O [ST93, SZ03, Hus04, Sil86]. That is to say we assume that there exists at least one rational point on the elliptic curve E and call that point O. Given this rational point at infinity we define O\u00E2\u0088\u0097O = O where the line at infinity meets the curve with multiplicity three at O [ST93, Sil86]. For some rational point P on E we also define O\u00E2\u0088\u0097P = P \u00E2\u0088\u0097O to be the third point of intersection between the vertical line at P and the elliptic curve E [ST93, Sil86]. Specifically, O \u00E2\u0088\u0097 P is the reflection of P about the x-axis [ST93]. We are now ready to introduce the binary group law operator + associ- ated with E(Q), where O is assumed to be the rational point at infinity on E. Assume that P and Q are rational points on E. Then we define P +Q to be the reflection of P \u00E2\u0088\u0097Q about the x-axis [ST93]. That is to say P +Q = O \u00E2\u0088\u0097 (P \u00E2\u0088\u0097Q), where O\u00E2\u0088\u0097(P \u00E2\u0088\u0097Q) is the third point of intersection on the vertical line through 20 3.3. Mordell\u00E2\u0080\u0099s Theorem P \u00E2\u0088\u0097Q with the elliptic curve E. The group law operator is visually depicted in Figure 3.7. Figure 3.7: Geometric interpretation of P +Q on the curve y2 = (x+ 1)(x2 \u00E2\u0088\u0092 4x+ 5). Furthermore, given rational points P,Q and R on the elliptic curve E we also have the following rules associated with the group law operator, previously defined [ST93]. P +Q = Q+ P (commutative) P +O = O + P = P (identity element O) P + (\u00E2\u0088\u0092P ) = (\u00E2\u0088\u0092P ) + P = O (inverses) (P +Q) +R = P + (Q+R) (associative), where \u00E2\u0088\u0092P is the notation for the reflection of P about the x-axis (i.e. \u00E2\u0088\u0092P = O \u00E2\u0088\u0097 P ). These properties can be easily proved using the definition of the + operator and the underlying \u00E2\u0088\u0097 operator (except the last one which requires a lengthy computation)[ST93]. It can also be shown that E(Q) forms a group under + [ST93]. Furthermore, Louis Mordell was able to elaborate on the structure of E(Q) [ST93]. 3.3 Mordell\u00E2\u0080\u0099s Theorem Given the preliminary background information in Abstract Algebra we recall the following theorem, which was proved in 1922 by Louis Mordell [tFE], Theorem 3.2. [ST93, Hem06, Cha06, SZ03, Hus04, Sil86, Mordell\u00E2\u0080\u0099s Theo- rem] Let E be the elliptic curve given by E : y2 = x3 + ax2 + bx, 21 3.3. Mordell\u00E2\u0080\u0099s Theorem where a and b are integers. Then the group of rational points E(Q) is a finitely generated abelian group, under the group law operator +. Proof. See [ST93, Chapter III], [Hus04, Chapter 6], or [Sil86, Chapter VIII]. In summary, Mordell\u00E2\u0080\u0099s Theorem states that for an elliptic curve of the form y2 = x3 + ax2 + bx, where a and b are integers, there exists a finite set of rational points that will generate all of the rational points on the elliptic curve using the group law operator, as defined in the previous section. As a result of Mordell\u00E2\u0080\u0099s Theorem, we can apply Theorem 1.8 to the group of rational points on an elliptic curve E, i.e. E(Q) ' Z \u00E2\u008A\u0095 \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 \u00E2\u008A\u0095 Z \u00E2\u008A\u0095 Zpv11 \u00E2\u008A\u0095 \u00C2\u00B7 \u00C2\u00B7 \u00C2\u00B7 \u00E2\u008A\u0095 Zpvss , where Z is a cyclic group with infinite order and Zpvii is a finite cyclic group of prime power order (for some prime pi and some positive integer vi) [ST93, Hem06, Cha06, SZ03, Hus04, Sil86]. This in turn leads us to the rank r of the elliptic curve E. Definition 3.3. Let E be an elliptic curve with an associated finitely gen- erated abelian group E(Q). The number of generators with infinite order in E(Q) is the rank r of the elliptic curve E. For example let E : y2 = x3 \u00E2\u0088\u0092 25x. Then using the MAGMA code in Appendix B we find that E(Q) ' Z \u00E2\u008A\u0095 Z2 \u00E2\u008A\u0095 Z2, which implies that the rank of E is 1. Note that the group of rational points on an elliptic curve is finite if and only if the rank of the elliptic curve is 0 [ST93]. Now we already stated in Theorem 2.2 that an integer n is a congruent number if the elliptic curve Y 2 = X3\u00E2\u0088\u0092n2X has non-trivial rational solutions. This in turn directly relates to the following lemma. Lemma 3.4. [Kob92, Hem06, DJS09, Ben02, Kra86, NW93, Nem98, Kob92] A positive integer n is a congruent number if and only if the elliptic curve E : y2 = x3 \u00E2\u0088\u0092 n2x has rank at least 1. 22 3.3. Mordell\u00E2\u0080\u0099s Theorem Proof. See [Kob92, Chapter I Section 9], [Hem06, Chapter 2 Section 10], [Cha06], or [Kob92, Chapter I Section 9]. We provide the following example as reinforcement, since Lemma 3.4 will be a useful tool in later chapters. Example 3.5. Consider, E := y2 = x3 \u00E2\u0088\u0092 62x. Using the MAGMA code in Appendix B we find that the rank of E is 1, thereby verifying that 6 is a congruent number by Lemma 3.4. In the next chapter we present the method that we will be using to determine the rank of an elliptic curve. 23 Chapter 4 Rank Calculation 4.1 Method of 2-descent Given an elliptic curve of the form y2 = x3 + ax2 + bx with a, b \u00E2\u0088\u0088 Z we can sometimes determine the rank of the curve using the 2-descent method described in this section [ST93]. We start by defining E : y2 = x3 + ax2 + bx and E : y2 = x3 + ax2 + bx where a = \u00E2\u0088\u00922a and b = a2 \u00E2\u0088\u0092 4b [ST93]. The process of determining the rank of E requires that we look at both curves E and E. So let \u00CE\u0093 be the rational points on E and let \u00CE\u0093 be the rational points on E. Then for the multiplicative group of non-zero rational numbers, denoted by Q\u00E2\u0088\u0097, and the subgroup of squares of Q\u00E2\u0088\u0097, namely Q\u00E2\u0088\u00972 = { u2 : u \u00E2\u0088\u0088 Q\u00E2\u0088\u0097}, define the homomorphisms [ST93] \u00CE\u00B1 : \u00CE\u0093 \u00E2\u0088\u0092\u00E2\u0086\u0092 Q \u00E2\u0088\u0097 Q\u00E2\u0088\u00972 and \u00CE\u00B1 : \u00CE\u0093 \u00E2\u0088\u0092\u00E2\u0086\u0092 Q \u00E2\u0088\u0097 Q\u00E2\u0088\u00972 such that \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y) and \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y). Then the rank r, of the elliptic curve E, satisfies the following equation [ST93]: 2r = |\u00CE\u00B1(\u00CE\u0093)| \u00C2\u00B7 |\u00CE\u00B1(\u00CE\u0093)| 4 , where | \u00C2\u00B7 | denotes the cardinality of set (i.e. the