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Computing elliptic curves over ℚ via Thue-Mahler equations and related problems Gherga, Adela 2019

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Computing elliptic curves over Q via Thue-Mahlerequations and related problemsbyAdela GhergaB.Sc. Mathematics, McMaster University, 2010M.Sc. Mathematics, McMaster University, 2012A THESIS SUBMITTED IN PARTIAL FULFILLMENTOF THE REQUIREMENTS FOR THE DEGREE OFDoctor of PhilosophyinTHE FACULTY OF GRADUATE AND POSTDOCTORALSTUDIES(Mathematics)The University of British Columbia(Vancouver)October 2019c© Adela Gherga, 2019The following individuals certify that they have read, and recommend to the Fac-ulty of Graduate and Postdoctoral Studies for acceptance, the thesis entitled:Computing elliptic curves over Q via Thue-Mahler equations and re-lated problemssubmitted by Adela Gherga in partial fulfillment of the requirements for the degreeof Doctor of Philosophy in Mathematics.Examining Committee:Michael A. Bennett, MathematicsSupervisorAndrew D. Rechnitzer, MathematicsSupervisory Committee MemberJoel Friedman, Computer ScienceUniversity ExaminerNicholas Harvey, Computer ScienceUniversity ExaminerAdditional Supervisory Committee Members:Greg Martin, MathematicsSupervisory Committee MemberJohn E. Cremona, MathematicsExternal ExamineriiAbstractWe present a practical and efficient algorithm for solving an arbitrary Thue-Mahlerequation. This algorithm uses explicit height bounds with refined sieves, combin-ing Diophantine approximation techniques of Tzanakis-de Weger with new geo-metric ideas. We begin by using methods of algebraic number theory to reducethe problem of solving the Thue-Mahler equation to the problem of solving a fi-nite collection of related Diophantine equations. In the first part of this thesis, weestablish the key results which allow us to drastically reduce the number of suchDiophantine equations and subsequently reduce the running time.In the second part of this thesis, we show that, by fixing one exponent, there ex-ists an effectively computable constant bounding the solutions of a Goormaghtighequation under certain conditions. For small values of this fixed exponent, we solvethe equation completely. For one such small exponent, we modify and specializeour Thue-Mahler algorithm to the resulting equation in order to fully resolve thiscase.In the third part, we discuss an algorithm for finding all elliptic curves over Qwith a given conductor. Though based on classical ideas derived from reducingthe problem to one of solving associated Thue-Mahler equations, our approach, inmany cases at least, appears to be reasonably efficient computationally. We providedetails of the output derived from running the algorithm, concentrating on the casesof conductor p or p2, for p prime, with comparisons to existing data.Finally, we specialize the Thue-Mahler algorithm to degree 3, applying an analogueiiiof Matshke-von Ka¨nel’s elliptic logarithm sieve to construct a global sieve, leadingto reduced search spaces. The algorithm is implemented in the Magma computeralgebra system, and is part of an ongoing collaborative project.ivLay SummaryConsider any collection of prime numbers {p1, . . . , pv} and any collection of inte-gers c, c0, . . . , cn. Our main result involves the Thue-Mahler equationF (x, y) = c0xn + c1xn−1y + · · ·+ cn−1xyn−1 + cnyn = cpz11 · · · pzvv ,where the values x, y, and z1, . . . , zv are unknown. In particular, for any suchequation, we know that there are only finitely many values of x, y, and z1, . . . , znwhich satisfy it. In our work, we develop an algorithm to find all of these solutionsfor any given collection of primes and coefficients ci. The solutions to these Thue-Mahler have many important mathematical applications, and we modify and refineour algorithm for use in those applications.vPrefaceThe work presented in Chapter 4 is joint work with Dr. M. Bennett and Dr. D.Kreso and has been submitted for publication [10]. I was responsible for modi-fying and specializing the Thue-Mahler algorithm to resolve the remaining cases,0 ≤ x ≤ 720, for n = 5. I implemented the resulting algorithm in Magma andperformed the tests on each remaining case, as well as wrote Section 4.5. The re-mainder of the work submitted for publication was originally drafted by M. Bennettand D. Kreso.Chapter 5 is work completed in collaboration with Dr. M. Bennett and Dr. A.Rechnitzer. A version of this chapter has been published and appears in M. A.Bennett, A. Gherga and A. Rechnitzer, Computing elliptic curves over Q, Math.Comp. 88 (2019), no. 317, 1341-1390. In this work, I modified and implementedall of the code needed to resolve the reducible and irreducible forms. Furthermore,I was responsible for running this code to generate all of the solutions and resultingelliptic curves to the forms in the section “Examples”. I drafted the majority of thissection, while the remainder of the paper was originally drafted by M. Bennett andA. Rechnitzer.Chapter 3 and Chapter 6 is part of an ongoing collaborative project, currently inpreparation [46] with Dr. B. Matshke, Dr. R. von Ka¨nel, and Dr. S. Siksek. Theideas presented in Section 3.3 and Section 3.4.1 are attributed to S. Siksek. Thework in Chapter 6 is joint work with R. von Ka¨nel, to whom the new ideas areattributed. Here, I helped to develop the theory and details behind these ideas, aswell as implemented and tested the algorithm presented in both chapters.viContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiLay Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viContents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAcknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Statement of the results . . . . . . . . . . . . . . . . . . . . . . . 72 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.1 Algebraic number theory . . . . . . . . . . . . . . . . . . . . . . 102.2 p-adic valuations . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 p-adic logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4 The Weil height . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.5 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.6 Cubic forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.7 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Algorithms for Thue-Mahler Equations . . . . . . . . . . . . . . . . 243.1 First steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.2 The relevant algebraic number field . . . . . . . . . . . . . . . . 27vii3.3 The prime ideal removing lemma . . . . . . . . . . . . . . . . . . 283.3.1 Computational remarks and refinements . . . . . . . . . . 343.4 Factorization of the Thue-Mahler equation . . . . . . . . . . . . . 353.4.1 Avoiding the class group Cl(K) . . . . . . . . . . . . . . 353.4.2 Using the class group Cl(K) . . . . . . . . . . . . . . . . 363.4.3 The S-unit equation . . . . . . . . . . . . . . . . . . . . 383.4.4 Computational remarks and comparisons . . . . . . . . . 393.5 A small upper bound for ul in a special case . . . . . . . . . . . . 413.6 Lattice-Based Reduction . . . . . . . . . . . . . . . . . . . . . . 463.6.1 The L3-lattice basis reduction algorithm . . . . . . . . . . 473.6.2 The Fincke-Pohst algorithm . . . . . . . . . . . . . . . . 493.6.3 Computational remarks and translated lattices . . . . . . . 514 Goormaghtigh Equations . . . . . . . . . . . . . . . . . . . . . . . . 554.1 Rational approximations . . . . . . . . . . . . . . . . . . . . . . 564.2 Pade´ approximants . . . . . . . . . . . . . . . . . . . . . . . . . 614.3 Proof of Theorem 4.0.1 . . . . . . . . . . . . . . . . . . . . . . . 674.3.1 Bounding δ . . . . . . . . . . . . . . . . . . . . . . . . . 674.3.2 Applying Proposition 4.2.3 . . . . . . . . . . . . . . . . . 684.4 Proof of Theorem 4.0.2 for x of moderate size . . . . . . . . . . 714.4.1 Case (1) : n = 3, d = 2, n0 = 1, x ≥ 40 . . . . . . . . . 724.4.2 Case (2) : n = 4, d = 3, n0 = 1, x ≥ 85 . . . . . . . . . 744.4.3 Case (3) : n = 5, d = 2, n0 = 2, x ≥ 720 . . . . . . . . 754.4.4 Case (4) : n = 5, d = 4, n0 = 1, x ≥ 300 . . . . . . . . 774.4.5 Treating the remaining small values of x for n ∈ {3, 4} . . 794.5 Small values of x for n = 5 . . . . . . . . . . . . . . . . . . . . . 814.5.1 First steps and small bounds . . . . . . . . . . . . . . . . 824.5.2 Bounding the∑vj=1 njaij . . . . . . . . . . . . . . . . . 894.5.3 A bound for |a1| . . . . . . . . . . . . . . . . . . . . . . 914.5.4 The reduction strategy . . . . . . . . . . . . . . . . . . . 944.5.5 The pl-adic reduction procedure . . . . . . . . . . . . . . 964.5.6 Computational conclusions . . . . . . . . . . . . . . . . . 1014.6 Bounding C(k, d) : the proof of Proposition 4.1.2 . . . . . . . . . 102viii4.7 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 1055 Computing Elliptic Curves over Q . . . . . . . . . . . . . . . . . . . 1075.1 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1085.2 Cubic forms : the main theorem and algorithm . . . . . . . . . . . 1105.2.1 Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 1135.2.2 The algorithm . . . . . . . . . . . . . . . . . . . . . . . . 1175.3 Proof of Theorem 5.2.1 . . . . . . . . . . . . . . . . . . . . . . . 1195.4 Finding representative forms . . . . . . . . . . . . . . . . . . . . 1285.4.1 Irreducible Forms . . . . . . . . . . . . . . . . . . . . . . 1295.4.2 Reducible forms . . . . . . . . . . . . . . . . . . . . . . 1295.4.3 Computing forms of fixed discriminant . . . . . . . . . . 1315.4.4 GL2(Z) vs SL2(Z) . . . . . . . . . . . . . . . . . . . . . 1325.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1325.5.1 Cases without irreducible forms . . . . . . . . . . . . . . 1335.5.2 Cases with fixed conductor (and corresponding irreducibleforms) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1365.5.3 Curves with good reduction outside {2, 3, 23} : an exam-ple of Koutsianis and of von Kanel and Matchke . . . . . 1475.5.4 Curves with good reduction outside {2, 3, 5, 7, 11} : an ex-ample of von Kanel and Matschke . . . . . . . . . . . . . 1515.6 Good reduction outside a single prime . . . . . . . . . . . . . . . 1525.6.1 Conductor N = p . . . . . . . . . . . . . . . . . . . . . . 1535.6.2 Conductor N = p2 . . . . . . . . . . . . . . . . . . . . . 1545.6.3 Reducible forms . . . . . . . . . . . . . . . . . . . . . . 1565.6.4 Irreducible forms : conductor p . . . . . . . . . . . . . . 1575.6.5 Irreducible forms : conductor p2 . . . . . . . . . . . . . . 1585.7 Computational details . . . . . . . . . . . . . . . . . . . . . . . . 1655.7.1 Generating the required forms . . . . . . . . . . . . . . . 1655.7.2 Complete solution of Thue equations : conductor p . . . . 1675.7.3 Non-exhaustive, heuristic solution of Thue equations . . . 1685.7.4 Conversion to curves . . . . . . . . . . . . . . . . . . . . 1695.7.5 Conductor p2 . . . . . . . . . . . . . . . . . . . . . . . . 169ix5.8 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1715.8.1 Previous work . . . . . . . . . . . . . . . . . . . . . . . 1715.8.2 Counts : conductor p . . . . . . . . . . . . . . . . . . . . 1725.8.3 Counts : conductor p2 . . . . . . . . . . . . . . . . . . . 1745.8.4 Thue equations . . . . . . . . . . . . . . . . . . . . . . . 1765.8.5 Elliptic curves with the same prime conductor . . . . . . . 1785.8.6 Rank and discriminant records . . . . . . . . . . . . . . . 1785.9 Completeness of our data . . . . . . . . . . . . . . . . . . . . . . 1805.10 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 1896 Towards Efficient Resolution of Thue-Mahler Equations . . . . . . . 1906.1 Decomposition of the Weil height . . . . . . . . . . . . . . . . . 1916.2 Initial height bounds . . . . . . . . . . . . . . . . . . . . . . . . 1976.3 Coverings of Σ . . . . . . . . . . . . . . . . . . . . . . . . . . . 1996.4 Construction of the ellipsoids . . . . . . . . . . . . . . . . . . . . 2006.4.1 The Archimedean ellipsoid: the real case . . . . . . . . . 2086.4.2 The non-Archimedean ellipsoid . . . . . . . . . . . . . . 2146.5 The Archimedean sieve: the real case . . . . . . . . . . . . . . . 2166.6 The non-Archimedean Sieve . . . . . . . . . . . . . . . . . . . . 218Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229xAcknowledgmentsI am indebted to Dr. Michael A. Bennett for the patient guidance, encouragementand advice he has provided throughout my time as his student. I would also liketo thank Dr. Andrew Rechnitzer and Dr. Greg Martin for the numerous commentsand suggestions that helped me to improve this thesis.This work would not have been possible without the insightful knowledge of mycollaborators, Dr. Rafael von Ka¨nel, Dr. Samir Siksek, and Dr. Benjamin Matschke.I would also like to thank my friends and family for supporting me throughout theyears. Special thanks to Aaron Berk, Celina Luther, and Matthew Coles. Finally, Iwould like to thank my parents, Marius and Monica, and my brother Andy.This research was funded in part by a National Sciences and Engineering ResearchCouncil Postgraduate Scholarship.xiChapter 1IntroductionA Diophantine equation is a polynomial equation in several variables defined overthe integers. The term Diophantine refers to the Greek mathematician Diophantusof Alexandria, who studied such equations in the 3rd century A.D. Let f(x1, . . . , xn)be a polynomial with integer coefficients. We wish to study the set of solutions(x1, . . . , xn) ∈ Zn to the equationf(x1, . . . , xn) = 0. (1.1)There are several different approaches for doing so, arising from three basic prob-lems concerning Diophantine equations. The first such problem is to determinewhether (1.1) has any solutions. Indeed, one of the most famous theorems in math-ematics states that for f(x, y, z) = xn + yn − zn, where n ≥ 3, there are nosolutions in the positive integers x, y, z. This equation is known as Fermat’s LastTheorem and was proven by Wiles in 1995. Qualitative questions of this type areoften studied using algebraic methods.Suppose now that (1.1) is solvable, that is, has at least one solution. The secondbasic problem is to determine whether the number of solutions is finite or infinite.For example, consider the Thue equation,f(x, y) = a, (1.2)1where f(x, y) is an integral binary form of degree n ≥ 3 and a is a fixed nonzerorational integer. In 1909, Thue [112] proved that this equation has only finitelymany solutions. This result followed from a sharpening of Liouville’s inequality[66], an observation that algebraic numbers do not admit very strong approximationby rational numbers. That is, if α is a real algebraic number of degree n ≥ 2 andp, q are integers, Liouville’s observation states that∣∣∣∣α− pq∣∣∣∣ > c1qn , (1.3)where c1 > 0 is a value depending explicitly on α. The finiteness of the number ofsolutions to (1.2) follows directly from a sharpening of (1.3) of the type∣∣∣∣α− pq∣∣∣∣ > λ(q)qn , λ(q)→∞. (1.4)Indeed, if α is a real root of f(x, 1) and α(i), i = 1, . . . , n are its conjugates, itfollows from (1.2) thatn∏i=1∣∣∣∣α(i) − xy∣∣∣∣ = a|a0||y|nwhere a0 is the leading coefficient of the polynomial f(x, 1). If the Thue equationhas integer solutions with arbitrarily large |y|, the product∏ni=1 |α(i) − x/y| musttake arbitrarily small values for solutions x, y of (1.2). As all the α(i) are different,x/y must be correspondingly close to one of the real numbers α(i), say α. Thuswe obtain ∣∣∣∣α− xy∣∣∣∣ < c2|y|nwhere c2 depends only on a0, n, and the conjugates α(i) (cf. Chapter 4 of [107]).Comparison of this inequality with (1.4) shows that |y| cannot be arbitrarily large,and so the number of solutions of the Thue equation is finite. Using this argument,an explicit bound can be constructed on the solutions of (1.2) provided that aneffective inequality (1.4) is known. The sharpening of the Liouville inequalityhowever, especially in effective form, proved to be very difficult.2In [112], Thue published a proof that∣∣∣∣α− pq∣∣∣∣ < 1q n2 +1+εhas only finitely many solutions in integers p, q > 0 for all algebraic numbers αof degree n ≥ 3 and any ε > 0. In essence, he obtained the inequality (1.4) withλ(q) = c3q12n−1−ε, where c3 > 0 depends on α and ε, thereby confirming thatall Thue equations have only finitely many solutions (cf. [107]). Unfortunately,Thue’s arguments do not allow one to find the explicit dependence of c3 on α andε, and so the bound for the heights of solutions of the Thue equation cannot begiven in explicit form either. That is, Thue’s proof is ineffective, meaning that itprovides no means to find the solutions to (1.2).Nonetheless, the investigation of Thue’s equation and its generalizations was cen-tral to the development of the theory of Diophantine equations in the early 20thcentury when it was discovered that many Diophantine equations in two unknownscould be reduced to it. In particular, the thorough development and enrichmentof Thue’s method led Siegel [103] to his theorem on the finiteness of the numberof integral points on an algebraic curve of genus greater than zero. However, asSiegel’s result relies on Thue’s rational approximation to algebraic numbers, it toois ineffective.Shortly following Thue’s result, Goormaghtigh [47] conjectured that the only non-trivial integer solutions of the exponential Diophantine equationxm − 1x− 1 =yn − 1y − 1 (1.5)satisfying x > y > 1 and n,m > 2 are31 =25 − 12− 1 =53 − 15− 1 and 8191 =213 − 12− 1 =903 − 190− 1 .These correspond to the known solutions (x, y,m, n) = (2, 5, 5, 3) and (2, 90, 13, 3)to what is nowadays termed Goormaghtigh’s equation. The Diophantine equation(1.5) asks for integers having all digits equal to one with respect to two distinct3bases, yet whether it has finitely many solutions is still unknown. By fixing theexponents m and n however, Davenport, Lewis, and Schinzel [40] were able toprove that (1.5) has only finitely many solutions. Unfortunately, this result rests onSiegel’s aforementioned finiteness theorem, and is therefore ineffective.In 1933, Mahler [69] published a paper on the investigation of the Diophantineequationf(x, y) = pz11 · · · pzvv , (x, y) = 1,in which S = {p1, . . . , pv} denotes a fixed set of prime numbers, x, y, zi ≥ 0,i = 1, . . . , v are unknown integers, and f(x, y) is an integral irreducible binaryform of degree n ≥ 3. Generalizing the classical result of Thue, Mahler proved thatthis equation has only finitely many solutions. Unfortunately, like Thue, Mahler’sargument is also ineffective.This leads us to the third basic problem regarding Diophantine equations and themain focus of this thesis: given a solvable Diophantine equation, determine all ofits solutions. Until long after Thue’s work, no method was known for the con-struction of bounds for the number of solutions of a Thue equation in terms ofthe parameters of the equation. Only in 1968 was such a method introduced byBaker [4], based on his theory of bounds for linear forms in the logarithms of al-gebraic numbers. Generalizing Baker’s ground-breaking result to the p-adic case,Sprindz˘uk and Vinogradov [108] and Coates [27] proved that the solutions of anyThue-Mahler equation,f(x, y) = cpz11 · · · pzvv , (x, y) = 1, (1.6)where c is a fixed integer, could, at least in principal, be effectively determined.The first practical method for solving the general Thue-Mahler equation (1.6) overZ is attributed to Tzanakis and de Weger (cf. [118], [114], [115], [116]), whoseideas were inspired in part by the method of Agrawal, Coates, Hunt, and van derPoorten [1] in their work to solve the specific Thue-Mahler equationx3 − x2y + xy2 + y3 = ±11z1 .4Using optimized bounds arising from the theory of linear forms in logarithms, arefined, automated version of this explicit method has since been implemented byHambrook [50] as a Magma package [19].As for Goormaghtigh’s equation, when m and n are fixed andgcd(m− 1, n− 1) > 1, (1.7)Davenport, Lewis, and Schinzel [40] were able to replace Siegel’s result by aneffective argument due to Runge. This result was improved by Nesterenko andShorey [83], and Bugeaud and Shorey [24] using Baker’s theory of linear formsin logarithms. In either case, in order to deduce effectively computable boundsupon the polynomial variables x and y, one must impose the constraints upon mand n that either m = n + 1, or that the assumption (1.7) holds. In the extensiveliterature on this problem, there are a number of striking results that go well beyondwhat we have mentioned here. By way of example, work of Balasubramanian andShorey [3] shows that equation (1.5) has at most finitely many solutions if we fixonly the set of prime divisors of x and y, while Bugeaud and Shorey [24] prove ananalogous finiteness result, under the additional assumption of (1.7), provided thequotient (m− 1)/(n− 1) is bounded above. Additional results on special cases ofequation (1.5) are available in, for example, [54], [62], [63], [101] and [64].A direct application of determining the solutions of a solvable Diophantine equa-tion is the computation of elliptic curves over Q. Let S be a finite set of ratio-nal primes. In 1963, Shafarevich [99] proved that there are at most finitely manyQ-isomorphism classes of elliptic curves defined over Q having good reductionoutside S. The first effective proof of this statement was provided by Coates [27]in 1970 using bounds for linear forms in p-adic and complex logarithms. Earlyattempts to make these results explicit for fixed sets of small primes overlap withthe arguments of [27], in that they reduce the problem to that of solving a numberof degree 3 Thue-Mahler equations of the formF (x, y) = cu,5where u is an integer whose prime factors all lie in S.In the 1950’s and 1960’s, Taniyama and Weil asked whether all elliptic curves overQ of a given conductor N are related to modular functions. While this conjectureis now known as the Modularity Theorem, until its proof in 2001 [20], attempts toverify it sparked a large effort to tabulate all elliptic curves over Q of given con-ductor N . In 1966, Ogg ([86], [87]) determined all elliptic curves defined overQ with conductor of the form 2a. Coghlan, in his dissertation [28], studied thecurves of conductor 2a3b independently of Ogg, while Setzer [98] computed allQ-isomorphism classes of elliptic curves of conductor p for certain small primes p.Each of these examples corresponds, via the [11] approach, to cases with reducibleforms. The first analysis on irreducible forms in (1.6) was carried out by Agrawal,Coates, Hunt and van der Poorten [1], who determined all elliptic curves of con-ductor 11 defined over Q to verify the (then) conjecture of Taniyama-Weil.There are very few, if any, subsequent attempts in the literature to find ellipticcurves of given conductor via Thue-Mahler equations. Instead, many of the ap-proaches involve a completely different method to the problem, using modularforms. This method relies upon the Modularity Theorem of Breuil, Conrad, Di-amond and Taylor [20], which was still a conjecture (under various guises) whenthese ideas were first implemented. Much of the success of this approach can beattributed to Cremona ([31], [32]) and his collaborators, who have devoted decadesof work to it. In fact, using this method, all elliptic curves over Q of conductor Nhave been determined for values of N as follows• Antwerp IV (1972): N ≤ 200• Tingley (1975): N ≤ 320• Cremona (1988): N ≤ 600• Cremona (1990): N ≤ 1000• Cremona (1997): N ≤ 5077• Cremona (2001): N ≤ 100006• Cremona (2005): N ≤ 130000• Cremona (2014): N ≤ 350000• Cremona (2015): N ≤ 364000• Cremona (2016): N ≤ 390000• Cremona (2019): N ≤ 4000001.1 Statement of the resultsThe novel contributions of this thesis concern the development and implementationof efficient algorithms to determine all solutions of certain Goormaghtigh equationsand Thue-Mahler equations. In particular, we follow [10] to prove that, in fact, un-der assumption (1.7), equation (1.5) has at most finitely many solutions which maybe found effectively, even if we fix only a single exponent.Theorem 1.1.1. If there is a solution in integers x, y, n and m to the equationxm − 1x− 1 =yn − 1y − 1 , where gcd(m− 1, n− 1) = d > 1, (1.8)thenx < (3d)4n/d ≤ 36n. (1.9)In particular, if n is fixed, there is an effectively computable constant c = c(n)such that max{x, y,m} < c.By refining our approach, new results from computational Diophantine approxi-mation enable us to achieve the complete solution of equation (1.8) for small fixedvalues of n.Theorem 1.1.2. If there is a solution in integers x, y and m to equation (1.8) withn ∈ {3, 4, 5}, then(x, y,m, n) = (2, 5, 5, 3) and (2, 90, 13, 3).7In the case n = 5 of Theorem 1.1.2, “off-the-shelf” techniques for finding integralpoints on models of elliptic curves or for solving Ramanujan-Nagell equations ofthe shape F (x) = zn (where F is a polynomial and z a fixed integer) do notapparently permit the full resolution of this problem in a reasonable amount oftime. Instead, we sharpen the existing techniques of [114] and [50] for solvingThue-Mahler equations and specialize them to this problem. The work presentedon Goormaghtigh equations is joint work with Dr. M. Bennett and Dr. D. Kresoand has been submitted for publication [10].A direct consequence and primary motivation for developing an efficient Thue-Mahler algorithm is the computation of elliptic curves over Q. In this thesis, wereturn to techniques based upon solving Thue-Mahler equations, using a numberof results from classical invariant theory. This work is in collaboration with Dr.M. Bennett and Dr. A. Rechnitzer and appears in [11]. We illustrate the connec-tion between elliptic curves over Q and cubic forms and subsequently describe aneffective algorithm for determining