number of elements in the set). Using the definition of \u00CE\u00B1(\u00CE\u0093) and \u00CE\u00B1(\u00CE\u0093), to find elements in each set we 24 4.1. Method of 2-descent need to find rational numbers on the respective elliptic curves where the x coordinates are distinct modulo squares. As another perspective, Silverman and Tate state that the group \u00CE\u00B1(\u00CE\u0093) consists, modulo Q\u00E2\u0088\u00972, of at least 1 and b. They also state that \u00CE\u00B1(\u00CE\u0093) contains all divisors b1 of b, as long as b1 6\u00E2\u0089\u00A1 1, b(modQ\u00E2\u0088\u00972) and the equation N2 = b1M 4 + aM2e2 + b2e 4 where b = b1b2 has an integral solution (N,M, e) \u00E2\u0088\u0088 Z, with the restriction that M 6= 0, e 6= 0 and gcd(M, e) = gcd(N, e) = gcd(b1, e) = gcd(b2,M) = gcd(M,N) = 1 [ST93]. Similarly, the group \u00CE\u00B1(\u00CE\u0093) consists, modulo Q\u00E2\u0088\u00972, of 1, b, and all divisors b1 of b, as long as b1 6\u00E2\u0089\u00A1 1, b(modQ\u00E2\u0088\u00972) and the equation N2 = b1M 4 + aM2e2 + b2e 4 where b = b1 b2 = b1M 4 \u00E2\u0088\u0092 2aM2e2 + b2e4 since a = \u00E2\u0088\u00922a has an integral solution (N,M, e) \u00E2\u0088\u0088 Z [ST93]. We also require that M 6= 0, e 6= 0 and gcd(M, e) = gcd(N, e) = gcd(b1, e) = gcd(b2,M) = gcd(M,N) = 1 [ST93]. Altogether, this gives us a method for determining the rank of E, pro- vided that we are able to determine if each of the curves generated by the divisors of b and b have solutions or not. It is also important to note that calculating the rank of an elliptic curve using the 2-descent method can be rather time consuming, depending on how many square free divisors there are of b and b. In addition, we can apply the following theorem to easily determine all of the rational points of finite order on E. Theorem 4.1. [ST93, Cha06, Sil86, Nagell-Lutz Theorem] Let E : y2 = f(x) = x3 + ax2 + bx+ c be an elliptic curve with integer coefficients; and let D be the discriminant of the cubic polynomial f(x) D = \u00E2\u0088\u00924a3c+ a2b2 + 18abc\u00E2\u0088\u0092 4b3 \u00E2\u0088\u0092 27c3. Let P = (x, y) be a rational point of finite order, also known as a torsion point. Then x and y are integers and either y = 0, in which case P has order two, or else y divides D. 25 4.2. Worked Example Proof. See [ST93, Chapter II], or [Sil86, Chapter VIII]. We note that the Nagell-Lutz\u00E2\u0080\u0099 Theorem is not an if and only if statement [ST93]. That is to say for each of the rational points P = (x, y) that satisfy y = 0 or y divides D we need to check that there exists an integer n \u00E2\u0089\u00A5 1 such that nP = 0, in order to verify that P has finite order [ST93]. All in all, using the 2-descent (if successful) and the possible points of finite order from Nagell-Lutz\u00E2\u0080\u0099 Theorem we can fully determine the group structure for E(Q). 4.2 Worked Example To re-enforce the 2-descent method we provide the following example. Example 4.2. Determine the rank of y2 = x3 \u00E2\u0088\u0092 232x. Solution: Using the 2-descent method we need to find all elements of \u00CE\u00B1(\u00CE\u0093) and \u00CE\u00B1(\u00CE\u0093) when a = 0, b = \u00E2\u0088\u0092232, a = 0 and b = 4 \u00C2\u00B7 (232). Given b = b1b2, we know that \u00CE\u00B1(\u00CE\u0093) contains b(modQ\u00E2\u0088\u00972), and b1(modQ\u00E2\u0088\u00972) when there exist N,M, e \u00E2\u0088\u0088 Z such that N2 = b1M 4 + aM2e2 + b2e 4 with M 6= 0. In our case b = \u00E2\u0088\u0092232 such that the divisors of b are \u00C2\u00B11,\u00C2\u00B123 and \u00C2\u00B1232. However, 232 \u00E2\u0089\u00A1 1(modQ\u00E2\u0088\u00972) and \u00E2\u0088\u0092232 \u00E2\u0089\u00A1 \u00E2\u0088\u00921(modQ\u00E2\u0088\u00972) so we need only consider b1 = 1,\u00E2\u0088\u00921, 23,\u00E2\u0088\u009223. Since, b(modQ\u00E2\u0088\u00972) and 1 are automatically in \u00CE\u00B1(\u00CE\u0093) we get that \u00E2\u0088\u00921, 1 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093). For the remaining b1 values consider the following equations: (i) N2 = \u00E2\u0088\u009223M4 + 23e4 (ii) N2 = 23M4 \u00E2\u0088\u0092 23e4. Clearly, the solution (N,M, e) = (0, 1, 1) satisfies both of the above equa- tions such that \u00CE\u00B1(\u00CE\u0093) = {1,\u00E2\u0088\u00921, 23,\u00E2\u0088\u009223} (4.1) \u00E2\u0087\u0092 |\u00CE\u00B1(\u00CE\u0093)| = 4 (4.2) Similarly, for b = b1 b2 we know that \u00CE\u00B1(\u00CE\u0093) contains b(modQ\u00E2\u0088\u00972), and b1(modQ\u00E2\u0088\u00972) when the equation N2 = b1M 4 + aM2e2 + b2e 4 26 4.2. Worked Example has an integral solution with M 6= 0. Now, upon removing squares we get that the square free divisors of b are \u00C2\u00B11,\u00C2\u00B12,\u00C2\u00B123 and \u00C2\u00B146. Since, b(modQ\u00E2\u0088\u00972) is automatically in \u00CE\u00B1(\u00CE\u0093) we get that 1 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093). We now consider the following equations: (i) N2 = \u00E2\u0088\u0092M4 \u00E2\u0088\u0092 4 \u00C2\u00B7 232e4 (ii) N2 = 2M4 + 2 \u00C2\u00B7 232e4 (iii) N2 = \u00E2\u0088\u00922M4 \u00E2\u0088\u0092 2 \u00C2\u00B7 232e4 (iv) N2 = 23M4 + 4 \u00C2\u00B7 23e4 (v) N2 = \u00E2\u0088\u009223M4 \u00E2\u0088\u0092 4 \u00C2\u00B7 23e4 (vi) N2 = 2 \u00C2\u00B7 23M4 + 2 \u00C2\u00B7 23e4 (vii) N2 = \u00E2\u0088\u00922 \u00C2\u00B7 23M4 \u00E2\u0088\u0092 2 \u00C2\u00B7 23e4. Clearly, N2 \u00E2\u0089\u00A5 0 and we requireM 6= 0 such that Equations (i), (iii), (v), (vii) will not have any integer solutions and can therefore be eliminated. Using the MAPLETM code in Appendix A we find that (N,M, e) = (410, 17, 1) is a valid solution for Equation (ii) such that 2 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093). So far we have that \u00CE\u00B1(\u00CE\u0093) = {1, 2} . However, the remaining equations, (iv) and (vi), are a little bit harder to solve, namely because they actually do not contain solutions. Our approach will be to show that one of the remaining equations does not have a solution. Then we can apply the fact that the |\u00CE\u00B1(\u00CE\u0093)| must be a power of 2 such that the other equation cannot have solutions either. From Equation (iv) we see that N2 \u00E2\u0089\u00A1 3M4(mod 4) \u00E2\u0087\u0092 N2 \u00E2\u0089\u00A1 3(mod 4), since the gcd(b2,M) = 1 we must have that M is odd and not divisible by 4 such that M4 \u00E2\u0089\u00A1 1(mod 4). However, N2 \u00E2\u0089\u00A1 3(mod 4) has no solutions, so we can conclude that N2 = 23M4+4 \u00C2\u00B723e4 has no solutions. This in turn allows us to conclude that Equation (vi) must also have no solutions. Therefore, \u00CE\u00B1(\u00CE\u0093) = {1, 2} (4.3) \u00E2\u0087\u0092 |\u00CE\u00B1(\u00CE\u0093)| = 2 (4.4) 27 4.2. Worked Example so that 2r = |\u00CE\u00B1(\u00CE\u0093)| \u00C2\u00B7 |\u00CE\u00B1(\u00CE\u0093)| 4 = 4 \u00C2\u00B7 2 4 = 21. Thus, the rank of y2 = x3 \u00E2\u0088\u0092 232x is 1. Note that we could have also determined rational points on E or E such that the x coordinate, modulo squares, would be in \u00CE\u0093 or \u00CE\u0093, respectively. 28 Chapter 5 Congruent Number Curves of Moderate Rank The purpose of this chapter is to present an infinite family of congruent number elliptic curves having moderate rank. In our case, the term moderate rank means rank at least three. In Section 5.1 we present the main theorem for this chapter, as well as some preliminary results, followed by Section 5.2 which contains the proof to the main theorem. Now, it may not seem very beneficial to find an infinite family of con- gruent number elliptic curve with rank at least three but it turns out that even general elliptic curves with rank at least three are rare. Using a typical sample set of all elliptic curves A. Brumer and O. McGuinness studied the ranks of 310716 elliptic curves to find that only 4.08% of these curves have rank at least 3 [BM90]. We also note that the largest known rank for a congruent number elliptic curve is 6 [DJS09], which re-enforces the signifi- cance of finding infinitely many congruent number elliptic curves with rank at least 3. 5.1 Rank Three Results To start, we present the main result for this chapter in the following theorem. Theorem 5.1. [JS] The curve w2 = t4 + 14t2 + 4 has infinitely many rational points. Let (t, w) with t 6= 0 be one of them. Set t = u/v where u and v are integers with gcd(u, v) = 1. Define the positive integer n by n = 6(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4). (5.1) Then the congruent number elliptic curve y2 = x(x2 \u00E2\u0088\u0092 n2) has rank at least three. 29 5.1. Rank Three Results We provide the following example, not as proof of Theorem 5.1 but to illustrate the results. Example 5.2. The point (t, w) = ( 1 2 , 11 4 ) is on the curve w2 = t4 + 14t2 + 4. Then by Theorem 5.1 let u = 1 and v = 2 such that n = 42486. Using the MAGMA code in Appendix B we find that the curve y2 = x3 \u00E2\u0088\u0092 424862x has rank of at least 3. Note that the actual output from MAGMA is \u00E2\u0080\u009CWarn- ing: rank computed (3) is only a lower bound (It may still be correct, though)\u00E2\u0080\u009D. However, since Theorem 5.1 is only stating a rank of at least three these results do coincide with the theorem. Before we prove Theorem 5.1 in Section 5.2 we will need the following helpful lemmas. Lemma 5.3. [JS] If u and v are integers with gcd(u, v) = 1 then the quan- tities (i) \u00C2\u00B16(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) (ii) \u00C2\u00B16(u4 + 2u2v2 + 4v4) (iii) \u00C2\u00B12(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) (iv) \u00C2\u00B12(u4 + 2u2v2 + 4v4) are not equal to squares in Q. Proof. The proof for each case is similar so we will prove Case i) in detail and summarize the other three cases accordingly. Case i) Assume that u and v are integers such that the gcd(u, v) = 1. Then we need to show that \u00C2\u00B1 6(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) (5.2) is not a square in Q. The only subcase that we need to consider is the subcase where u is even and v is odd, since we require that gcd(u, v) = 1 and we see that the other three subcases either cannot 30 5.1. Rank Three Results occur or by properties of even and odd addition and multiplication will result in 2 \u00E2\u0080\u0096 \u00C2\u00B16(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4), which clearly implies that \u00C2\u00B16(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) is not a square in Q. So, assuming that u is even and v is odd we can let u = 2k for some k \u00E2\u0088\u0088 Z and apply properties of even and odd numbers such that (5.2) becomes \u00C2\u00B16(24k4 + 23k2v2 + 4v4)(24k4 + 25k2v2 + 4v4). Hence, 25 \u00E2\u0080\u0096 \u00C2\u00B16(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4), since we can factor out 25 and the remaining factors 3, (22k4 + 2k2v2 + v4) and (22k4 + 23k2v2 + v4) are odd, so that \u00C2\u00B13(22k4 + 2k2v2 + v4)(22k4 + 23k2v2 + v4) must also be odd. Therefore, \u00C2\u00B16(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) is not a square in Q. Case ii) Similarly, for integers u and v assume that the gcd(u, v) = 1 with u even and v odd. Then 23 \u00E2\u0080\u0096 \u00C2\u00B16(u4 + 2u2v2 + 4v4), so that \u00C2\u00B16(u4 + 2u2v2 + 4v4) is not a square in Q. Case iii) Once again, assume that u and v are integers such that gcd(u, v) = 1 with u even and v odd. Then 25 \u00E2\u0080\u0096 \u00C2\u00B12(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4), so that \u00C2\u00B12(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) is not a square in Q. Case iv) Lastly, assume that u and v are integers such that the gcd(u, v) = 1 with u even and v odd. Then 23 \u00E2\u0080\u0096 \u00C2\u00B12(u4 + 2u2v2 + 4v4), so that \u00C2\u00B12(u4 + 2u2v2 + 4v4) is not a square in Q. 31 5.1. Rank Three Results Lemma 5.4. [JS] If u and v are nonzero integers with gcd(u, v) = 1 then the quantities (i) \u00C2\u00B1(u4 + 2u2v2 + 4v4) (ii) \u00C2\u00B1(u4 + 8u2v2 + 4v4) are not equal to squares in Q. Proof. Clearly, if u and v are nonzero integers with gcd(u, v) = 1 then the only possible way for any of the quantities (i) \u00C2\u00B1(u4 + 2u2v2 + 4v4) (ii) \u00C2\u00B1(u4 + 8u2v2 + 4v4) to be squares would occur in the positive cases of (i) and (ii). So, by way of contradiction assume that one of the quantities is a square such that (u4 + 2u2v2 + 4v4) = z21 or (u4 + 8u2v2 + 4v4) = z22 , for some z1, z2 \u00E2\u0088\u0088 Z. Equivalently, either u2 v6 (u4 + 2u2v2 + 4v4) = z21 u2 v6 \u00E2\u0087\u0092 u 6 v6 + 2u4 v4 + 4u2 v2 = z21u 2 v6 or u2 v6 (u4 + 8u2v2 + 4v4) = z22 u2 v6 \u00E2\u0087\u0092 u 6 v6 + 8u4 v4 + 4u2 v2 = z22u 2 v6 . Now, let x = u 2 v2 and yi = ziu v3 , for i = 1, 2, be the rational points on the above curves. Then by substitution either x3 + 2x2 + 4x = y21 (5.3) 32 5.1. Rank Three Results or x3 + 8x2 + 4x = y22, (5.4) respectively, such that (5.3) and (5.4) are elliptic curves by definition. Using the MAGMA code in Appendix B we find that the rank of both (5.3) and (5.4) is 0. This in turn gives us that there is a finite number of rational points on (5.3) and (5.4), by Theorem 