all elliptic curves over Q having good reduc-tion outside S. This result can be summarized as follows. If we wish to find anelliptic curves E of conductor N = pa11 · · · pavv for some ai ∈ N, by Theorem 1 of[11], there exists an integral binary cubic form F of discriminant N0 | 12N andrelatively prime integers u, v satisfyingF (u, v) = w0u3 + w1u2v + w2uv2 + w3v3 = 2α13β1∏p|N0pκpfor some α1, β1, κp. Then E is isomorphic over Q to the elliptic curve ED, whereED is determined by the form F and (u, v). It is worth noting that Theorem 1of [11] very explicitly describes how to generate ED; once a solution (u, v) tothe Thue-Mahler equation F is known, a quick computation of the Hessian andJacobian discriminant of F evaluated at (u, v) yields the coefficients of ED. Us-ing this theorem, all E/Q of conductor N may be computed by generating all ofthe relevant binary cubic forms, solving the corresponding Thue-Mahler equations,and outputting the elliptic curves that arise. The first and last steps of this processare straightforward. Indeed, Bennett and Rechnitzer [12] describe an efficient al-gorithm for carrying out the first step. In fact, they have carried out a one-time8computation of all irreducible forms that can arise in Theorem 1 of absolute dis-criminant bounded by 1010. The bulk of the work is therefore concentrated in step2, solving a large number of degree 3 Thue-Mahler equations.Unfortunately, despite many refinements, the Magma implementation of a Thue-Mahler [50] solver encounters a multitude of bottlenecks which often yield un-avoidable timing and memory problems, even when parallelization is considered.As our aim is to use the results of [11] to generate all elliptic curves over Q ofconductor N < 106, in its current state, this Magma implementation is insuffi-cient for this task. The main novel contributions of this thesis are new theoreticalresults towards the efficient resolution of an arbitrary degree 3 Thue-Mahler equa-tion and the implementation of these results as a Magma package. This work isbased on ideas of [58] and is part of an ongoing collaborative project, currently inpreparation [46] with Dr. B. Matshke, Dr. R. von Ka¨nel, and Dr. S. Siksek.9Chapter 2Preliminaries2.1 Algebraic number theoryIn this section we recall some basic results from algebraic number theory thatwe use throughout the remaining chapters. We refer to [73] and [85] for full de-tails.Let K be a finite algebraic extension of Q of degree n = [K : Q]. Take g to bethe minimal polynomial of some θ ∈ K such that K = Q(θ). The polynomialg has n distinct roots in C. Each such root is called a conjugate of θ over Q.Each conjugate determines a unique embedding of K in to C. Conversely, everyembedding σ : K → C must arise in this way since θ must be sent to one of itsconjugates. Thus, there are precisely n embeddings of K into C.Let s denote the number of real embeddings of K and let t denote the number ofconjugate pairs of complex embeddings of K, where n = s + 2t. By Dirichlet’sUnit Theorem, the group of units of K is the direct product of a finite cyclic groupconsisting of the roots of unity in K and a free abelian group of rank r = s+ t−1.Equivalently, there exists a system of r independent units ε1, . . . , εr such that the10group of units of K is given by{ζ · εa11 · · · εarr : ζ a root of unity, ai ∈ Z for i = 1, . . . , r} .Any set of independent units that generate the torsion-free part of the unit group iscalled a system of fundamental units.An element α ∈ K is called an algebraic integer if its minimal polynomial over Zis monic. The set of algebraic integers in K forms a ring, denotedOK . We refer tothis ring as the ring of integers or number ring corresponding to the number fieldK. For any α ∈ K, we define the norm of α asNK/Q(α) =∏σ:K→Cσ(α)where the product is taken over all embeddings σ of K. For algebraic integers,NK/Q(α) ∈ Z. The units are precisely the elements of norm ±1. Two elementsα, β of K are called associates if there exists a unit ε such that α = εβ. Let(α)OK denote the ideal generated by α. Associated elements generate the sameideal, and distinct generators of an ideal are associated. There exist only finitelymany non-associated algebraic integers in K with given norm.Any element of the ring of integers can be written as a product of irreducible el-ements. These are non-zero non-unit elements of OK which have no integral di-visors but their own associates. Unfortunately, number rings are not alway uniquefactorization domains: this decomposition into irreducible elements may not beunique. However, every number ring is a Dedekind domain. This means that everyideal can be decomposed into a product of prime ideals and this decomposition isunique. A principal ideal is an ideal generated by a single element α. Two frac-tional ideals are called equivalent if their quotient is principal. It is well known thatthere are only finitely many equivalence classes of fractional ideals and the set ofall such classes forms a finite abelian group called the ideal class group, Cl(K).The number of ideal classes, # Cl(K), is called the class number of OK and isdenoted by hK . For an ideal a of OK , it is always true that ahK is principal. Thenorm of the (integral) ideal a is defined by NK/Q(a) = # (OK/a). If a = (α)OK11is a principal ideal, then NK/Q(a) =∣∣NK/Q(α)∣∣.Let L be a finite field extension of K with ring of integers OL. Hereafter, a primeideal of OK is denoted by p, while a prime ideal of OL is denoted by P, unlessotherwise stated. Every prime ideal P of OL lies over a unique prime ideal p inOK . That is, P divides p. The ramification index e(P|p) is the largest power towhich P divides p. The field OL/P is an extension of finite degree f(P|p) overOK/p. We call f(P|p) the inertial degree ofP over p. For p lying over the rationalprime p, this is the integer such thatNK/Q(p) = pf(p|p).The ramification index and inertial degree are multiplicative in a tower of fields. Inparticular, if P lies over p which lies over the rational prime p, thene(P|p) = e(P|p)e(p|p) and f(P|p) = f(P|p)f(p|p).Let P1, . . . ,Pm be the primes ofOL lying over a prime ideal p ofOK . Denote bye(P1|p), . . . , e(Pm|p) and f(P1|p), . . . , f(Pm|p) the corresponding ramificationindices and inertial degrees. Thenm∑i=1e(Pi|p)f(Pi|p) = [L : K].If L is normal over K and Pi and Pj are two prime ideals lying over the primeideal p of OK , then e(Pi|p) = e(Pj |p) and f(Pi|p) = f(Pj |p). In this case, pfactors aspOL = (P1 · · ·Pm)ein OL, where the Pi are distinct prime ideals all having the same ramificationdegree e and inertial degree f over p. It follows thatmef = [L : K].122.2 p-adic valuationsIn this section we give a brief exposition of p-adic valuations. We refer to [18],[25], [52], [60], and [82] as references for this material.Let K be an arbitrary number field. A homomorphism v : K× → R≥0 of the mul-tiplicative group of K into the group of positive real numbers is called a valuationif it satisfies the conditionv(x+ y) ≤ v(x) + v(y).This definition may be extended to all of K by setting v(0) = 0. Ifv(x+ y) ≤ max(v(x), v(y))holds for all x, y ∈ K, then v is called a non-Archimedean valuation. All remainingvaluations on K are called Archimedean.Every valuation v induces on K the structure of a metric topological space whichmay or may not be complete. We say that two valuations are equivalent if theydefine the same topology and we call an equivalence class of absolute values aplace of K. It is an elementary result of topology that every metric space has aunique (up to isometry) completion, which is a complete metric space that containsthe given space as a dense subset. For the field K, the resulting complete metricspace may be given a field structure. Equivalently, there exists a field L with avaluation w such that L is complete in the topology induced by w. The field K iscontained inL and the valuations v andw coincide inK. Moreover, the completionL of K is unique up to topological isomorphism.For any non-zero prime ideal p ofOK , let ordp(a) denote the exact power to whichp divides the ideal a. For fractional ideals a this number may be negative. Forα ∈ K, we write ordp(α) for ordp ((α)OK). Every prime ideal defines a discretenon-Archimedean valuation on K viav(x) :=(1NK/Q(p))ordp(x).13Furthermore, every embedding ofK into the complex field defines an Archimedeanvaluation. Conversely, every discrete valuation on K arises in this way by a primeideal of OK , while every Archimedean valuation of K is equivalent to |σ(x)|,where σ is an embedding of K into C. Valuations defined by different prime ide-als are non-equivalent, and two valuations defined by different embeddings of Kinto C are equivalent if and only if those embeddings are complex conjugates. Thetopology induced in K by a prime ideal p of OK is called the p-adic topology.The completion of K under this valuation is denoted by Kp or Kv and is calledthe p-adic field. Let V be the set of all valuations of an algebraic number field K.Then for every non-zero element α ∈ K we have∏v∈Vv(α) = 1.In the ring of integers of Q, the prime ideals are generated by the rational primesp, and the resulting topology in the field Q is called the p-adic topology. Thecompletion of Q under this valuation is denoted by Qp. If v(x) is a non-trivialvaluation ofQ, then either v(x) is equivalent to the ordinary absolute value |x|, or itis equivalent to one of the p-adic valuations induced by rational primes. Analogousto ordp, for any prime pwe define the p-adic order of x ∈ Q as the largest exponentof p dividing x. Then, the p-adic valuation v is defined asv(x) = p− ordp(x).Thus, for any nonzero x ∈ Qp we can writex =∞∑i=kuipiwhere k = ordp(x) and the p-adic digits ui are in {0, . . . , p − 1} with uk 6= 0. Ifordp(x) ≥ 0 then x is called a p-adic integer. The set of p-adic integers is denotedby Zp. A p-adic unit is an x ∈ Qp having ordp(x) = 0. For any p-adic integerx ∈ Zp and µ ∈ N0, there exists a unique rational integer x(µ) =∑µ−1i=0 uipi such14thatordp(x− x(µ)) ≥ µ, and 0 ≤ x(µ) ≤ pµ − 1.For ordp(x) ≥ k we also write x ≡ 0 mod pk.Let Qp be the algebraic closure of Qp. If L is a p-adic field, it is necessarily afinite extension of a certain Qp. Furthermore, the p-adic valuation v of Qp extendsuniquely to L asv(x) = |NL/Qp(x)|1/[L:Qp].Here, we define the p-adic order of x ∈ L byordp(x) =1[L : Qp]ordp(NL/Qp(x)).This definition is independent of the field L containing x. Since each element ofQp is by definition contained in some finite extension of Qp, the above definitionmay be used to define the p-adic valuation v of any x ∈ Qp. Every finite extensionof Qp is complete with respect to v, but Qp is not. The completion of Qp withrespect to v is denoted by Cp.Consider now a finite field extension K of Q. Let g(t) ∈ Q[t] denote the minimalpolynomial of some θ ∈ K such that K = Q(θ). Let p be a rational prime andsuppose g(t) = g1(t) · · · gm(t) is the decomposition of g(t) into irreducible poly-nomials gi(t) ∈ Qp[t] of degree ni = deg gi(t). The prime ideals in K dividing pare in one-to-one correspondence with g1(t), . . . , gm(t). More precisely, we havein K the following decomposition of (p)OK :(p)OK = pe(p1|p)1 · · · pe(pm|p)m .Here, p1, . . . , pm are distinct prime ideals of OK . Then, for i = 1, . . . ,m, wehave ni = e(pi|p)f(pi|p) = [Kpi : Qp], where e(pi|p), f(pi|p) are the ramifica-tion index and inertial degree of pi over p, respectively, and Kpi ' Qp(θi) whereg(θi) = 0.There are n embeddings ofK intoQp, and each one fixesQ and maps θ to a root ofg in Qp. Let θ(1)i , . . . , θ(ni)i denote the roots of gi(t) in Qp. For i = 1, . . . ,m and15j = 1, . . . , ni, let σij be the embedding of K into Qp(θ(j)i ) defined by θ 7→ θ(j)i .Them classes of conjugate embeddings are {σi1, . . . , σini} for i = 1, . . . ,m. Eachmap σij coincides with the embedding K ↪→ Kpi ' Qp(θi) ' Qp(θ(j)i ).The m extensions of the p-adic valuation on Q to K are multiples of the pi-adicvaluation on K:ordp(x) =1eiordpi(x) for i = 1, . . . ,m.We also view these extensions as arising from various embeddings of K into Qp.Indeed, the extension to Qp(θ(j)i ) of the p-adic valuation on Qp induces a p-adicvaluation on K via the embedding σij asv(x) = |NKpi/Qp(σij(x))|1/ni .Here, as before, ni = deg gi(t) = [Kpi : Qp]. Furthermore,ordp(x) = ordp(σij(x)),and we haveordp(σij(x)) =1eiordpi(x) for i = 1, . . . ,m, j = 1, . . . , ni.2.3 p-adic logarithmsWe have seen how to extend p-adic valuations to algebraic extensions of Q. Forany z ∈ Cp with ordp(z − 1) > 0, we can also define the p-adic logarithm of zbylogp(z) = −∞∑i=1(1− z)ii.This series converges precisely in the region where ordp(z− 1) > 0. Three impor-tant properties of the p-adic logarithm are:161. logp(xy) = logp(x) + logp(y) whenever x, y ∈ Cp with ordp(x − 1) > 0and ordp(y − 1) > 0.2. logp(zk) = k logp(z) whenever z ∈ Cp with ordp(z − 1) > 0 and k ∈ Z.3. ordp(logp(z)) = ordp(z−1) whenever z ∈ Cp with ordp(z−1) > 1/(p−1).Proofs of the first and last property can be found in [52] (pp. 264-265). The secondproperty follows from the first.We will use the following lemma to extend the definition of the p-adic logarithmto all p-adic units in Qp.Lemma 2.3.1. Let z be a p-adic unit belonging to a finite extensions L of Qp. Lete and f be the ramification index and inertial degree of L.