3.2. Once again using the MAGMA code in Appendix B we find that the only finite rational point on (5.3) is (x, y) = (0, 0). This, however, is not a valid solution for x = u 2 v2 since u is required to be a nonzero integer. Similarly, we also have that the only finite rational points on (5.4) are (0, 0), (\u00E2\u0088\u00922, 4) and (\u00E2\u0088\u00922,\u00E2\u0088\u00924), of which none correspond to a nonzero x = u2/v2 with u and v being integers. Hence, there does not exist z1, z2 \u00E2\u0088\u0088 Z such (u4 + 2u2v2 + 4v4) or (u4 + 8u2v2 + 4v4) are squares in Q. Lemma 5.5. [JS] For integers u, v such that the gcd(u, v) = 1 we have (i) 3 - (u4 + 8u2v2 + 4v4), (ii) 3 - (u4 + 2u2v2 + 4v4). Proof. Let u, v be integers such that the gcd(u, v) = 1. Then we need to show that 3 does not divide either of the quantities (u4 + 8u2v2 + 4v4) or (u4+2u2v2+4v4). First, consider each of the quantities modulo 3 such that (u4 + 8u2v2 + 4v4) \u00E2\u0089\u00A1 (u4 + 2u2v2 + v4)(mod 3) \u00E2\u0089\u00A1 (u2 + v2)2(mod 3) and (u4 + 2u2v2 + 4v4) \u00E2\u0089\u00A1 (u4 + 2u2v2 + v4)(mod 3) \u00E2\u0089\u00A1 (u2 + v2)2(mod 3). Note that both of the quantities are congruent to (u2 + v2)2 modulo 3. It remains to show that 3 does not divide (u2 + v2)2. 33 5.1. Rank Three Results By way of contradiction assume that 3 does divide (u2 + v2)2. Then (u2 + v2)2 \u00E2\u0089\u00A1 0(mod 3). (5.5) Considering squares modulo 3, for some x \u00E2\u0088\u0088 {0, 1, 2}, we have that x2 \u00E2\u0089\u00A1 0(mod 3) when x = 0 and x \u00E2\u0089\u00A1 1(mod 3) when x = 1, 2. Therefore, (u2 + v2)2 \u00E2\u0089\u00A1 0(mod 3) \u00E2\u0087\u0092 u2 + v2 \u00E2\u0089\u00A1 0(mod 3) \u00E2\u0087\u0092 u2 \u00E2\u0089\u00A1 v2 \u00E2\u0089\u00A1 0(mod 3). This however yields a contradictions since gcd(u, v) = 1, by assumption. Therefore 3 does not divide (u2 + v2)2 which is enough to show that 3 does not divide (u4 + 8u2v2 + 4v4) or (u4 + 2u2v2 + 4v4). Lemma 5.6. [JS] For integers u, v with gcd(u, v) = 1, neither of the fol- lowing quantities is equal to a square in Q. \u00C2\u00B1(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4). Proof. Clearly, \u00E2\u0088\u0092(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) cannot be a square in Q. It remains to show that (u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) (5.6) is not a square in Q. We start by showing that if (5.6) is a square in Q then this would require that each of (u4 + 2u2v2 + 4v4) (5.7) and (u4 + 8u2v2 + 4v4) (5.8) must be squares in Q. In order to show that each of the factors is a square we consider the resultant of (5.7) and (5.8), since we can use the resultant of two factors to find all possible greatest common divisors (gcds) of the two factors (as seen in [Wal05]). We can then use the possible gcds to show that each of the factors only have squares in common. Using the MAPLETM code in Appendix A we find that the resultant of these factors is 20736v16, with respect to u, and 20736u16, with respect to 34 5.2. Proof of the Main Theorem v. This in turn implies that either 2 and/or 3 must divide (5.7) and (5.8), since 20736 = 2834 and gcd(u, v) = 1. Recall from Lemma 5.5 that 3 does not divide (5.7) or (5.8). Also, if 2 does divide (5.7) and (5.8) then this would imply that u is even and v is odd, which directly gives us that 4 exactly divides (5.7) and 4 exactly divides (5.8). However, 4 is a square which is enough to show that if (5.6) is a square in Q then each of (5.7) and (5.8) must be squares in Q. Finally, using Lemma 5.4 we see that each of (u4 + 2u2v2 + 4v4) and (u4 + 8u2v2 + 4v4) cannot be squares in Q so that (u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) is not a square in Q. Lemma 5.7. [JS] There exist infinitely many pairs of rational numbers (t, w) such that w2 = t4 + 14t2 + 4. Proof. We need to show that there exist infinitely many rational points on the curve w2 = t4 + 14t2 + 4. (5.9) Equivalently, we can show that there exist infinitely many rational points on the elliptic curve Y 2 = X3 \u00E2\u0088\u0092 6588X + 39312, since the MAPLETM code in Appendix A shows that the two curves are bi- rationally equivalent. Recall that an elliptic curve contains infinitely many rational points if the rank of the elliptic curve is at least 1. Using the MAGMA code in Appendix B we find that rank of Y 2 = X3\u00E2\u0088\u00926588X+39312 is 1. Hence, w2 = t4 + 14t2 + 4 contains infinitely many rational points. 5.2 Proof of the Main Theorem We are now prepared to prove Theorem 5.1. Proof. Recall from Lemma 5.7 that the curve w2 = t4 + 14t2 + 4 (5.10) 35 5.2. Proof of the Main Theorem has infinitely many rational points. Now, let (t, w) be a rational point on (5.10) with t 6= 0, and set t = u/v where u and v are integers with gcd(u, v) = 1. Then for n = 6(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) it remains to show that the curve y2 = x3 \u00E2\u0088\u0092 n2x (5.11) has rank, r, at least three. Using the 2-descent method described in Chapter 4 it is enough to show that 2r \u00E2\u0089\u00A5 8, since this would imply that r is at least 3. More specifically, since 2r = |\u00CE\u00B1(\u00CE\u0093)| \u00C2\u00B7 |\u00CE\u00B1(\u00CE\u0093)| 4 we will show that |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 32. As described in Chapter 4 let \u00CE\u0093 be the group of rational points on the elliptic curve y2 = x3 \u00E2\u0088\u0092 n2x and let \u00CE\u0093 be the group of rational point on the elliptic curve y2 = x3 + 4n2x. Then we need to find as many elements of \u00CE\u00B1(\u00CE\u0093) and \u00CE\u00B1(\u00CE\u0093) where \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y) and \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y), as defined in Chapter 4. By definition, \u00CE\u00B1(\u00CE\u0093) contains b(modQ\u00E2\u0088\u00972). Also recall that, for b = b1b2, \u00CE\u00B1(\u00CE\u0093) contains b1(modQ\u00E2\u0088\u00972) when the equation N2 = b1M 4 + aM2e2 + b2e 4 has a solution with M 6= 0. In our case b = \u00E2\u0088\u0092n2 such that the divisors of b include \u00C2\u00B11,\u00C2\u00B1n and \u00C2\u00B1n2, plus the divisors of n. However, n2 \u00E2\u0089\u00A1 1(modQ\u00E2\u0088\u00972) and \u00E2\u0088\u0092n2 \u00E2\u0089\u00A1 \u00E2\u0088\u00921(modQ\u00E2\u0088\u00972) such that \u00E2\u0088\u00921, 1 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093), and we need to only consider b1 = n,\u00E2\u0088\u0092n and all other divisors of n. So, for b1 = n and b1 = \u00E2\u0088\u0092n consider the following equations: 36 5.2. Proof of the Main Theorem (i) N2 = nM4 \u00E2\u0088\u0092 ne4 (ii) N2 = \u00E2\u0088\u0092nM4 + ne4. Clearly, the solution (N,M, e) = (0, 1, 1) satisfies both of the above equa- tions. Furthermore, Lemma 5.3(i) shows that n is not a square such that n,\u00E2\u0088\u0092n \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093). So far we have \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 {1,\u00E2\u0088\u00921, n,\u00E2\u0088\u0092n} . To find more elements of \u00CE\u00B1(\u00CE\u0093) we consider the following non-torsion points, that satisfy y2 = x3 \u00E2\u0088\u0092 n2x, P1 = (x, y) = (\u00E2\u0088\u009236u2v2(u4 + 8u2v2 + 4v4), 36uv(u2 \u00E2\u0088\u0092 2v2)(u4 + 8u2v2 + 4v4)2) , P2 = (x, y) = ( 12(u4 + 2u2v2 + 4v4)2, 36(u4 \u00E2\u0088\u0092 4v4)(u4 + 2u2v2 + 4v4)2) , P3 = (x, y) = (\u00E2\u0088\u009236u2v2(u4 + 2u2v2 + 4v4), 36uv3(u4 + 2u2v2 + 4v4)2 w). It is easy to show that P1 and P2 are on y 2 = x3 \u00E2\u0088\u0092 n2x, just make direct substitutions for x and y. For P3 we need to recall that w 2 = t4 + 14t2 + 4 and t = u/v such that y2 = (\u00E2\u0088\u009236u2v2(u4 + 8u2v2 + 4v4))3 \u00E2\u0088\u0092 n2(\u00E2\u0088\u009236u2v2(u4 + 8u2v2 + 4v4)) \u00E2\u0087\u0092 y2 = 1296u2v2(u4 + 14u2v2 + 4v4)(u4 + 2u2v2 + 4v4)4 \u00E2\u0087\u0092 y2v4 = 362u2v2 ((u v )4 + 14 (u v )2 + 4 ) (u4 + 2u2v2 + 4v4)4 \u00E2\u0087\u0092 y2v4 = 362u2v2(t4 + 14t2 + 4)(u4 + 2u2v2 + 4v4)4 \u00E2\u0087\u0092 y2 = 362u2v6w2(u4 + 2u2v2 + 4v4)4 \u00E2\u0087\u0092 y = \u00C2\u00B136uv3w(u4 + 2u2v2 + 4v4)2. It remains to show that P1, P2 and P3 are not congruent modulo Q\u00E2\u0088\u00972 to each other or any of the other points already in \u00CE\u00B1(\u00CE\u0093). To start, consider \u00CE\u00B1(P1) \u00E2\u0089\u00A1 \u00E2\u0088\u0092(u4 + 8u2v2 + 4v4)(modQ\u00E2\u0088\u00972). Then using Lemma 5.4 (ii) we see that \u00C2\u00B1(u4 + 8u2v2 + 4v4) is not a square in Q such that \u00E2\u0088\u0092(u4+8u2v2+4v4) \u00E2\u008A\u0086 \u00CE\u00B1(\u00CE\u0093), since clearly \u00C2\u00B1(u4+8u2v2+4v4) is not congruent to \u00C2\u00B11 or \u00C2\u00B1n. Using Lemma 5.3(ii) we see that \u00C2\u00B1\u00CE\u00B1(P1)n is not a square in Q either such that \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 S1 + {\u00C2\u00B11,\u00C2\u00B1n,\u00C2\u00B1(u4 + 8u2v2 + 4v4),\u00C2\u00B1n(u4 + 8u2v2 + 4v4)} . 37 5.2. Proof of the Main Theorem Now, consider \u00CE\u00B1(P2) \u00E2\u0089\u00A1 3(modQ\u00E2\u0088\u00972). Then clearly 3 is not a square in Q, however it remains to show that \u00CE\u00B1(P2) 6\u00E2\u0089\u00A1 s(modQ\u00E2\u0088\u00972) for all s \u00E2\u0088\u0088 S1. So, consider the case where the congruence holds then there would exist integers c1, c2 with c1, c2 \u00E2\u0088\u0088 {0, 1} such that \u00CE\u00B1(P2) \u00E2\u0089\u00A1 \u00C2\u00B1nc1 ( u4 + 8u2v2 + 4v4 )c2 (modQ\u00E2\u0088\u00972) \u00E2\u0087\u0092 3 \u00E2\u0089\u00A1 \u00C2\u00B1nc1 (u4 + 8u2v2 + 4v4)c2 (modQ\u00E2\u0088\u00972). Upon comparing powers of 3 on both sides of the congruence we see, by Lemma 5.5(i), that c1 = 1. Hence, 3 \u00E2\u0089\u00A1 \u00C2\u00B1n(modQ\u00E2\u0088\u00972) when c2 = 0 (5.12) and/or 3 \u00E2\u0089\u00A1 \u00C2\u00B1n (u4 + 8u2v2 + 4v4) (modQ\u00E2\u0088\u00972) when c2 = 1. (5.13) For congruence (5.12) we deduce that 2(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) must be a square modulo Q, since n = 6(u4+2u2v2+4v4)(u4+8u2v2+4v4). However, by Lemma 5.3(iii) we know that 2(u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) is not a square modulo Q. Similarly, in order for congruence (5.13) to hold we would require that 2(u4 + 2u2v2 + 4v4) be a square modulo Q, which contradicts Lemma 5.3(iv). Altogether, this gives us that \u00CE\u00B1(P2) 6\u00E2\u0089\u00A1 s(modQ\u00E2\u0088\u00972) for all s \u00E2\u0088\u0088 S1 such that \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 S2 +< 1, n, (u4 + 8u2v2 + 4v4), 3 >, where the elements of S2 are generators for the subgroup of \u00CE\u00B1(\u00CE\u0093) and |S2| = 16. Finally, consider \u00CE\u00B1(P3) = \u00E2\u0088\u0092(u4 + 2u2v2 + 4v4)(modQ\u00E2\u0088\u00972). In this case, we need to show that \u00CE\u00B1(P3) 6\u00E2\u0089\u00A1 s(modQ\u00E2\u0088\u00972) for all s \u00E2\u0088\u0088 S2. So, consider the case where the congruence holds then there would exist integers e1, e2 and e3 with e1, e2, e3 \u00E2\u0088\u0088 {0, 1} such that \u00CE\u00B1(P3) \u00E2\u0089\u00A1 \u00C2\u00B13e1ne2(u4 + 8u2v2 + 4v4)e3(modQ\u00E2\u0088\u00972) \u00E2\u0087\u0092 \u00E2\u0088\u0092(u4 + 2u2v2 + 4v4) \u00E2\u0089\u00A1 \u00C2\u00B13e1ne2(u4 + 8u2v2 + 4v4)e3(modQ\u00E2\u0088\u00972). 38 5.2. Proof of the Main Theorem Upon comparing powers of 2 on both sides of this congruence, as in the proof of Lemma 5.3, we deduce that 22m \u00E2\u0080\u0096 3e1ne2(u4 + 8u2v2 + 4v4)e3 , for some non-negative integer m. However, as determined in the proof of Lemma 5.3, there is an odd power of 2 exactly dividing n, such that e2 must be 0. Therefore, \u00E2\u0088\u0092 (u4 + 2u2v2 + 4v4) \u00E2\u0089\u00A1 \u00C2\u00B13e1(u4 + 8u2v2 + 4v4)e3(modQ\u00E2\u0088\u00972). (5.14) Comparing powers of 3 on both sides of (5.14) implies that e1 = 0, by Lemma 5.5. So, either \u00E2\u0088\u0092 (u4 + 2u2v2 + 4v4) \u00E2\u0089\u00A1 \u00C2\u00B11(modQ\u00E2\u0088\u00972) when e3 = 0 (5.15) and/or \u00E2\u0088\u0092(u4+2u2v2+4v4) \u00E2\u0089\u00A1 \u00C2\u00B1(u4+8u2v2+4v4)(modQ\u00E2\u0088\u00972) when e3 = 1. (5.16) For congruence (5.15) we deduce that \u00E2\u0088\u0092(u4 + 2u2v2 + 4v4) must be a square modulo Q. However, by Lemma 5.4(i) we know that \u00E2\u0088\u0092(u4 + 2u2v2 + 4v4) is not a square modulo Q. Similarly, in order for congruence (5.16) to hold we would require that (u4 + 2u2v2 + 4v4)(u4 + 8u2v2 + 4v4) be a square modulo Q, which is contradicted by Lemma 5.6. Altogether, this gives us that \u00CE\u00B1(P3) 6\u00E2\u0089\u00A1 s(modQ\u00E2\u0088\u00972) for all s \u00E2\u0088\u0088 S2 such that \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 S2 \u00E2\u0088\u00AA {\u00E2\u0088\u0092(u4 + 2u2v2 + 4v4)} , which contains at least 17 elements. Then since |\u00CE\u00B1(\u00CE\u0093)| must be a power of 2 we have that |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 32, which is enough to show that the rank of y2 = x3 \u00E2\u0088\u0092 n2x is at least 3. 