(a) There is a positive integer r such that ordp(zr − 1) > 0.(b) If r is the smallest positive integer having ordp(zr − 1) > 0, then r dividespf −1, and an integer q satisfies ordp(zq−1) > 0 if and only if it is a multipleof r.(c) If r is a nonzero integer with ordp(zr − 1) > 0, and if k is an integer withpk(p− 1) > e, thenordp(zrpk − 1) > 1p− 1 .For z a p-adic unit in Qp we definelogp z =1qlogp zq,where q is an arbitrary non-zero integer such that ordp(zq − 1) > 0. To seethat this definition is independent of q, let r be the smallest positive integer withordp(zr−1) > 0, note that q/r is an integer, and use the second property of p-adiclogarithms above to write1qlogp zq =1r(q/r)logp zr(q/r) =1rlogp zr.17Choosing q such that ordp(zq − 1) > 1/(p − 1) helps to speed up and controlthe convergence of the series defining logp (cf. [105] (pp. 28-30) and [30] (pp.263-265)).It is straightforward to see that Properties 1 and 2 above extend to the case wherex, y, z are p-adic units. Combining this with Property 3, we obtainLemma 2.3.2. Let z1, . . . , zm ∈ Qp be p-adic units and let b1, . . . , bm ∈ Z. Ifordp(zb11 · · · zbmm − 1) >1p− 1thenordp(b1 logp z1 + · · ·+ bm logp zm) = ordp(zb11 · · · zbmm − 1).2.4 The Weil heightLet K be a number field and at each place v of K, let Kv denote the completion ofK at v. Then ∑v|p[Kv : Qv] = [K : Q]for all places p ofQ. We will use two absolute values | · |v and ‖·‖v onK which wenow define. If v|∞, then ‖ · ‖v restricted to Q is the usual Archimedean absolutevalue; if v|p for a rational prime p, then ‖ · ‖v restricted to Q is the usual p-adicvaluation. We then set| · |v = ‖ · ‖[Kv :Qv ]/[K:Q]v .Let x ∈ K× and let log+(·) denote the real-valued function max{log(·), 0} onR≥0. We define the logarithmic Weil height h(x) byh(x) =1[K : Q]∑vlog+ |x|v,where the sum is take over all places v of K. This definition may be extendedto all of K by setting h(0) = 0. If x is an algebraic unit, then |x|v = 1 for allnon-Archimedean places v, and therefore h(x) can be taken over the Archimedean18places only. In particular, if x ∈ Q, we may write h(x) = log max{|p|, |q|}for x = p/q with p, q ∈ Z such that gcd(p, q) = 1. Finally, if x ∈ Z thenh(x) = log |x|.2.5 Elliptic curvesLet K be a field of characteristic char(K) 6= 2, 3. An elliptic curve E over K is anonsingular curve of the formE : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6, (2.1)where ai ∈ K, together with a specified base point, O ∈ E. An equation ofthe form (2.1) is called a Weierstrass equation. This equation is unique up to acoordinate transformation of the formx = u2x′ + r, y = u3y′ + su2x′ + t,with r, s, t, u ∈ K,u 6= 0. In fact, applying several linear changes of variables, wemay write E asE : y2 = x3 − 27c4x− 54c6,whereb2 = a21 + 4a2, b4 = a1a3 + 2a4, b6 = a23 + 4a6,b8 = a21a6 + 4a2a6 − a1a3a4 + a2a23 − a24,c4 = b22 − 24b4, and c6 = −b32 + 36b2b4 + 9b2b4b6.Associated to this curve are the quantities∆E = −b22b8 − 8b34 − 27b26 + 9b2b4b6 and jE = c34/∆E ,where ∆E is called the discriminant of the Weierstrass equation and the quantityjE is called the j-invariant of the elliptic curve. The condition of being nonsingularis equivalent to ∆E being non-zero. Two elliptic curves are isomorphic over K¯,19the algebraic closure of K, if and only if they have the same j-invariant.When K = Q, the Weierstrass model (2.1) can be chosen so that ∆E has minimalnonnegative p-adic order for each rational prime p and ai ∈ Z. Suppose (2.1)is such a global minimal model for an elliptic curve E over Q. Reducing thecoefficients modulo a rational prime p yields a (possibly singular) curve over FpE˜ : y2 + a˜1xy + a˜3y = x3 + a˜2x2 + a˜4x+ a˜6, (2.2)where a˜i ∈ Fp. This “reduced” curve E˜/Fp is called the reduction of E modulop. It is nonsingular provided that ∆E 6≡ 0 mod p, in which case it is an ellipticcurve defined over Fp. The curve E is said to have good reduction modulo p ifE˜/Fp is nonsingular, otherwise, we say E has bad reduction modulo p.The reduction type of E at a rational prime p is measured by the conductor,N =∏ppηpwhere the product runs over all primes p and ηp = 0 for all but finitely manyprimes. In particular, ηp 6= 0 if p does not divide ∆E . Equivalently, E has badreduction at p if and only if p | N . Suppose E has bad reduction at p so thatηp 6= 0. The reduction type of E at p is said to be multiplicative (E has a node overRp) or additive (E has a cusp over Rp) depending on whether ηp = 1 or ηp ≥ 2,respectively. The ηp, hence the conductor, are invariant under isogeny.2.6 Cubic formsLet a, b, c and d be integers and consider the binary cubic formF (x, y) = ax3 + bx2y + cxy2 + dy3. (2.3)Two such forms F1 and F2 are called equivalent if they are equivalent under theGL2(Z)-action. That is, F1 and F2 are equivalent if there exist a1, a2, a3, a4 ∈ Z20such that a1a4 − a2a3 = ±1 andF1(a1x+ a2y, a3x+ a4y) = F2(x, y)for all x, y. In this case, we write F1 ∼ F2. The discriminant DF of a cubic form(2.3) is given byDF = −27a2d2 + b2c2 + 18abcd− 4ac3 − 4b3d = a4∏i<j(αi − αj)2,where α1, α2 and α3 are the roots of the polynomial F (x, 1). We observe that ifF1 ∼ F2, then DF1 = DF2 .Associated to F is the Hessian HF (x, y), given byHF (x, y) = −14(∂2F∂x2∂2F∂y2−(∂2F∂x∂y)2)= (b2 − 3ac)x2 + (bc− 9ad)xy + (c2 − 3bd)y2,and the Jacobian determinant ofF andHF , a cubic formGF (x, y) defined byGF (x, y) =∂F∂x∂HF∂y− ∂F∂y∂HF∂x= (−27a2d+ 9abc− 2b3)x3 + (−3b2c− 27abd+ 18ac2)x2y++ (3bc2 − 18b2d+ 27acd)xy2 + (−9bcd+ 2c3 + 27ad2)y3.2.7 LatticesAn n-dimensional lattice is a discrete subgroup of Rn of the formΓ ={n∑i=1xibi : xi ∈ Z},where b1, . . . ,bn are vectors forming a basis for Rn. We say that the vectorsb1, . . . ,bn form a basis for Γ, or that they generate Γ. Let B denote the matrix21whose columns are the vectors b1, . . . ,bn. Any lattice element v may be ex-pressed as v = Bx for some x ∈ Zn. We call v the embedded vector and x thecoordinate vector.A bilinear form on a lattice Γ is a function Φ : Γ× Γ→ Z satisfying1. Φ(u,v +w) = Φ(u,v) + Φ(u,w)2. Φ(u+ v,w) = Φ(u,w) + Φ(v,w)3. Φ(au,w) = aΦ(u,w)4. Φ(u, aw) = aΦ(u,w)for all u,v, and w in Γ and any a ∈ Z.Given a basis, we can define a specific bilinear form on our lattice Γ as part of itsstructure. This form describes a kind of distance between elements u and v andwe say the lattice is defined by Φ. Associated to this bilinear form is a quadraticform Q : Γ→ Z defined by Q(v) = Φ(v,v). A lattice is called positive definite ifits quadratic form is positive definite.The bilinear forms (and their associated quadratic forms) that we will be usingcome from the usual inner product on vectors in Rn. This is simply the dot productΦ(u,v) = u · v for embedded vectors, u,v. For the coordinate vectors x,yassociated to these vectors, this translates to multiplication with the basis matrix.Precisely, if u = Bx and v = By, we have Φ(u,v) = xTBTBy.If v = Bx, the norm of the vector v ∈ Γ is defined to be the inner product Φ(v,v).In terms of the corresponding coordinate vector x, this isvTv = xTBTBx.Equivalently, we write xTAx where A = BTB is the Gram matrix of Γ with basisB and bilinear form Φ. The entries of the matrix A are aij = Φ(bi,bj).Two basis matrices B1 and B2 define the same lattice Γ if and only if there isa unimodular matrix U such that B1U = B2. The bilinear form on Γ can be22written with respect to either embedded or coordinate vectors. Using another basisto express the lattice elements is possible, and sometimes preferable. However, theGram matrix is specific to the bilinear form on the lattice and should not changewhen operating on embedded vectors. If it is operating on coordinate vectors, thechange of basis must be accounted for.23Chapter 3Algorithms for Thue-MahlerEquationsIn this chapter, we give some of the primary algorithms needed to solve an arbitraryThue-Mahler equation. The methods presented here follow somewhat [50] and[116], with new results and modifications from [46].3.1 First stepsFix a nonzero integer c and let S = {p1, . . . , pv} be a set of rational primes.LetF (X,Y ) = c0Xn + c1Xn−1Y + · · ·+ cn−1XY n−1 + cnY nbe an irreducible binary form over Z of degree n ≥ 3. We want to solve theThue–Mahler equationF (X,Y ) = cpZ11 · · · pZvv (3.1)for unknowns X,Y, Z1, . . . , Zv with gcd(X,Y ) = 1 and Zi ≥ 0 for i = 1, . . . , v.To do so, we first reduce (3.1) to the special case where c0 = 1 and gcd(c, pi) = 1for i = 1, . . . , v, loosely following [50].24LetD be the set of all positive integersm dividing c0 such that ordp(m) ≤ ordp(c)for each rational prime p /∈ S.Lemma 3.1.1. D is precisely the set of all d ∈ Z>0 such that d = gcd(c0, Y ).Proof. To see this, let q1, . . . , qw denote the distinct prime divisors of c not con-tained in S. Thenc =w∏i=1qbii ·v∏i=1pordpi (c)ifor some integers bi > 0. If (X,Y, Z1, . . . , Zv) is a solution of the Thue-Mahlerequation in question, it follows thatF (X,Y ) = cpZ11 . . . pZvv =w∏i=1qbii ·v∏i=1pordpi (c)+Zii .Suppose gcd(c0, Y ) = d. Since d divides F (X,Y ), it necessarily dividesw∏i=1qbii ·v∏i=1pordpi (c)+Zii .In particular,d =w∏i=1qsii ·v∏i=1ptiifor some non-negative integers s1, . . . , sw, t1, . . . , tv such thatsi ≤ min{ordqi(c), ordqi(c0)} and ti ≤ min{ordpi(c) + Zi, ordpi(c0)}.From here, it is easy to see that ordp(d) ≤ ordp(c) for each rational prime p /∈ Sso that d ∈ D.Conversely, suppose d ∈ D so that ordp(d) ≤ ordp(c) for all p /∈ S. That is, theright-hand side ofordp(d) ≤ ordp(c) = ordp(w∏i=1qbii ·v∏i=1pordpi (c)i)25is non-trivial only at the primes {q1, . . . , qw}. In particular,d =w∏i=1qsii ·v∏i=1ptiifor non-negative integers s1, . . . , sw, t1, . . . , tv such thatsi ≤ min{ordqi(c), ordqi(c0)} and ti ≤ ordpi(c0).It follows that d = gcd(c0, Y ).For any d ∈ D, we define the rational numbersud = cn−10 /dn and cd = sgn(udc)∏p/∈Spordp(udc).On using that d ∈ D, we see that the rational number cd is in fact an integer coprimeto S. That is, gcd(cd, pi) = 1 for all pi ∈ S.Suppose (X,Y, Z1, . . . , Zv) is a solution of (3.1) with gcd(X,Y ) = 1 and d ∈ D.Define the homogeneous polynomial f(x, y) ∈ Z[x, y] of degree n byf(x, y) = xn + C1xn−1y + · · ·+ Cn−1xyn−1 + Cnyn,wherex = c0Xd , y =Yd and Ci = cici−10 for i = 1, . . . , n.Since gcd(X,Y ) = 1, the numbers x and y are also coprime integers by definitionof d. We observe thatf(x, y) = udF (X,Y ) = udcv∏i=1pZii = cd∏p∈SpZi+ordp(udc).Setting zi = Zi + ordp(udc) for all i ∈ {1, . . . , v}, we obtainf(x, y) = xn + C1xn−1y + · · ·+ Cn−1xyn−1 + Cnyn = cdpz11 · · · pzvv , (3.2)26where gcd(x, y) = 1 and gcd(cd, pi) = 1 for all i = 1, . . . , v.Since there are only finitely many choices for d = gcd(c0, Y ), there are onlyfinitely many choices for {cd, ud, d}. Then, solving (3.1) is equivalent to solvingthe finitely many Thue-Mahler equations (3.2) for each choice of {cd, ud, d}. Foreach such choice, the solution {x, y, z1, . . . , zv} is related to {X,Y, Z1, . . . , Zv}viaX =dxc0, Y = dy and Zi = zi − ordp(udc).Lastly, we observe that the polynomial f(x, y) of (3.2) remains the same for anychoice of {cd, ud, d}. Thus, to solve the family of equations (3.2), we need only toenumerate over every possible cd. Now, if C denotes the set of all {cd, ud, d} andd1, d2 ∈ D, we may have {cd1 , ud1 , d1}, {cd2 , ud2 , d2} ∈ C where cd1 = cd2 . Inother words, d1, d2 may yield the same value of cd, reiterating that we need onlysolve (3.2) for each distinct cd.3.2 The relevant algebraic number fieldFor the remainder of this chapter, we consider the Thue-Mahler equationf(x, y) = xn + C1xn−1y + · · ·+ Cn−1xyn−1 + Cnyn = cpz11 · · · pzvv (3.3)where gcd(x, y) = 1 and gcd(c, pi) = 1 for i = 1, . . . , pv.Following [116], putg(t) = f(t, 1) = tn + C1tn−1 + · · ·+ Cn−1t+ Cnand note that g(t) is irreducible in Z[t]. Let K = Q(θ) with g(θ) = 0. Now (3.3)is equivalent to the norm equationNK/Q(x− yθ) = cpz11 . . . pzvv . (3.4)27Let pi be any rational prime and let(pi)OK =mi∏j=1pe(pij |pi)ijbe the factorization of pi into prime ideals in the ring of integers OK of K. Letf(pij |pi) be the inertial degree of pij over pi. Since N(pij) = pfiji , (3.4) leads tofinitely many ideal equations of the form(x− yθ)OK = am1∏j=1pz1j1j · · ·mv∏j=1pzvjvj (3.5)where a is an ideal of norm |c| and the zij are unknown integers related to zibymi∑j=1f(pij |pi)zij = zifor i ∈ {1, . . . , v}.Our first task is to cut down the number of variables appearing in (3.5). We willdo this by showing that only a few prime ideals can