39 Chapter 6 The Distribution of Congruent Numbers Previously, it has been shown that every congruence class modulo eight contains infinitely many congruent numbers [Cha06]. That is to say, modulo 8 there exist infinitely many positive integers n such that the rank of the associated elliptic curve, y2 = x3 \u00E2\u0088\u0092 n2x, is at least 1. In general, it has been shown that every congruence class modulo m contains infinitely many congruent numbers for any integer m greater than 1 [Ben02]. The goal of this chapter is to prove a similar result involving congruent numbers where the rank of the associated elliptic curve is at least 2. 6.1 Rank Two Results We begin by presenting the main theorem that will be proved in Section 6.2. Theorem 6.1. [JS10] If m > 1 is an integer then any congruence class modulo m contains infinitely many congruent numbers n, inequivalent mod- ulo squares, such that the rank of y2 = x(x2 \u00E2\u0088\u0092 n2) is greater than or equal to 2. We note that in Theorem 6.1, n is any positive integer not just a square- free integer. Before we prove Theorem 6.1 we need to recall the following theorem, that was conjectured by Mordell and then later proved by Faltings in 1983, where the term genus is a well-defined quantity associated with any algebraic curve [Hus04]. Theorem 6.2. [Hus04, Remark 6.4] [Faltings Theorem] Let E be a non- singular curve (i.e. all the roots are distinct) of genus strictly greater than 1. Then the set E(Q) of rational points on E is finite. Proof. See [Fal83]. We also need to prove the following lemma. 40 6.1. Rank Two Results Lemma 6.3. [JS10] Let t 6\u00E2\u0088\u0088 {0,\u00E2\u0088\u00921,\u00E2\u0088\u00921/3} be a rational number and define f(t) by f(t) = t(t+ 1)(3t+ 1)(9t4 + 24t3 + 26t2 + 8t+ 1). (6.1) Then the elliptic curve y2 = x3 \u00E2\u0088\u0092 f(t)2x (6.2) has rank greater than or equal to 2, for all but finitely many values of t. Proof. Given a rational number t 6= 0,\u00E2\u0088\u00921,\u00E2\u0088\u00921/3 let r be the rank of the elliptic curve y2 = x3 \u00E2\u0088\u0092 f(t)2x, (6.3) where f(t) is as described in Equation (6.1). As in Chapter 4, let \u00CE\u0093 be the group of rational points on the elliptic curve y2 = x3 \u00E2\u0088\u0092 f(t)2x and let \u00CE\u0093 be the group of rational point on the elliptic curve y2 = x3 + 4f(t)2x. Then we need to find as many elements of \u00CE\u00B1(\u00CE\u0093) and \u00CE\u00B1(\u00CE\u0093) where \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y) and \u00CE\u00B1(P ) = \u00EF\u00A3\u00B1\u00EF\u00A3\u00B2\u00EF\u00A3\u00B3 1(modQ\u00E2\u0088\u00972), for P = O, b(modQ\u00E2\u0088\u00972), for P = (0, 0), x(modQ\u00E2\u0088\u00972), for P = (x, y). Using the 2-descent method, as described in Chapter 4, we need to show that 2r \u00E2\u0089\u00A5 4, since this would imply that the rank of y2 = x3 \u00E2\u0088\u0092 f(t)2x is at least 2. More specifically, since 2r = |\u00CE\u00B1(\u00CE\u0093)| \u00C2\u00B7 |\u00CE\u00B1(\u00CE\u0093)| 4 we will show that |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 8 and |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 2. To start, recall that \u00CE\u00B1(\u00CE\u0093) contains 1 and b(modQ\u00E2\u0088\u00972), and for b = b1b2 \u00CE\u00B1(\u00CE\u0093) also contains b1(modQ\u00E2\u0088\u00972) when the equation N2 = b1M 4 + aM2e2 + b2e 4 has a solution with M 6= 0. In our case b = \u00E2\u0088\u0092f(t)2 such that 1,\u00E2\u0088\u00921 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093) by definition, since \u00E2\u0088\u0092f(t)2 \u00E2\u0089\u00A1 \u00E2\u0088\u00921(modQ\u00E2\u0088\u00972). Then we need to only consider b1 = f(t),\u00E2\u0088\u0092f(t) plus all other divisors of f(t). Now, for b1 = f(t) and b1 = \u00E2\u0088\u0092f(t) consider the following equations: 41 6.1. Rank Two Results (i) N2 = f(t)M4 \u00E2\u0088\u0092 f(t)e4 (ii) N2 = \u00E2\u0088\u0092f(t)M4 + f(t)e4. Clearly, the solution (N,M, e) = (0, 1, 1) satisfies both of the above equa- tions. It remains to show that \u00C2\u00B1f(t) is not a square in Q. So, consider Y 2 = \u00C2\u00B1f(t) = \u00C2\u00B1t(t+ 1)(3t+ 1)(9t4 + 24t3 + 26t2 + 8t+ 1). (6.4) Using the MAPLETM code in Appendix A we find that Equation (6.4) is a genus 3 curve such that by Theorem 6.2 there are only finitely many solutions. Therefore, except for finitely many t values, we have \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 {1,\u00E2\u0088\u00921, f(t),\u00E2\u0088\u0092f(t)} . To find another element of \u00CE\u00B1(\u00CE\u0093) we consider the following non-torsion point P1 = (x1, y1), that satisfies y 2 = x3 \u00E2\u0088\u0092 f(t)2x, where x1 = \u00E2\u0088\u00924t 2(t+ 1)2(3t+ 1)2(9t4 + 24t3 + 26t2 + 8t+ 1) (3t2 + 2t+ 1)2 y1 = 2t2(t+ 1)2(3t+ 1)2(3t2 \u00E2\u0088\u0092 1)(9t4 + 24t3 + 26t2 + 8t+ 1)2 (1 + 2t+ 3t2)3 . It remains to show that P1 is distinct modulo Q\u00E2\u0088\u00972, in \u00CE\u00B1(\u00CE\u0093). So, consider \u00CE\u00B1(P1) \u00E2\u0089\u00A1 \u00E2\u0088\u0092(9t4 + 24t3 + 26t2 + 8t+ 1)(modQ\u00E2\u0088\u00972). Then clearly P1 is not congruent to \u00C2\u00B11 or \u00C2\u00B1f(t) modulo Q\u00E2\u0088\u00972 as long as \u00CE\u00B1(P1) is not a square itself. However, if \u00CE\u00B1(P1) is a square in Q then we would have Y 2 = \u00C2\u00B1(9t4 + 24t3 + 26t2 + 8t+ 1). (6.5) Now, we need to only consider the positive case of (6.5), since the negative case yields no rational solutions for all rational values of t. So, consider Y 2 = (9t4 + 24t3 + 26t2 + 8t+ 1), which using the MAPLETM code in Appendix A we find is bi-rationally equivalent to Y 2 = t3 \u00E2\u0088\u0092 5616t+ 120960. (6.6) Using the MAGMA code in Appendix B we find that the rank of (6.6) is zero. Therefore, by definition, we have that there are only finitely many 42 6.2. Proof of the Main Theorem points that satisfy (6.6). This in turn gives us that, except for finitely many t values, \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 {1,\u00E2\u0088\u00921, f(t),\u00E2\u0088\u0092f(t), (9t4 + 24t3 + 26t2 + 8t+ 1), \u00E2\u0088\u0092(9t4 + 24t3 + 26t2 + 8t+ 1)}. Then since |\u00CE\u00B1(\u00CE\u0093)| must be a power of 2 we have that |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 8. Next consider b = 4f(t)2 such that 1 \u00E2\u0088\u0088 \u00CE\u00B1(\u00CE\u0093) by definition. Then we need to only consider b1 = 4f(t),\u00E2\u0088\u00924f(t), as well as the other divisors of 4f(t). We start with the following non-torsion point, P2 = (x2, y2), that satisfies y2 = x3 + 4f(t)2x, (x2, y2) = ((4t 2(9t4 + 24t3 + 26t2 + 8t+ 1), 4t2(9t4 + 24t3 + 26t2 + 8t+ 1)2). It remains to show that \u00CE\u00B1(P2) is not a square in Q. So, consider \u00CE\u00B1(P2) \u00E2\u0089\u00A1 (9t4 + 24t3 + 26t2 + 8t+ 1)(modQ\u00E2\u0088\u00972). This, however, is the same as \u00E2\u0088\u0092\u00CE\u00B1(P1) which we already showed was a square in Q\u00E2\u0088\u00972 for only finitely many rational t values. Hence, \u00CE\u00B1(\u00CE\u0093) \u00E2\u008A\u0087 {1, (9t4 + 24t3 + 26t2 + 8t+ 1)} such that |\u00CE\u00B1(\u00CE\u0093)| \u00E2\u0089\u00A5 2. Altogether, we have shown that the rank of y2 = x3 \u00E2\u0088\u0092 f(t)2x is at least 2, for finitely many rational t values, since 2r = |\u00CE\u00B1(\u00CE\u0093)| \u00C2\u00B7 |\u00CE\u00B1(\u00CE\u0093)| 4 \u00E2\u0089\u00A5 8 \u00C2\u00B7 2 4 = 4. 6.2 Proof of the Main Theorem We are now ready to prove Theorem 6.1. 43 6.2. Proof of the Main Theorem Proof. Let m be a positive integer grater than 1. Then we need to show that every congruence class modulo m contains infinitely many congruent numbers n, inequivalent modulo squares, such that the associated elliptic curve, y2 = x3 \u00E2\u0088\u0092 n2x, has rank at least 2. So, consider the integer a \u00E2\u0088\u0088 {1, 2, . . . ,m}, which is a representative of a congruence class modulo m. Then we need to show that there exist infinitely many positive integers n such that n \u00E2\u0089\u00A1 a(modm), the rank of the associated elliptic curve, y2 = x3\u00E2\u0088\u0092n2x, is at least 2 and the n\u00E2\u0080\u0099s are inequivalent modulo squares. To do this we define the integer n to be n + f(am 2x2) m2x2 , (6.7) where f is as defined in (6.1) and x = 1, 2, . . .. So, n = a(am2x2 + 1)(3am2x2 + 1)(9(am2x2)4 +24(am2x2)3 + 26(am2x2)2 + 8am2x2 + 1), which is clearly a positive integer greater than zero and clearly n \u00E2\u0089\u00A1 a(modm). Moreover, by scaling Lemma 6.3 we have that the rank of y2 = x3 \u00E2\u0088\u0092 n2x is at least 2, which also implies that n is a congruent number by Lemma 3.4. So far, since a and m are arbitrary and x = 1, 2, . . . we have shown that there exist infinitely many congruent numbers n in each congruence class modulo m such that the associated elliptic curve has rank at least 2. It remains to show that there exist infinitely many n\u00E2\u0080\u0099s that are inequivalent modulo squares. So, by way of contradiction assume that there exists a finite set of nonzero rational numbers di where i = 1, . . . k that are inequivalent modulo squares such that for each value of x in (6.7) we have f(am2x2) m2x2 = diY 2, (6.8) where Y is a rational number dependent on x (i.e. there are only finitely many n\u00E2\u0080\u0099s that are incongruent modulo squares). However, since there are an infinite number of distinct x values this would imply that there exists an infinite set of distinct points on the set of algebraic curves defined in (6.8). Using the MAPLETM code in Appendix A we find that the algebraic curves defined in (6.8) have genus 5, of which we know by Theorem 6.2 that for each di there are only finitely many points satisfying each equation. Therefore, there must exist infinitely many numbers n that are inequivalent modulo squares. 44 6.2. Proof of the Main Theorem Altogether, we have that there exist infinitely many congruent numbers n, inequivalent modulo squares, in each congruence class modulo m such that the associated elliptic curve has rank at least 2. 45 Chapter 7 Future Work Future work in this area would involve congruent numbers, elliptic curves or congruent number elliptic curves. In the field of congruent numbers there are several topics that people may find interesting. For instance, one might look at finding another equiv- alent definition for a congruent number, as in Theorem 2.2. As we saw in Chapter ?? the more we know about congruent numbers the closer we are to solving the congruent number problem. The definition of a congruent number involving elliptic curves has brought us significantly closer to solv- ing the congruent number problem and without this discovery we may not know nearly as much about congruent numbers [Cip09]. Another interesting topic of consideration would be to find more families of congruent numbers, similar to Chapter 2 Section 2.2. One might also be interested in looking closer at the statistical distribution of congruent numbers, as Rubinstein does in Chapter 1. Specifically, given the results found in this thesis one might try to prove that for an integer m > 1 any congruence class modulo m contains infinitely many congruent numbers n, inequivalent modulo squares, such that the rank of y2 = x(x2 \u00E2\u0088\u0092 n2) is greater than or equal to 3. Suffice it to say, there are still many open questions surrounding congruent numbers making this field of study interesting and beneficial to any inquisitive person. When discussing elliptic curves it is important to remember that we do not necessarily need to talk about congruent number elliptic curves. A lot of work has been done with elliptic curves in general, however there are still some open problems surrounding this topic, including the BSD conjecture [Cip09]. The BSD conjecture is widely believed to be true and would provide significant results, including the verification of Tunnell\u00E2\u0080\u0099s Theorem [Hem06]. Now for some future work involving congruent number elliptic curves. Based on the results found in Chapter 5 one might be interested in finding more families of congruent number elliptic curves with moderate rank, as well as individual congruent number elliptic curves with significant rank. Currently, the highest known rank of a congruent number elliptic curve is only 7 but it would be interesting to find higher ranks and possibly work 46 Chapter 7. Future Work with those individual curves to find families of congruent number elliptic curves with equivalent rank [DJS09]. In closing, I would like to say that the study of congruent numbers and congruent number elliptic curves is significant to many researchers for the simple reason that they are interesting, and that there are still a lot of unanswered questions [Cip09]. I thoroughly enjoyed working with congru- ent numbers and congruent number elliptic curves, and I look forward to continuing my work in this area. 47 Bibliography [AC74] R. Alter and T. B. Curtz. A note on congruent numbers. Math. Comp., 28(125):303\u00E2\u0080\u0093305, 1974. \u00E2\u0086\u0092 pages 2, 7, 8, 12, 13, 14, 15, 16 [Alt80] R. Alter. The congruent number problem. Amer. Math. 