divide (x− yθ)OK to a largepower.3.3 The prime ideal removing lemmaIn this section, we establish some key results that will allow us to cut down thenumber of prime ideals that can appear to a large power in the factorization of(x − yθ)OK . It is of particular importance to note that we do not appeal to thePrime Ideal Removing Lemma of Tzanakis and de Weger ([116]) here and insteadapply the following results of [46].Let p ∈ {p1, . . . , pv}. We will produce the following two finite lists Lp and Mp.The list Lp will consist of certain ideals b of OK supported at the prime idealsabove p. The list Mp will consist of certain pairs (b, p) where b is supported at theprime ideals above p and p is a prime ideal lying over p such that e(p|p) = 1 and28f(p|p) = 1. These lists will satisfy the following property: if (x, y, z1, . . . , zv) isa solution to the Thue-Mahler equation (3.3) then(i) either there is some b ∈ Lp such thatb | (x− yθ)OK , (x− yθ)OK/b is coprime to (p)OK ; (3.6)(ii) or there is a pair (b, p) ∈Mp and a non-negative integer vp such that(bpvp) | (x−yθ)OK , (x− yθ)OK/(bpvp) is coprime to (p)OK . (3.7)To generate the lists Mp, Lp we consider two affine patches, p - y and p | y. Webegin with the following lemmata.Lemma 3.3.1. Let (x, y, z1, . . . , zv) be a solution of (3.3) with p - y, let t be apositive integer, and suppose x/y ≡ u (mod pt), where u ∈ {0, 1, 2, . . . , pt − 1}.If q is a prime ideal of OK lying over p, thenordq(x− yθ) ≥ min{ordq(u− θ), t · e(q|p)}.Moreover, if ordq(u− θ) < t · e(q|p), thenordq(x− yθ) = ordq(u− θ).Lemma 3.3.2. Let (x, y, z1, . . . , zv) be a solution of (3.3) with p | y (and thusp - x), let t be a positive integer, and suppose y/x ≡ u (mod pt), where u ∈{0, 1, 2, . . . , pt − 1}. If q is a prime ideal of OK lying over p, thenordq(x− yθ) ≥ min{ordq(1− θu), t · e(q|p)}.Moreover, if ordq(1− θu) < t · e(q|p), thenordq(x− yθ) = ordq(1− θu).29Proof of Lemmas 3.3.1 and 3.3.2. Suppose p - y. Thus ordq(y) = 0 and henceordq(x− yθ) = ordq(x/y − θ).Since x/y − θ = u− θ + x/y − u, we haveordq(x/y − θ) = ordq(u− θ + x/y − u)≥ min{ordq(u− θ), ordq(x/y − u)}.By assumption,ordq(x/y − u) ≥ ordq(pt) = t · e(q|p),completing the proof of Lemma 3.3.1. The proof of Lemma 3.3.2 is similar.The following algorithm computes the lists Lp and Mp that come from the firstpatch p - y. We denote these respectively by Lp andMp.Algorithm 3.3.3. To compute Lp andMp:Step (1) LetLp ← ∅, Mp ← ∅,t← 1, U ← {w : w ∈ {0, 1, . . . , p− 1}}.Step (2) LetU ′ ← ∅.Loop through the elements u ∈ U . LetPu = {q lying above p : ordq(u− θ) ≥ t · e(q|p)}andbu =∏q|pqmin{ordq(u−θ),t·e(q|p)} = (u− θ)OK + ptOK .(i) If Pu = ∅ thenLp ← Lp ∪ {bu}.30(ii) Else if Pu = {p} with e(p|p) = f(p|p) = 1 and there is at least oneZp-root α of g(t) satisfying α ≡ u (mod pt), thenMp ←Mp ∪ {(bu, p)}.(iii) ElseU ′ ← U ∪ {u+ ptw : w ∈ {0, . . . , p− 1}}.Step (3) If U ′ 6= ∅ then lett← t+ 1, U ← U ′,and return to Step (2). Else output Lp,Mp.Lemma 3.3.4. Algorithm 3.3.3 terminates.Proof. Suppose otherwise. Write t0 = 1 and ti = t0 + i for i = 1, 2, 3, . . . .Then there is an infinite sequence of congruence classes ui mod pti such thatui+1 ≡ ui mod pti , and such that the ui fail the hypotheses of both (i) and (ii).This means that Pui is non-empty for every i ∈ N>0. By the pigeon-hole prin-ciple, some prime ideal p of OK appears in infinitely many of the Pui . Thusordp(ui − θ) ≥ ti · e(p|p) infinitely often. However, the sequence {ui}∞i=1 con-verges to some α ∈ Zp so that α = θ in Kp. This forces e(p|p) = f(p|p) = 1 andα to be a Zp-root of g(t). In this case, p corresponds to the factor (t − α) in thep-adic factorisation of g(t). There can be at most one such p, forcing Pui = {p}for all i. In particular, the hypothesis of (ii) are satisfied and we reach a contradic-tion.Lemma 3.3.5. Let p ∈ {p1, . . . , pv} and letLp,Mp be as given by Algorithm 3.3.3.Let (x, y, z1, . . . , zv) be a solution to (3.3). Then• either there is some b ∈ Lp such that (3.6) is satisfied;• or there is some (b, p) ∈ Mp with e(p|p) = f(p|p) = 1 and integer vp ≥ 0such that (3.7) is satisfied.31Proof. Lett0 = 1 and U0 = {w : w ∈ {0, 1, . . . , p− 1}}be the initial values for t and U in the algorithm. Then x/y ≡ u0 (mod pt0) forsome u0 ∈ U0. Write Ui for the value of U after i iterations of the algorithm andlet ti = t0 + i. As the algorithm terminates, Ui = ∅ for some sufficiently largei. Hence there is some i such that x/y ≡ ui mod pti where ui ∈ Ui, but thereis no element in Ui+1 congruent to x/y modulo pti+1 . In other words, ui mustsatisfy the hypotheses of either step (i) or (ii) of algorithm 3.3.3. Write u = ui andt = ti for x/y ≡ u mod pt and consider the ideal bu generated in this step. ByLemma 3.3.1, bu divides (x − yθ)OK . Furthermore, by definition of Pu, if q is aprime ideal of OK not contained in Pu, then (x− yθ)OK/bu is not divisible by q.Suppose first that the hypothesis of (i) is satisfied: Pu = ∅. The algorithm adds buto the set Lp, with the above remarks ensuring that (3.6) is satisfied.Suppose next that the hypothesis of (ii) is satisfied: Pu = {p} where e(p|p) = 1,f(p|p) = 1, and there is a unique Zp root α of g(t) such that α ≡ u mod pt.The algorithm adds (bu, p) to the set Mp. By the above, (x − yθ)OK/bu is anintegral ideal, not divisible by any prime ideal q 6= p lying over p. Thus there issome positive integer vp ≥ 0 such that (3.7) is satisfied, concluding the proof.Having computed the lists arising from the affine patch p - y, we initialize Lp andMp as Lp and Mp, respectively, and append to these lists the elements from thesecond patch, p | y, using the following algorithm.Algorithm 3.3.6. To compute Lp and Mp.Step (1) LetLp ← Lp, Mp ←Mp,where Lp,Mp are computed by Algorithm 3.3.3.Step (2) Lett← 2, U ← {pw : w ∈ {0, 1, . . . , p− 1}}.32Step (3) LetU ′ ← ∅.Loop through the elements u ∈ U . LetPu = {q lying above p : ordq(1− uθ) ≥ t · e(q|p)},andbu =∏q|pqmin{ordq(1−uθ),t·e(q|p)} = (1− uθ)OK + ptOK .(i) If Pu = ∅ thenLp ← Lp ∪ {bu}.(ii) ElseU ′ ← U ′ ∪ {u+ ptw : w ∈ {0, . . . , p− 1}}.Step (4) If U ′ 6= ∅ then lett← t+ 1, U ← U ′,and return to Step (3). Else output Lp, Mp.Lemma 3.3.7. Algorithm 3.3.6 terminates.Proof. Suppose that the algorithm does not terminate. Let t0 = 2 and ti = t0 + ifor i ∈ N. Then there is an infinite sequence of congruence classes {ui}∞i=0 andcorresponding sets {Pui}∞i=0 such that ui+1 ≡ ui mod ti and Pui 6= ∅ for alli. Moreover, p | u0. Let α be the limit of {ui}∞i=0 in Zp. By the pigeon-holeprinciple, there is some ideal q in OK above p which appears in infinitely manysets Pui . It follows that ordq(1 − uiθ) ≥ ti · e(q|p) and thus 1 − αθ = 0 in Kq.But as p | u0, we have ordp(α) ≥ 1, and so ordq(θ) < 0. This contradicts the factthat θ is an algebraic integer. Therefore the algorithm must terminate.Lemma 3.3.8. Let p ∈ {p1, . . . , pv} and letLp,Mp be as given by Algorithm 3.3.6.Let (x, y, z1, . . . , zv) be a solution to (3.3). Then33• either there is some b ∈ Lp such that (3.6) is satisfied;• or there is some (b, p) ∈ Mp with e(p|p) = f(p|p) = 1 and integer vp ≥ 0such that (3.7) is satisfied.Proof. Let (x, y, z1, . . . , zv) be a solution to (3.3). In view of Lemma 3.3.5 wemay suppose p | y. Then ordq(x) = 0 and ordq(x − yθ) = ordq(1 − (y/x)θ)for any prime ideal q lying over p. The remainder of the proof is analogous to theproof of Lemma 3.3.5.3.3.1 Computational remarks and refinementsIn implementing Algorithms 3.3.3 and 3.3.6, we reduce the number of prime idealsappearing to a large power in the factorization of (x − yθ)OK . The Prime IdealRemoving Lemma, as originally stated in Tzanakis - de Weger outlines a similarprocess by comparing the valuations of (x − yθ)OK at two prime ideals p1 andp2 above p. Of course if p1 | (x − yθ)OK , we restrict the possible values for xand y modulo p. However any choice of x and y modulo p affects the valuationsof (x − yθ)OK at all prime ideals above p. In the present refinement outlined byLemma 3.3.1 and Lemma 3.3.2, we instead study the valuations of (x − yθ)OKat all prime ideals above p simultaneously. This presents us with considerably lessideal equations of the form (3.5) to resolve.Moreover, this variant of the Prime Ideal Removing Lemma permits the followingadditional refinements:• Let b ∈ Lp. If there exists a pair (b′, p) ∈Mp with b′ | b and b/b′ = pw forsome w ≥ 0, then we may delete b from Lp. In doing so, the conclusion toLemma 3.3.8 continues to hold.• Suppose (b, p), (b′, p) ∈ Mp with b′ | b, and b/b′ = pw for some w ≥ 0.Then, we may delete (b, p) from Mp without affecting the conclusion toLemma 3.3.8.343.4 Factorization of the Thue-Mahler equationAfter applying Algorithm 3.3.3 and Algorithm 3.3.6, we are reduced to solvingfinitely many ideal equations of the form(x− yθ)OK = apu11 · · · puνν (3.8)in integer variables x, y, u1, . . . , uν with ui ≥ 0 for i = 1, . . . , ν, where 0 ≤ ν ≤ v.Here• for i ∈ {1, . . . , ν}, pi is a prime ideal of OK arising from Algorithm 3.3.3and Algorithm 3.3.6 applied to p ∈ {p1, . . . , pv}, such that (b, pi) ∈Mp forsome ideal b;• for i ∈ {ν + 1, . . . , v}, the corresponding rational prime pi ∈ S yieldsMpi = ∅, in which case we set ui = 0;• a is an ideal of OK of norm |c| · pt11 · · · ptvv such that ui + ti = zi.For each choice of a and prime ideals p1, . . . , pν , we reduce equation (3.8) to anumber of so-called “S-unit equations”. We present two different algorithms fordoing so and outline the advantages and disadvantages of each. In practice, we donot know a priori which of these two options is more efficient. Instead, we imple-ment and use both algorithms simultaneously, selecting the most computationallyefficient option as it appears.3.4.1 Avoiding the class group Cl(K)For i = 1, . . . , ν let hi be the smallest positive integer for which phii is principaland let ri be a positive integer satisfying 0 ≤ ri < hi. Letai = (a1i, . . . , aνi).where aii = hi and aji = 0 for j 6= i. We let A be the matrix with columnsa1, . . . ,aν . Hence A is a ν × ν diagonal matrix over Z with diagonal entries hi.35Now, if (3.8) has a solution u = (u1, . . . , uν), it necessarily must be of the formu = An + r, where n = (n1, . . . , nν) and r = (r1, . . . , rν). The vector n iscomprised of integers ni which we solve for. The vector r is comprised of thevalues ri satisfying 0 ≤ ri < hi for i = 1, . . . , ν.Using the above notation, we letci = p˜ai = pa1i1 · pa2i2 · · · paνiν = phiifor all i ∈ {1, . . . , ν}.Thus, we can write (3.8) as(x− yθ)OK = ap˜u = (a · p˜r) · cn11 · · · cnνν .By definition of hi, each i ∈ {1, . . . , ν} yields an element γi ∈ K× such thatci = (γi)OK .Furthermore, if u is a solution of (3.8) with corresponding vectors n, r, there existssome α ∈ K× such thata · p˜r = (α)OK .3.4.2 Using the class group Cl(K)Let u = (u1, . . . , uν) be a solution of (3.8) and consider the mapφ : Zν → Cl(K), (x1, . . . , xν) 7→ [p1]x1 · · · [pν ]xν ,where [q] denotes the equivalence class of the fractional ideal q. Since the productof a and pu11 · · · puνν defines a principal ideal, the map φ impliesφ(u) = [a]−1.36In particular, if [a]−1 does not belong to the image of φ then (3.8) has no solutions.We therefore suppose that [a]−1 belongs to the image. Let r = (r1, . . . , rν) denotea preimage of [a]−1 and observe that u−r belongs to the kernel of φ. The kernel isa subgroup of Zv of rank ν. Let a1, . . . ,aν be a basis for the kernel, whereai = (a1i, . . . , aνi) for i = 1, . . . , ν.Letu− r = n1a1 + · · ·+ nνaνfor some integers ni ∈ Z and let A denote the ν × ν matrix over Z with columnsa1, . . . ,aν . It follows that u = An+ r where n = (n1, . . . , nν).For ai = (a1i, . . . , aνi) ∈ Zν , we adopt the notationp˜a := pa1i1 · pa2i2 · · · paνiν .Letc1 = p˜a1 , . . . , cν = p˜aν .Thus, we can rewrite (3.8) as(x− yθ)OK = ap˜u = (a · p˜r) · cn11 · · · cnνν .Consider the ideal equivalence class of (a · p˜r) in Cl(K) and note that[a · p˜r] = [a] · [p1]r1 · · · [pν ]rν = [a] · φ(r1, . . . , rν) = [1]as φ(r1, . . . , rν) = [a]−1 by construction. This meansa · p˜r = (α)OKfor some α ∈ K×. Furthermore,[ci] = [p˜ai ] = φ(ai) = [1] for i = 1, . . . , ν,37as the ai are a basis for the kernel of φ. For all i ∈ {1, . . . , ν}, we thereforehaveci = (γi)OKfor some γi ∈ K×.3.4.3 The S-unit equationSection 3.4.1 and Section 3.4.2 outline two different algorithms which will allowus to reduce the ideal equation (3.8) to a number of certain “S-unit equations”.Regardless of which method we use, under both algorithms outlined above, equa-tion (3.8) becomes(x− yθ)OK = (α · γn11 · · · γnνν )OK (3.9)for some vector n = (n1, . . . , nν) ∈ Zν . The ideal generated by α in K hasnorm|c| · pt1+r11 · · · ptν+rνν ptν+1ν+1 · · · ptvvand the ni are related to the zi viazi = ui + ti =ν∑j=1njaij + ri + ti for i = 1, . . . , v,where ui = ri = 0 for all i ∈ {ν + 1, . . . , v}.Fix a complete set of fundamental units {ε1, . . . , εr} of OK . Here r = s + t − 1,where s denotes the number of real embeddings of K into C and t denotes thenumber of complex conjugate pairs of non-real embeddings of K into C. Then,under either method, equation (3.8) reduces to a finite number of equations in Kof the formx− yθ = αζεa11 · · · εarr γn11 · · · γnνν (3.10)with unknowns ai ∈ Z, ni ∈ Z, and ζ in the set T of roots of unity in OK . SinceT is finite, we treat ζ as another parameter.38Let p ∈ {p1, . . . , pv,∞}. Recall that g(t) is an irreducible polynomial in Z[t]arising from (3.3) such thatg(t) = f(t, 1) = tn + C1tn−1 + · · ·+ Cn−1t+ Cn.Denote the roots of g(t) in Qp (where Q∞ = R = C) by θ(1), . . . , θ(n). Leti0, j, k ∈ {1, . . . , n} be distinct indices and consider the three embeddings of KintoQp defined by θ 7→ θ(i0), θ(j), θ(k). We use z(i) to denote the image of z underthe embedding θ 7→ θ(i). From the Siegel identity(θ(i0)−θ(j))(x−yθ(k))+(θ(j)−θ(k))(x−yθ(i0))+(θ(k)−θ(i0))(x−yθ(j)) = 0,applying the embeddings to β = x − yθ yields the so-called “S-unit equation”δ1r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni− 1 = δ2r∏i=1(ε(i0)iε(j)i)ai ν∏i=1(γ(i0)iγ(j)i)ni, (3.11)whereδ1 =θ(i0) − θ(j)θ(i0) − θ(k) ·α(k)ζ(k)α(j)ζ(j), δ2 =θ(j) − θ(k)θ(k) − θ(i0) ·α(i0)ζ(i0)α(j)ζ(j)are constants.To summarize, our original problem of solving (3.3) for (x, y, z1, . . . , zv) has beenreduced to solving finitely many equations of the form (3.11) for the variables(x, y, n1, . . . , nν , a1, . . . , ar).3.4.4 Computational remarks and comparisonsIn Section 3.4.1, we follow closely the method of [116] to reduce the ideal equa-tion (3.8) to the S-unit equation (3.11). To implement this reduction, we beginby computing all hi for which phii is principal for i = 1, . . . , ν. In doing so, wegenerate all possible values for ri, the non-negative integer satisfying 0 ≤ ri < hi.We then generate every possible vector r = (r1, . . . , rν) and test the correspond-39ing ideal product a · p˜r for principality. Those vectors which pass this test yieldan S-unit equation (3.11). In the worst case scenario, this method reduces to hνKsuch equations, where hK is the class number of K. Moreover, this process needsto be applied to every ideal equation (3.8), yielding what may be a very large num-ber of principalization tests and subsequent large number of S-unit equations tosolve.In contrast, the method in Section 3.4.2 reduces (3.8) to only #T/2 S-unit equa-tions to solve, where T is the set of roots of unity in K. In particular, the sum totalof S-unit equations does not drastically increase. If [a]−1 is not in the image of φ,we reach a contradiction. If [a]−1 is in the image of φ then we obtain only #T/2corresponding equations (3.11). In particular, the number of principalization testsin this method is limited by the number of ideal equations (3.8), where each suchequation yields only (1 + ν) tests.However, when generating the vectors r = (r1, . . . , rν) using the class group, weobserve that some of the integers ri may be negative, so we do not expect α to be analgebraic integer in general. This can be problematic later in the algorithm whenwe compute the embedding of K into our p-adic fields. In those instances, theprecision on our p-adic fields may not be high enough, and as a result, some non-zero elements of K may be erroneously mapped to 0. To avoid this, we force theri to be positive by adding sufficiently many multiples of the class number.In most cases, the method described in Section 3.4.2 is far more efficient than thatof Section 3.4.1. However, computing the class group may be a very costly com-putation. Indeed, for some Thue-Mahler equations, this may be the bottle-neckof the algorithm. In this case, it may happen that computing the class group willtake longer than directly checking each potential S-unit equation arising from thealternative method. Unfortunately, we cannot know a priori how long computingCl(K) will take in so much that we cannot know a priori how long solving all S-unit equations from the other algorithm will take. In practice, generating the classgroup in Magma is a process which cannot be terminated without exiting the pro-gram. For this reason, we cannot simply apply a timeout in Magma if computingCl(K) is exceeding what we deem a reasonable amount of time. Adding to this,40Magma does not support parallelization, so we cannot implement both algorithmssimultaneously. Our compromise to solve a single Thue-Mahler equation is to runtwo separate instances of Magma in parallel, each generating the S-unit equationsusing the two aforementioned algorithms. When one of these instances finishes,the other is forced to terminate. Though this method is far from ideal, in this way,we are able to select the most computationally efficient option.3.5 A small upper bound for ul in a special caseWe now restrict our attention to those p ∈ {p1, . . . , pν} and study the p-adic val-uations of the numbers appearing in (3.11). In particular, for l ∈ {1, . . . , ν}, weidentify conditions in which∑νj=1 njalj can be bounded by a small explicit con-stant, where alj is the (l, j)th entry of the matrix A derived in either Section 3.4.1or Section 3.4.2. Recall that ul + rl =∑νj=1 njalj , where rl is known, so that abound on∑νj=1 njalj yields a bound on the exponent ul in (3.8).Fix a rational prime pl ∈ {p1, . . . , pν} and recall that z ∈ Cpl having ordpl(z) = 0is called a pl-adic unit. Part (i) of the Corollary of Lemma 7.2 of [116] tells us thatε(i0)1ε(j)1, . . . , ε(i0)rε(j)rand ε(k)1ε(j)1, . . . , ε(k)rε(j)rare pl-adic units.Let gl(t) be the irreducible factor of g(t) inQpl [t] corresponding to the prime idealpl. Since pl has ramification index and residue degree equal to 1, deg(gl(t)) = 1.We now choose i0 ∈ {1, . . . , n} so that θ(i0) is the root of gl(t). We fix this choiceof index i0 for the remainder of this chapter. The indices of j, k are fixed, butarbitrary.Lemma 3.5.1.(i) Let i ∈ {1, . . . , ν}. Then γ(k)iγ(j)iare pl-adic units.(ii) Let i ∈ {1, . . . , ν}. Then ordpl(γ(i0)iγ(j)i)= ali, where ai = (a1i, . . . , avi) isthe ith column of the matrix A of either Section 3.4.1 or Section 3.4.2.Proof. Consider the factorization g(t) = g1(t) · · · gm(t) of g(t) in Qpl [t]. Note41that θ(j) is a root of some gh(t) 6= gl(t). Let ph be the corresponding prime idealabove pl and e(ph|pl) be its ramification index. Then p 6= pl and since(γi)OK = pa1i1 · · · paviv ,we haveordpl(γ(j)i ) =1e(ph|pl) ordph(γi) = 0.An analogous argument gives ordpl(γ(k)i ) = 0. On the other hand,ordpl(γ(i0)i ) =1e(pl|pl) ordpl(γi) = ordpl(pa1i1 · · · paviv ) = ali.The next lemma deals with a special case in which the sum∑νj=1 njalj can becomputed directly. This lemma is analogous to Lemma 7.3 of [116].Recall the constantsδ1 =θ(i0) − θ(j)θ(i0) − θ(k) ·α(k)ζ(k)α(j)ζ(j), δ2 =θ(j) − θ(k)θ(k) − θ(i0) ·α(i0)ζ(i0)α(j)ζ(j)of (3.11).Lemma 3.5.2. Let l ∈ {1, . . . , v}. If ordpl(δ1) 6= 0, thenν∑i=1niali = min{ordpl(δ1), 0} − ordpl(δ2).Proof. Apply the Corollary of Lemma 7.2 of [116] and Lemma 3.5.1 to both ex-pressions of λ in (3.11). On the one hand, we obtain thatordpl(λ) = min{ordpl(δ1), 0},42and on the other hand,ordpl(λ) = ordpl(δ2) +ν∑i=1ordpl(γ(i0)iγ(j)i)ni= ordpl(δ2) +ν∑i=1niali.For the remainder of this section, we assume ordpl(δ1) = 0. Here, it is convenientto use the notationb1 = 1, b1+i = ni for i ∈ {1, . . . , ν},andb1+ν+i = ai for i ∈ {1, . . . , r}.Putα1 = logpl δ1, α1+i = logpl(γ(k)iγ(j)i)for i ∈ {1, . . . , ν},andα1+ν+i = logpl(ε(k)iε(j)i)for i ∈ {1, . . . , r}.DefineΛl =1+ν+r∑i=1biαi.Let L be a finite extension of Qpl containing δ1,γ(k)1γ(j)1, . . . , γ(k)νγ(j)ν, and ε(k)1ε(j)1, . . . , ε(k)rε(j)r.Since finite pl-adic fields are complete, αi ∈ L for i = 1, . . . , 1 + ν + r as well.Choose φ ∈ Qpl such that L = Qpl(φ) and ordpl(φ) > 0. Let G(t) be theminimal polynomial of φ over Qpl and let s be its degree. For i = 1, . . . , 1 + ν+ rwriteαi =s∑h=1αihφh−1, αih ∈ Qpl .43ThenΛl =s∑h=1Λlhφh−1, (3.12)withΛlh =1+ν+r∑i=1biαihfor h = 1, . . . , s.Lemma 3.5.3. For every h ∈ {1, . . . , s}, we haveordpl(Λlh) > ordpl(Λl)−12ordpl(Disc(G(t))).Proof. For h = 1, . . . , s, taking the images of (3.12) under conjugation φ 7→ φ(h)yields Λ(1)l...Λ(s)l =1 φ(1) · · · φ(1)s−1.........1 φ(s) · · · φ(s)s−1Λl1...Λls .The s× s matrix (φ(h)i−1) above is invertible, with inverse1∏1≤j<k≤s(φ(k) − φ(j))γ11 · · · γ1s......γs1 · · · γss ,where γjk is an integral polynomial in the entries of (φ(h)i−1). As ordpl(φ) > 0and ordpl(φ(h)) = ordpl(φ) for all h = 1, . . . , s, it follows that ordpl(γjk) > 0 forevery γjk. Therefore, asΛlh =1∏1≤j<k≤s(φ(k) − φ(j))s∑i=1γhiΛ(i)l ,44we haveordpl(Λlh) = min1≤i≤s{ordpl(γhi) + ordpl(Λ(i)l )}− 12ordpl(Disc(G(t)))≥ min1≤i≤sordpl(Λ(i)l ) + min1≤i≤sordpl(γhi)−12ordpl(Disc(G(t)))= ordpl Λl + min1≤i≤sordpl(γhi)−12ordpl(Disc(G(t)))for every h ∈ {1, . . . , s}.Lemma 3.5.4. Ifν∑i=1niali >1pl − 1 − ordpl(δ2),thenordpl(Λl) =ν∑i=1niali + ordpl(δ2).Proof. Immediate from Lemma 2.3.2.Lemma 3.5.5. Letwl =⌊1pl − 1 − ordpl(δ2)⌋.(i) If ordpl(α1) < min2≤i≤1+ν+rordpl(αi), thenν∑i=1niali ≤ max{wl,⌈min2≤i≤1+ν+rordpl(αi)− ordpl(δ2)⌉− 1}(ii) For all h ∈ {1, . . . , s}, if ordpl(α1h) < min2≤i≤1+ν+rordpl(αih), thenν∑i=1niali ≤ max{wl,⌈min2≤i≤1+ν+rordpl(αih)− ordpl(δ2) + dl⌉− 1},wheredl =12ordpl(Disc(G(t))).45Proof.(i) We prove the contrapositive. Supposeν∑i=1niali >1pl − 1 − ordpl(δ2),andν∑i=1niali ≥ min2≤i≤1+ν+rordpl(αi)− ordpl(δ2).Observe thatordpl(α1) = ordpl(Λl −1+ν+r∑i=2biαi)≥ min{ordpl(Λl), min2≤i≤1+ν+rordpl(biαi)}.Therefore, it suffices to show thatordpl(Λl) ≥ min2≤i≤1+ν+rordpl(biαi).By Lemma 2.3.2, the first inequality impliesordpl(Λl) =ν∑i=1niali + ordpl(δ2),from which the result follows.(ii) Similar to the proof of (i).3.6 Lattice-Based ReductionAt this point in solving the Thue-Mahler equation, we proceed to solve each S-unitequation (3.11) for the exponents (n1, . . . , nν , a1, . . . , ar). To do so, we generate46a very large upper bound on the exponents and reduce this bound via Diophantineapproximation computations. The specific details of this process are described inChapter 6 and Chapter 4. In general, from each S-unit equation, we generate sev-eral linear forms in logarithms to which we associate an integral lattice Γ. It will beimportant in this reduction process to enumerate all short vectors in these lattices.In this section, we describe two algorithms used in the short vector enumerationprocess.3.6.1 The L3-lattice basis reduction algorithmLet Γ be an n-dimensional lattice with basis vectors b1, . . . ,bn equipped with abilinear form Φ : Γ×Γ→ Z. Recall that Φ defines a norm on Γ via the usual innerproduct on Rn. For i = 1, . . . , n, define the vectors b∗i inductively byb∗i = bi −i−1∑j=1µijb∗j , µij =Φ(bi,b∗j )Φ(b∗j ,bj),where µij ∈ R for 1 ≤ j < i ≤ n. This is the usual Gram-Schmidt process. Thebasis b1, . . . ,bn is called LLL-reduced if|µij | ≤ 12for 1 ≤ j < i ≤ n,34|b∗i−1|2 ≤ |b∗i + µii−1b∗i−1|2 for 1 < i ≤ n,where | · | is the usual Euclidean norm in Rn,|v| = Φ(v,v) = vTv.These properties imply that an LLL-reduced basis is approximately orthogonal,and that, generically, its constituent vectors are roughly of the same length. Everyn-dimensional lattice has an LLL-reduced basis and such a basis can be computedvery quickly using the LLL algorithm [65]. This algorithm takes as input an arbi-trary basis for a lattice and outputs an LLL-reduced basis. The algorithm is typi-47cally modified to additionally output a unimodular matrix U such that A = BU ,whereB is the matrix whose column-vectors are the input basis andA is the matrixwhose column-vectors are the LLL-reduced output basis. Several versions of thisalgorithm are implemented in Magma, including de Weger’s exact integer version[118].We remark that a lattice may have more than one reduced basis, and that the order-ing of the basis vectors is not arbitrary. The properties of reduced bases that are ofmost interest to us are the following. Let v a vector in Rn and denote by l(Γ,v)the distance from v to the nearest point in the lattice Γ, viz.l(Γ,v) = minu∈Γ\{v}|u− v|.From an LLL-reduced basis for Γ, we can compute lower bounds for l(Γ,v), ac-cording to the following results.Lemma 3.6.1. Let Γ be a lattice with LLL-reduced basis c1, . . . , cn and let v be avector in Rn.(a) If v = 0, then l(Γ,v) ≥ 2−(n−1)/2|c1|.(b) Assume v = s1c1 + · · ·+ sncn, where s1, . . . , sn ∈ R with not all si ∈ Z. PutJ = {j ∈ {1, . . . , n} : sj /∈ Z}.For j ∈ J , setδ(j) =maxi>j ‖si‖|ci| if j < n0 if j = n,where ‖ · ‖ denotes the distance to the nearest integer. We havel(Γ,v) ≥ maxj∈J(2−(n−1)/2‖sj‖|c1| − (n− j)δ(j)).Lemma 4.5.8 (a) is Proposition 1.11 in [65]; proofs can be found in [65], [118](Section 3.4), or [105] (Section V.3). Lemma 4.5.8 (b) is a combination of Lemmas3.5 and 3.6 in [118]. Note that the assumption in Lemma 4.5.8 (b) is equivalent to48v /∈ Γ.We see that the vector c1 in a reduced basis is, in a very precise sense, not too farfrom being the shortest non-zero vector of Γ. As has already been mentioned, whatmakes this result so valuable is that there is a very simple and efficient algorithmto find a reduced basis in a lattice, namely the LLL algorithm.3.6.2 The Fincke-Pohst algorithmSometimes it is not sufficient to have a lower bound for l(Γ,v) only. It may beuseful to know exactly all vectors u ∈ Γ such that |u| = Φ(u,u) ≤ C for a givenconstant C. This can be done efficiently using an algorithm of Fincke-Pohst (cf.