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Birch and swinnerton-dyer conjecture. http://en.wikipedia.org/wiki/Birch and Swinnerton- Dyer conjecture. \u00E2\u0086\u0092 pages 21 [Tun83] J. Tunnell. A classical diophantine problem and modular forms of weight 3/2. Invent. Math., 72:323\u00E2\u0080\u0093334, 1983. \u00E2\u0086\u0092 pages 2, 3, 7, 8 [Wal05] P.G. Walsh. Squares in lucas sequences with rational roots. Inte- gers, 5(A15):1\u00E2\u0080\u00938, 2005. \u00E2\u0086\u0092 pages 34 50 Appendix A MAPLE Code The calculations in this section were executed using MAPLE 12 where > is used to reference MAPLE input and center text is used to reference MAPLE output. A.1 Worked Example Calculations from Section4.2 Finding the rank of y2 = x(x2 \u00E2\u0088\u0092 232). > e3\u00CE\u0093 := N 2 + 23M4 \u00E2\u0088\u0092 23e4 : > seq(seq(isolve(e3\u00CE\u0093]),M = 1..10), e = 1..10); {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} , {N = 0} > soln := subs(N = 0, e3\u00CE\u0093]); soln := 23M4 \u00E2\u0088\u0092 23 e4 > seq(isolve(soln), e = 1..10); {M = 1} , {M = \u00E2\u0088\u00921} , {M = 2} , {M = \u00E2\u0088\u00922} , {M = 3} , {M = \u00E2\u0088\u00923} , {M = \u00E2\u0088\u00924} , {M = 4} , {M = \u00E2\u0088\u00925} , {M = 5} , {M = \u00E2\u0088\u00926} , {M = 6} , {M = \u00E2\u0088\u00927} , {M = 7} , {M = \u00E2\u0088\u00928} , {M = 8} , {M = 9} , {M = \u00E2\u0088\u00929} , {M = \u00E2\u0088\u009210} , {M = 10} Hence there exists a soln satisfying the gcd conditions, namely (0, 1, 1). > with(numtheory): > divisors(2116); {1, 2, 4, 23, 46, 92, 529, 1058, 2116} 51 A.2. Chapter 5 Calculations > e32 := N 2 \u00E2\u0088\u0092 2M4 \u00E2\u0088\u0092 1058e4 : > seq(seq(isolve(e32),M = 1..50), e = 1..50); {N = \u00E2\u0088\u0092410} , {N = 410} , {N = \u00E2\u0088\u00921640} , {N = 1640} , {N = \u00E2\u0088\u00929430} , {N = 9430} , {N = \u00E2\u0088\u009237720} , {N = 37720} > soln1 := subs(N = 410, e32); soln1 := 168100 \u00E2\u0088\u0092 2 M 4 \u00E2\u0088\u0092 1058 e4 > seq(isolve(soln1), e = 1..10); {M = \u00E2\u0088\u009217} , {M = 17} > factor(subs(M = 17, soln1)); \u00E2\u0088\u00921058 (e\u00E2\u0088\u0092 1) (e+ 1) (e2 + 1) Hence, the solution (410, 17, 1) satisfies the equation and the gcd conditions. A.2 Chapter 5 Calculations Finding the resultants for the factors defined in Lemma 5.6. > f := u4 + 2u2v2 + 4v4 : > g := u4 + 8u2v2 + 4v4 : > resultant(f, g, u); 20736v16 > resultant(f, g, v); 20736u16 > ifactor(20736); (2)8(3)4 Converting w2 = t4 + 14t2 + 4 to y2 = x3 \u00E2\u0088\u0092 6588x + 39312 in the proof of Lemma 5.7. > with(algcurves) : > f := t4 + 14t2 + 4\u00E2\u0088\u0092 w2 : > genus(f, t, w); 1 52 A.3. Chapter 6 Calculations > newf := expand(subs(t = 3t, f)); 81t4 + 126t2 + 4\u00E2\u0088\u0092 w2 > sol := Weierstrassform(newf, t, w, x, y);[ x3 \u00E2\u0088\u0092 6588x+ y2 \u00E2\u0088\u0092 39312, \u00E2\u0088\u00922(21 t2+4\u00E2\u0088\u00922w) t2 , \u00E2\u0088\u00928(4+63 t 2\u00E2\u0088\u00922w) t3 , \u00E2\u0088\u00924y 468+84x+x2 , \u00E2\u0088\u009212 \u00E2\u0088\u009230096\u00E2\u0088\u0092672x+4x2 468+84x+x2 ] > subs(x = \u00E2\u0088\u0092x, sol[1]); \u00E2\u0088\u0092x3 + 6588x+ y2 \u00E2\u0088\u0092 39312 Similarly, we can convert Y 2 = (9t4 + 24t3 + 26t2 + 8t+ 1) to Y 2 = t3 \u00E2\u0088\u0092 5616t+120960 by first scaling t by 1/3 and then applying theWeierstrassform command in MAPLETM. A.3 Chapter 6 Calculations Genus calculations corresponding to Equation (6.4) and Equation (6.8, re- spectively. > with(algcurves); > genus(x(x+ 1)(3x+ 1)(9x4 + 24x3 + 26x2 + 8x+ 1) + y2, x, y); 3 > genus(dy 2\u00E2\u0088\u0092am2x2(am2x2+1)(3am2x2+1)(9(am2x2)4+24(am2x2)3+26(am2x2)2+8am2x2+1) (m2x2) , x, y); 5 53 Appendix B MAGMA Code The calculations in this section were done using an online trial version of MAGMA Version 2.16-10, which can be found at http://magma.maths.usyd.edu.au/calc/. B.1 Elliptic Curve Calculations Calculating the rank of the elliptic curve y2 = x3 \u00E2\u0088\u0092 25x defined on page 22. Input: E := EllipticCurve([\u00E2\u0088\u009225, 0]); MordellWeilGroup(E); Rank(E); Output: Abelian Group isomorphic to Z/2 + Z/2 + Z Defined on 3 generators Relations: 2 \u00E2\u0088\u0097 $.1 = 0 2 \u00E2\u0088\u0097 $.2 = 0 1 Calculating the rank of the elliptic curve y2 = x3 \u00E2\u0088\u0092 62x in Example 3.5. Input: E := EllipticCurve([\u00E2\u0088\u009262, 0]); Rank(E); Output: 1 Calculating the rank of the elliptic curve y2 = x3\u00E2\u0088\u0092424862x in Example 5.2. Input: E := EllipticCurve([\u00E2\u0088\u0092424862, 0]); Rank(E); Output: 54 B.1. Elliptic Curve Calculations Warning: rank computed (3) is only a lower bound (It may still be correct, though) 3 Calculating the rank of x3 + 2x2 + 4x = y21 and x 3 + 8x2 + 4x = y22 in the proof of Lemma 5.4. Input: E1 := EllipticCurve([0, 2, 0, 4, 0]); Rank(E1); E2 := EllipticCurve([0, 8, 0, 4, 0]); Rank(E2); Output: 0 0 Finding the rational points on x3 + 2x2 + 4x = y21 and x 3 + 8x2 + 4x = y22 in the proof of Lemma 5.4. Input: E1 := EllipticCurve([0, 2, 0, 4, 0]); MordellWeilGroup(E1); RationalPoints(E1 : Bound := 1000); E2 := EllipticCurve([0, 8, 0, 4, 0]); MordellWeilGroup(E2); RationalPoints(E2 : Bound := 1000); Output: Abelian Group isomorphic to Z/2 Defined on 1 generator Relations: 2 \u00E2\u0088\u0097 $.1 = 0 {@(0 : 1 : 0), (0 : 0 : 1)@} Abelian Group isomorphic to Z/4 Defined on 1 generator Relations: 4 \u00E2\u0088\u0097 $.1 = 0 {@(0 : 1 : 0), (0 : 0 : 1), (\u00E2\u0088\u00922 : 4 : 1), (\u00E2\u0088\u00922 : \u00E2\u0088\u00924 : 1)@} Note that the torsion subgroup of E1 is isomorphic to Z/2 which implies that there are only 2 points of finite order, namely the point at infinity and (0, 0). Also note that the torsion subgroup of E2 is isomorphic to Z/4 which implies that there are only 4 points of finite order, namely the point at in- finity, (0, 0), (\u00E2\u0088\u00922, 4) and (\u00E2\u0088\u00922,\u00E2\u0088\u00924). 55 B.1. Elliptic Curve Calculations Calculating the rank of the elliptic curve Y 2 = X3 \u00E2\u0088\u0092 6588X + 39312 in the proof of Lemma 5.7. Input: E := EllipticCurve([0, 0, 0,\u00E2\u0088\u00926588, 39312]); Rank(E); Output: 1 Calculating the rank of the elliptic curve Y 2 = t3 \u00E2\u0088\u0092 5616t + 120960 in the proof of Lemma 6.3. Input: E := EllipticCurve([0, 0, 0,\u00E2\u0088\u00925616, 120960]); Rank(E); Output: 0 56"@en . "Thesis/Dissertation"@en . "2010-11"@en . "10.14288/1.0071062"@en . "eng"@en . "Mathematics"@en . "Vancouver : University of British Columbia Library"@en . "University of British Columbia"@en . "Attribution-NonCommercial-NoDerivatives 4.0 International"@en . "http://creativecommons.org/licenses/by-nc-nd/4.0/"@en . "Graduate"@en . "Congruent numbers and elliptic curves"@en . "Text"@en . "http://hdl.handle.net/2429/26993"@en .