[45], [29]). A version of this algorithm with some improvements due to Stehle´ isimplemented in Magma. As input this algorithm takes a matrix B, whose columnsspan the lattice Γ, and a constant C > 0. The output is a list of all lattice pointsu ∈ Γ with |u| ≤ C, apart from u = 0. In this section, we outline the main stepsin this algorithm.We begin by letting B denote the basis matrix associated to the lattice Γ, withcorresponding bilinear form Φ. We call a vector u ∈ Γ small if its norm Φ(u,u) isless than a constant C. As an element of the lattice, u = Bx for some coordinatevector x ∈ Zn. Let Q be the quadratic form associated to Φ and let A = BTB.Now finding the short vectors u ∈ Γ is equivalent to solvingQ(x) = xTAx ≤ C. (3.13)Let x = (x1, . . . , xn). To solve this inequality, we first rearrange the terms of thequadratic form via quadratic completion. Here we assume that Γ is positive definiteso that every nonzero element of the lattice has a positive norm. With this, we findthe Cholesky decomposition A = RTR, where R is an upper triangular matrix,49and express Q asQ(x) =n∑i=1qiixi + n∑j=i+1qijxj2 .The coefficients qij are defined from R and stored in a matrix Q˜ for convenience.In particular,qij =rijriiif i < jr2ii if i = j.(3.14)Since R is upper triangular, the matrix Q˜ is as well. This yields the followingreformulation of (3.13)n∑i=1qiixi + n∑j=i+1qijxj2 ≤ C.From here we observe that the individual term qnnx2n must also be less than C.Specifically,x2n ≤Cqnnso that xn is bounded above by√C/qnn and below by−√C/qnn. This illustratesthe first step in establishing bounds on a specific entry xi. Adding more terms fromthe outer sum to this sequence, a pattern emerges. LetUk =n∑j=k+1qkjxj ,where Un = 0, and rewrite Q(x) asQ(x) =n∑i=1qiixi + n∑j=i+1qijxj2 = n∑i=1qii (xi + Ui)2 .50In general,qkk(xk + Uk)2 ≤ C −n∑i=k+1qii(xi + Ui)2.Let Tk denote the bound on the right-hand side,Tk = C −n∑i=k+1qii(xi + Ui)2.We set Tn = C and find each subsequent Tk by subtracting the next term from theouter summand,Tk = Tk+1 − qk+1,k+1(xk+1 + Uk+1)2.This yields the upper boundqkk(xk + Uk)2 ≤ Tkso that xk is bounded above by√Tk/qkk−Uk and below by−√Tk/qkk − Uk. Inthis way, we iteratively enumerate all vectors x satisfying Q(x) ≤ C, beginningwith the entry xn of x and working down towards x1.3.6.3 Computational remarks and translated latticesRecall that the Cholesky decomposition of A = BTB yields the upper triangularmatrix R where A = RTR. It is noted in the [45] that if we label the columns ofR by ri and the rows of R−1 by r′i, thenx2k =(r′ Tk ·n∑i=1xiri)2≤ r′ Tk rk(xTRTRx) ≤ |r′k|2C.To reduce the search space, it is thus beneficial to reduce the rows ofR−1. Further-more, rearranging the columns of R so that the shortest column vector is first helpsreduce the total running time of the Fincke-Pohst algorithm. In particular, doing soleads to progressively smaller intervals in which xk may exist.51We express this reduction with a unimodular matrix V −1 so that R−11 = V−1R−1.Applying an appropriate permutation matrix P , we then reorder the columns ofR1.Since R1 = RV , this yields R2 = (RV )P . Finally, we compute the solutions y toyTRT2 R2y ≤ C and recover the short vectors x satisfying the original inequality(3.13) via x = V Py.As before, let Γ be an n-dimensional lattice with basis matrix B, quadratic formΦ, and associated bilinear form Q. In Section 3.6.2, it is noted that an implemen-tation of the Fincke-Pohst algorithm is available in Magma. Unfortunately, thisimplementation does not support translated lattices, a variant of the Fincke-Pohstalgorithm which we will need in Chapter 6. By a translated lattice, we mean thediscrete subgroup of Rn of the formΓ +w ={n∑i=1xibi +w : xi ∈ Z},where b1, . . . ,bn form the columns of B and w ∈ Rn. In the remainder of thissection, we describe how to modify the Fincke-Pohst algorithm and its refinementsto support translated lattices.Analogous to the non-translated case, any embedded vector u of Γ + w may beexpressed as u = Bx + w for a corresponding coordinate vector x. In this case,we call the vector u ∈ Γ +w small if(x− c)TBTB(x− c) ≤ C (3.15)for some C ≥ 0, where c = −w.As in the usual short vectors process, we begin by applying Cholesky decomposi-tion to the positive definite matrix A = BTB to obtain an upper triangular matrixR satisfyingA = RTR. We then generate the matricesR1, R2, V, and P describedearlier in this section. This allows us to write A = UTGU for a unimodular matrixU and Gram matrix G given byU = P−1V −1 and G = RT2 R2.52Thus the inequality (3.15) becomes(y − d)TG(y − d) ≤ C (3.16)wherey = Ux and d = Uc.To enumerate the vectors y which satisfy this inequality, we consider the bilinearform Q associated to the lattice Γ. We express this form asQ(y − d) =n∑i=1qiiyi − di + n∑j=i+1qij(yj − dj)2 .As in the usual Fincke-Pohst algorithm, the coefficients qij are defined from thematrix R via equation (3.14). LetUk = −dk +n∑j=k+1qkj(yj − dj),where Un = −dn, and rewrite Q(y − d) asQ(y − d) =n∑i=1qiiyi − di + n∑j=i+1qij(yj − dj)2 = n∑i=1qii (yi + Ui)2 .From here, we proceed as in the usual Fincke-Pohst algorithm described in Sec-tion 3.6.2. Once we compute all vectors y which satisfy (3.16), we recover x usingx = U−1y.As a final remark about Fincke-Pohst for translated lattices, it is worth noting thatone could use the variant implemented in Magma simply by increasing the di-mension of the lattice Γ and appropriately redefining the basis vectors bi. This ishighly ill-advised as it increases the search space and subsequent running time ofthe algorithm.Generally speaking, the use of Fincke-Pohst in our applications poses one of the53main bottlenecks in solving Thue-Mahler and Thue-Mahler-like equations. Specif-ically, this algorithm often yields upwards of hundreds of millions of short vectors,each one needing to be stored and, in our case, appropriately manipulated. Thiscreates both timing and memory problems, often leading to gigabytes of data us-age. Deleting these vectors does not release the memory and, as with the classgroup function, Magma’s built-in Fincke-Pohst process cannot be terminated with-out exiting the program. The primary advantage of implementing and using ourown version of Fincke-Pohst, as described in this section, is therefore the ability toadd a fail-stop should the number of vectors found become too large.54Chapter 4Goormaghtigh EquationsLet m and n be integers such that m > n > 2, where either m = n+ 1 orgcd(m− 1, n− 1) = d > 1 (4.1)and consider the Goormaghtigh equationxm − 1x− 1 =yn − 1y − 1 , y > x > 1, m > n > 2. (4.2)In this chapter, we prove that, in fact, under assumption (4.1), equation (4.2) has atmost finitely many solutions which may be found effectively, even if we fix only asingle exponent.Theorem 4.0.1. If there is a solution in integers x, y, n and m to equation (4.2),satisfying (4.1), thenx < (3d)4n/d ≤ 36n. (4.3)In particular, if n is fixed, there is an effectively computable constant c = c(n)such that max{x, y,m} < c.We note that the latter conclusion here follows immediately from (4.3), in con-junction with, for example, work of Baker [5]. The constants present in our upperbound (4.3) may be sharpened somewhat at the cost of increasing the complexity55of our argument. By refining our approach, in conjunction with some new resultsfrom computational Diophantine approximation, we are able to achieve the com-plete solution of equation (4.2), subject to condition (4.1), for small fixed values ofn.Theorem 4.0.2. If there is a solution in integers x, y and m to equation (4.2), withn ∈ {3, 4, 5} and satisfying (4.1), then(x, y,m, n) = (2, 5, 5, 3) and (2, 90, 13, 3).Essentially half of the current chapter is concerned with developing Diophantineapproximation machinery for the case n = 5 in Theorem 4.0.2. Here, “off-the-shelf” techniques for finding integral points on models of elliptic curves or forsolving Ramanujan-Nagell equations of the shape F (x) = zn (where F is a poly-nomial and z a fixed integer) do not apparently permit the full resolution of thisproblem in a reasonable amount of time. Instead, we specialize the Thue-Mahlersolver refinements of Chapter 3 to the case of Ramanujan-Nagell equations, andto introduce some further sharpenings which enable us to complete the proof ofTheorem 4.0.2.4.1 Rational approximationsIn what follows, we will always assume that x, y,m and n are integers satisfying(4.2) with (4.1), and writem− 1 = dm0 and n− 1 = dn0. (4.4)We note, for future use, that an appeal to The´ore`me II of Karanicoloff [59] (which,in our notation, states that the only solution to (4.2) with n0 = 1 and m0 = 2in (4.4) is given by (x, y,m, n) = (2, 5, 5, 3)) allows us to suppose that either(x, y,m, n) = (2, 5, 5, 3), or that m0 ≥ 3 and n0 ≥ 1.Our starting point, as in, for example, [24] and [83], is the observation that the ex-istence of a solution to (4.2) with (4.1) implies a number of unusually good rational56approximations to certain irrational algebraic numbers. One such approximationarises from rewriting (4.2) asxxdm0x− 1 − yydn0y − 1 =1x− 1 −1y − 1 ,whereby ∣∣∣∣∣ d√y(x− 1)x(y − 1) −xm0yn0∣∣∣∣∣ < 1ydn0 . (4.5)The latter inequality was used, in conjunction with lower bounds for linear formsin logarithms (in [83]) and with machinery based upon Pade´ approximation to bi-nomial functions (in [24]), to derive a number of strong restrictions upon x, y andd satisfying equation (4.2).Our argument will be somewhat different, as we consider instead a rational ap-proximation to d√(x− 1)/x that is, on the surface, much less impressive than thatto d√y(x−1)x(y−1) afforded by (4.5). The key additional idea is that we are able to takeadvantage of the arithmetic structure of our approximations to obtain very stronglower bounds for how well they can approximate d√(x− 1)/x. This argument hasits genesis in work of Beukers [14], [15].For the remainder of this section, we will always assume that x ≥ 40. Fromyn − 1y − 1 = ydn0(1 +1y+ · · ·+ 1ydn0)andxm − 1x− 1 = xdm0(1 +1x+ · · ·+ 1xdm0),we thus haveydn0 <yn − 1y − 1 =xm − 1x− 1 <xx− 1 xdm0andyy − 1 xdm0 ≤ x+ 1xxdm0 <xm − 1x− 1 =yn − 1y − 1 <yy − 1 ydn0 ,57so thatxm0 < yn0 <(xx− 1)1/dxm0 ≤√40/39xm0 < 1.013xm0 . (4.6)We will rewrite (4.2) asxdm0 − (x− 1)xdn0∑j=0yj =1x.From this equation, we will show that d√(x− 1)/x is well approximated by arational number whose numerator is divisible by xm0 .If we define, as in Nesterenko and Shorey [83], Ak(d) via(1− 1X)−1/d=∞∑k=0Ak(d)X−k =∞∑k=0d−1(d−1 + 1) · · · (d−1 + k − 1)k!X−k,then we can writedn0∑j=0yj =(n0∑k=0Ak(d)yn0−k)d+(d−1)n0−1∑j=0Bj(d)yj .Here, the Bj are positive, monotone increasing in j, and satisfyB(d−1)n0−1(d) =nn0 + 1An0(d),while, for the Ak(d), we have the inequalitiesd+ 1kd2≤ Ak(d) ≤ d+ 12d2,valid provided k ≥ 2 (see displayed equation (14) of [83]).58We thus havexdm0 − (x− 1)x(n0∑k=0Ak(d)yn0−k)d=1x+x− 1x(d−1)n0−1∑j=0Bj(d)yj (4.7)and so0 < xdm0−(x− 1)x(n0∑k=0Ak(d)yn0−k)d<(dn0 + 1)(d+ 1)2(n0 + 1)d2yy − 1 y(d−1)n0−1.(4.8)Since(dn0 + 1)(d+ 1)2(n0 + 1)d2<d+ 12d≤ 34,from the fact that n0 ≥ 1 and d ≥ 2, and since y > x ≥ 40, we may conclude that0 < xdm0 − (x− 1)x(n0∑k=0Ak(d)yn0−k)d< 0.769 y(d−1)n0−1. (4.9)Applying the Mean Value Theorem,0 < xm0 − d√x− 1xn0∑k=0Ak(d)yn0−k < 0.769y(d−1)n0−1dY d−1, (4.10)where Y lies in the interval(d√x− 1xn0∑k=0Ak(d)yn0−k, xm0).We thus haveY d−1 >(x− 1x)(d−1)/dy(d−1)n0and so, from (4.10) and the fact that d ≥ 2 and x ≥ 40,0 < xm0 − d√x− 1xn0∑k=0Ak(d)yn0−k <0.779dy. (4.11)59Let us defineC(k, d) = dk∏p|dpordp(k!),where by ordp(z) we mean the largest power of p that divides a nonzero integer z.Here, k and d positive integers with d ≥ 2. Then we haveC(k, d) = dk∏p|dp[kp]+[kp2]+···and hence it follows thatC(k, d) <d∏p|dp1/(p−1)k . (4.12)Further (see displayed equation (18) of Nesterenko and Shorey [83]), and criti-cally for our purposes, C(k, d)Ak(d) is an integer. Multiplying equation (4.7) byC(n0, d) and settingP = C(n0, d)xm0 and Q = C(n0, d)n0∑k=0Ak(d)yn0−k, (4.13)then P and Q are integers and, defining = P − d√x− 1xQ, (4.14)we thus have, from (4.11), that the following result holds.Proposition 4.1.1. Suppose that (x, y,m, n) is a solution in integers to equation(4.2), with (4.1) and x ≥ 40. If we define  via (4.14), then0 <  <0.779C(n0, d)dy. (4.15)Our next goal will be to construct a second linear form δ, in 1 and d√(x− 1)/x,with the property that a particular linear combination of  and δ is a (relativelylarge) nonzero integer, a fact we will use to derive a lower bound on . This argu-60ment, which will employ off-diagonal Pade´ approximants to the binomial functiond√1 + z, follows work of Beukers [14], [15].To apply Proposition 4.1.1 and for our future arguments, we will have use ofbounds upon the quantity C(k, d).Proposition 4.1.2. If k is a positive integer, then2k ≤ C(k, 2) < 4kanddk ≤ C(k, d) < (2d log d)k,for d > 2.We will postpone the proof of this result until Section 4.6; the upper bound here forlarge dmay be sharpened somewhat, but this is unimportant for our purposes.4.2 Pade´ approximantsIn this section, we will define Pade´ approximants to (1+z)1/d, for d ≥ 2. Supposethat m1 and m2 are nonnegative integers, and setIm1,m2(z) =12pii∫γ(1 + zv)m2(1 + zv)1/dvm1+1(1− v)m2+1 dv,where γ is a closed, counter-clockwise contour, containing v = 0 and v = 1. Ap-plying Cauchy’s residue theorem, we may write Im1,m2(z) asR0+R1, whereRi = Resv=i((1 + zv)m2(1 + zv)1/dvm1+1(1− v)m2+1).NowR0 =1m1!limv→0dm1dvm1(1 + zv)m2(1 + zv)1/d(1− v)m2+1 = Pm1,m2(z)61andR1 =1m2!limv→1dm2dvm2(1 + zv)m2(1 + zv)1/dvm1+1= −Qm1,m2(z) (1 + z)1/d,wherePm1,m2(z) =m1∑k=0(m2 + 1/dk)(m1 +m2 − km2)zk (4.16)andQm1,m2(z) =m2∑k=0(m1 − 1/dk)(m1 +m2 − km1)zk. (4.17)Note that there are typographical errors in the analogous statement given in dis-played equation (2.3) of [6]. We take z = −1/x. Arguing as in the proof ofLemma 4.1 of [6], we find that|Im1,m2(−1/x)| =sin(pi/d)pi xm1+m2+1∫ 10vm2+1/d(1− v)m1−1/ddv(1− (1− v)/x)m2+1 . (4.18)Upon multiplying the identityPm1,m2(−1/x)−Qm1,m2(−1/x) d√x− 1x= Im1,m2(−1/x)through by xm2C(m2, d), and settingδ = C0P1 − d√x− 1xQ1,where we write m0 = m2 −m1,C0 = xm0C(m2, d)/C(m1, d), P1 = xm1C(m1, d)Pm1,m2(−1/x)andQ1 = xm2C(m2, d)Qm1,m2(−1/x), (4.19)62it follows, from Lemma 3.1 of Chudnovsky [26], that C0, P1 and Q1 are integers.Further, from (4.18),|δ| = sin(pi/d)C(m2, d)pi xm1+1∫ 10vm2+1/d(1− v)m1−1/ddv(1− (1− v)/x)m2+1 . (4.20)Recall that P and Q are defined as in (4.13). Here and henceforth, we will assumethatm2 −m1 = m0. (4.21)We haveLemma 4.2.1. If m1 and m2 are nonnegative integers satisfying (4.21), then itfollows that PQ1 6= C0P1Q.Proof. Let p be a prime with p | d. Thenordp(P ) = n0 ordp(d) + ordp(n0!) +m0 ordp(x),ordp(P1) = ordp(Q1) = ordp(Q) = 0andordp(C0) = m0 ordp(d) + ordp(m2!)− ordp(m1!) +m0 ordp(x).Since m2 −m1 = m0 > n0, we haveordp(C0P1QPQ1)= (m0 − n0) ordp(d) + ordp(m2!m1!n0!)> 0so thatordp(PQ1 − C0P1Q) = ordp(PQ1) = n0 ordp(d) + ordp(n0!) +m0 ordp(x)and, in particular, PQ1 − C0P1Q 6= 0.63It follows from Lemma 4.2.1 and its proof that PQ1−C0P1Q is a nonzero integermultiple of C(n0, d)xm0 , so that, from the definitions of  and δ,|Q1 − δQ| = |PQ1 − C0P1Q| ≥ C(n0, d)xm0 . (4.22)NowQ = C(n0, d)n0∑k=0Ak(d)yn0−k <yy − 1 C(n0, d) yn0 ≤ 1.025C(n0, d) yn0 ,since y > x ≥ 40, and hence, from (4.6),Q < 1.039C(n0, d)xm0 . (4.23)Combining (4.6), (4.15), (4.22) and (4.23), we thus haveProposition 4.2.2. Suppose that (x, y,m, n) is a solution in integers to equation(4.2), with (4.1) and x ≥ 40. If m0, n0 and d are defined as in (4.4), and m1and m2 are nonnegative integers satisfying (4.21), then for Q1 and |δ| as given in(4.19) and (4.20), we may conclude that|Q1| > 1.28 d (1− 1.039|δ|)xm0+m0/n0 . (4.24)In the other direction, we will deduce two upper bounds upon |Q1|; we will useone or the other depending on whether or not m1 is “large”, relative to x. The firstresult is valid for all choices of x.Proposition 4.2.3. If m1,m2 and x are integers with m2 > m1 ≥ 1 and x ≥ 2,define α = m2/m1 and |δ| as in (4.20). Then|Q1| < d√xx− 1((α+ 1)2α(e (α+ 1))m1 xm2 C(m2, d) + |δ|). (4.25)If x ≥ m1, we will have use of the following slightly sharper bound.Proposition 4.2.4. Ifm1 andm2 are integers withm2 > m1 ≥ 0 and x ≥ m1m2m1+m2 ,64then|Q1| < xx− 1(m1 +m2m1)C(m2, d)xm2 .Proof of Proposition 4.2.3. Let us write α = m2/m1 > 1 and definer(α, u) =12u((α+ 1)− (α− 1)u−√((α+ 1)− (α− 1)u)2 − 4u),(4.26)andM(α, x) =(1− r(α, 1/x)/x)α(1− r(α, 1/x))αr(α, 1/x) . (4.27)Via the Mean Value Theorem,1α+ 1< r(α, 1/x) <x(x− 1)(α+ 1) (4.28)and so, from calculus,M(α, x) <((x− 1) (α+ 1)− 1(x− 1)(α+ 1)− x)α· (α+ 1) < e (α+ 1) (4.29)andM(α, x) >(1 +x− 1xα)α (x− 1x)(α+ 1). (4.30)Arguing as in the proof of Lemma 3.1 of [6], we find that|C0P1| ≤ (1− r(α, 1/x)/x)1/dr(α, 1/x)(1− r(α, 1/x)) M(α, x)m1 xm2 C(m2, d),whereby inequalities (4.28) and (4.29) imply that|C0P1| < (α+ 1)2α(e (α+ 1))m1 xm2 C(m2, d).Since C0P1 = d√x−1x Q1 + δ, we conclude as desired.65Proof of Proposition 4.2.4. To bound Q1 from above, we begin by noting thatxm2 |Qm1,m2(−1/x)| =∣∣∣∣∣m2∑k=0(m1 − 1/dk)(m1 +m2 − km1)(−1)kxm2−k∣∣∣∣∣ .(4.31)Definingf(k) =(m1 − 1/dk)(m1 +m2 − km1),it follows that, for 0 ≤ k ≤ m2 − 1,f(k + 1)/f(k) =(m1 − 1/d− k)(m2 − k)(k + 1)(m1 +m2 − k) .If k ≤ m1 − 1, we thus have that0 < f(k + 1)/f(k) <(m1 − k)(m2 − k)(k + 1)(m1 +m2 − k) ≤m1m2m1 +m2. (4.32)If instead k ≥ m1,(m1 − k − 1)(m2 − k)(k + 1)(m1 +m2 − k) < f(k + 1)/f(k) < 0. (4.33)It follows via calculus, in this case, that|f(k + 1)/f(k)| < (m2 −m1 + 1)2(m2 +m1 + 1)2.We thus have that xm2 |Qm1,m2(−1/x)| is bounded above by(m1 +m2m1)xm2 +∣∣∣∣(m1 − 1/dm1)∣∣∣∣ (m2m1) m2∑k=m1+1xm2−kwhich implies the desired result.664.3 Proof of Theorem 4.0.1To prove Theorem 4.0.1, we will work with Pade´ approximants to (1 + z)1/d, as inSection 4.2, of degrees m1 and m2 where we choosem1 =[m02n0]and m2 = m0 +[m02n0], (4.34)for m0, n0 and d as given in (4.4). Here [x] denotes the greatest integer less than orequal to x. Let us assume further that x ≥ (3d)4n/d ≥ 66. We will make somewhatdifferent choices later, when we prove Theorem 4.0.2.Our strategy will be as follows. We begin by showing that δ as given in (4.20)satisfies |δ| < 11.039 , so that the lower bound upon |Q1| in Proposition 4.2.2 isnontrivial. From there, we will appeal to Proposition 4.2.3 to contradict Proposition4.2.2.4.3.1 Bounding δFrom the aforementioned The´ore`me II of Karanicoloff [59], we may suppose thatm0 ≥ 3 and hence, arguing crudely, since m2 ≥ m0 ≥ 3 and m1 ≥ 0, wehave ∫ 10vm2+1/d(1− v)m1−1/ddv(1− (1− v)/x)m2+1 < 1and hence, from (4.20),|δ| < sin(pi/d)C(m2, d)pi xm1+1≤ C(m2, d)pi xm1+1. (4.35)From (4.34), m1 + 1 > m02n0 and so, the assumption that x ≥ (3d)4n/d yields theinequalityxm1+1 > (3d)2m0 .67Applying Proposition 4.1.2, if d = 2, it follows from m1 ≤ m02n0 that|δ| < 1pi4m1 3−2m0 ≤ 8729pi< 0.01,since m0 ≥ 3 and n0 ≥ 1. Similarly, if d ≥ 3,|δ| < (2d log d)m0+m1(3d)2m0≤ (2d log d)m0+m02n0(3d)2m0=((2d log d)1+ 12n09d2)m0< 0.01,again from m0 ≥ 3 and n0 ≥ 1. Appealing to Proposition 4.2.2, we thus have, ineither case,|Q1| > 1.25 d xm0+m0/n0 . (4.36)4.3.2 Applying Proposition 4.2.3We will next apply Proposition 4.2.3 to deduce an upper bound upon |Q1|. To usethis result, we must first separately treat the case when m1 = 0. In this situation,Proposition 4.2.4 implies that|Q1| < xx− 1 C(m0, d)xm0 .Inequality (4.36) and x ≥ (3d)4n/d > (3d)4n0 thus lead to the inequalitiesC(m0, d) > dxm0/n0 > (3d)4m0 ,contradicting Proposition 4.1.2 in all cases.Assuming now that m1 ≥ 1, combining Proposition 4.2.3 with (4.36), d ≥ 2 andthe fact that α = 1 +m0/m1 ≥ 3, implies thatxm0n0−m1 < αC(m2, d) (e (α+ 1))m1 .Since m1 ≤ m0/2n0, x ≥ (3d)4n/d > (3d)4n0 and α = 1 + m0/m1, it follows68that(3d)2m0 < (1 +m0/m1)C(m0 +m1, d) (e (2 +m0/m1))m1and so9d2 < (1 +m0/m1)1/m0 C(m0 +m1, d)1/m0 (e (2 +m0/m1))m1/m0 . (4.37)If d = 2, Proposition 4.1.2 yields36 < (1 +m0/m1)1/m0 41+m1/m0 (e (2 +m0/m1))m1/m0 , (4.38)contradicting the fact that m0 ≥ max{3, 2m1}.If d ≥ 3, (4.37) and Proposition 4.1.2 lead to the inequality9d2 < (1 +m0/m1)1/m0 (2d log d)1+m1/m0 (e (2 +m0/m1))m1/m0 ,whence2.744 <9√d2√2(log d)3/2< (1 +m0/m1)1/m0 (e (2 +m0/m1))m1/m0 . (4.39)If n0 ≥ 3, then m0 ≥ 6m1 and hence(1 +m0/m1)1/m0 (e (2 +m0/m1))m1/m0 < 2.4,a contradiction, while, from the second inequality in (4.39), we find that d ≤ 1112or d ≤ 64, if n0 = 1 or n0 = 2, respectively.For these remaining values, we will argue somewhat more carefully. From (4.12)and (4.37),9d2 < (1 +m0/m1)1/m0d∏p|dp1/(p−1)1+m1/m0 (e (2 +m0/m1))m1/m0 .(4.40)69If n0 = 2 (so that m0 ≥ 4m1), we thus haved3/4 < 0.34∏p|dp1/(p−1)5/4 ,and hence, for 3 ≤ d ≤ 64, a contradiction. Similarly, if n0 = 1, we have fromm0 ≥ 3 that either (m0,m1) = (3, 1) or m0 ≥ 4. In the first case,d2/3 < 0.43∏p|dp1/(p−1)4/3 ,contradicting the fact that d ≤ 1112. If m0 ≥ 4 (so that m1 ≥ 2), then (5.30)implies the inequalityd1/2 <e1/2 · 2 · 31/2m19∏p|dp1/(p−1)3/2and hence, after a short computation and using that d ≤ 1112, either d = 6, m0 =2m1 and m1 ≤ 15, or d = 30 and (m0,m1) = (4, 2). In this last case,x6Q2,6(−1/x) =6∑k=0(2− 1/30k)(8− k2)(−x)6−kand so x6Q2,6(−1/x) is equal to28x6−41310x5+1711120x4+171116200x3+530413240000x2+3235501972000000x+294430591524880000000< 28x6,since x ≥ 66. From C(6, 30) = 52488000000, we have that|Q1| < 1.47 · 1013 x6.On the other hand, (4.36) implies that |Q1| > 37.5 · x8, so that x < 6.3 · 105,contradicting x ≥ (3d)4n/d > 904.70For d = 6, 2 ≤ m1 ≤ 15 and m0 = 2m1, we argue in a similar fashion, explicitlycomputing Qm1,m2(z) and finding that|Q1| < κm1x3m1 ,wherem1 κm1 m1 κm1 m1 κm12 1.89 · 108 7 1.35 · 1032 12 1.60 · 10573 2.30 · 1013 8 1.24 · 1037 13 1.89 · 10614 9.86 · 1017 9 1.29 · 1042 14 1.79 · 10665 1.09 · 1022 10 6.02 · 1046 15 1.28 · 10716 5.88 · 1027 11 1.13 · 1052With (4.36), we thus havexm1 <215κm1 ,and sox <(215κm1)1/m1< 5.5 · 104,contradicting our assumption that x ≥ 182n/3 ≥ 1814/3 > 7.2·105. This completesthe proof of Theorem 4.0.1.4.4 Proof of Theorem 4.0.2 for x of moderate sizeAs can be observed from the proof of Theorem 4.0.1, the upper bound x < (3d)4n/dmay, for fixed values of n (and hence d), be improved with a somewhat more care-ful argument. By way of example, for small choices of n, we may derive boundsof the shape x < x0(n), provided we assume that m ≥ m0(n) for effectivelycomputable m0, where we haven x0(n) n x0(n) n x0(n) n x0(n)3 38 5 676 7 11647 9 1957124 80 6 230 8 492 10 72043.71To prove Theorem 4.0.2, we will begin by deducing slightly weaker versions ofthese bounds, for n ∈ {3, 4, 5}, where the corresponding values m0 are amenableto explicit computation. Our arguments will closely resemble those of the pre-ceding section, with slightly different choices of m1 and m2, and with a certainamount of additional care. Note that, from Theorem 4.0.1, we may assume that weare in one of the following cases1. n = 3, d = 2, n0 = 1, 2 ≤ x ≤ 46655,2. n = 4, d = 3, n0 = 1, 2 ≤ x ≤ 122826,3. n = 5, d = 2, n0 = 2, 2 ≤ x ≤ 60466175,4. n = 5, d = 4, n0 = 1, 2 ≤ x ≤ 248831.Initially, we will suppose that x ≥ 40 and, in all cases, thatm1 andm2 are nonneg-ative integers satisfying (4.21). We will always, in fact, choose m1 positive. Againsetting m2 = αm1, via calculus, we may bound the integral∫ 10vm2+1/d(1− v)m1−1/ddv(1− (1− v)/x)m2+1in (4.20) by(maxv∈[0,1]v(α+1)/d(1− (1− v)/x)(α+d)/d)M(α, x)1/d−m1 < M(α, x)1/d−m1 .From (4.20), it thus follows that|δ| < sin(pi/d)C(m2, d)pi xm1+1M(α, x)1/d−m1 . (4.41)4.4.1 Case (1) : n = 3, d = 2, n0 = 1, x ≥ 40In this case, we will takem1 =⌈2m07⌉and m2 = m0 +⌈2m07⌉,72where by dxe we mean the least integer that is ≥ x, so that m1 ≥ 2m2/9, i.e.α ≤ 9/2. From (4.41) and Proposition 4.1.2,|δ| < M(α, x)1/2pix(4αxM(α, x))m1.Appealing to (4.30), since x ≥ 40 and α ≤ 9/2, it follows that4αxM(α, x)≤ 4α(1 + 3940α)α39 (α+ 1)< 1,whence, from (4.29),|δ| < M(α, x)1/2pix<(e (α+ 1))1/2pix< 0.031.We may therefore apply Proposition 4.2.2 to conclude that|Q1| > 2.477x2m0 . (4.42)From (4.25), Proposition 4.1.2, α ≤ 9/2 and x ≥ 40, we have|Q1| < 6.81 · 14.951m1 (4x)m0+m1and sox <(2.75 · 14.951m1 4m0+m1) 1m0−m1 . (4.43)We may check that m0 > 3.4m1 (so that α > 4.4) whenever m0 ≥ 96 and hence,since the right hand side of (4.43) is monotone decreasing in m0, may concludethat x < 40, a contradiction.For m0 ≤ 95, we note thatm1m2m1 +m2≤ m1 =⌈2m07⌉≤⌈2 · 957⌉= 28 < xand hence may appeal to Proposition 4.2.4. It follows from (4.42) and x ≥ 4073thatx <(C(m2, 2)2.415(m1 +m2m1)) 1m0−m1.A short computation leads to the conclusion that x < 40, unless m0 = 4 (in whichcase x ≤ 108) or m0 = 18 (whence x ≤ 40). In the last case, we therefore havex = 40 and m = 37, and we may easily check that there are no correspondingsolutions to equation (4.2). If m0 = 4 (so that m = 9) and 40 ≤ x ≤ 108, thereare, similarly, no solutions to (4.2) with n = 3.4.4.2 Case (2) : n = 4, d = 3, n0 = 1, x ≥ 85We argue similarly in this case, choosingm1 =⌈ m03.23⌉and m2 = m0 +⌈ m03.23⌉,so that α ≤ 4.23. From (4.41) and Proposition 4.1.2,|δ| <√3M(α, x)1/32pix(33α/2xM(α, x))m1.Applying (4.30), x ≥ 85 and α ≤ 4.23,33α/2xM(α, x)≤ 33α/2(1 + 8485α)α84 (α+ 1)< 1and so|δ| <√3M(α, x)1/32pix<√3 (e (α+ 1))1/32pix< 0.008.Proposition 4.2.2 thus implies|Q1| > 3.808x2m0 (4.44)while (4.25), Proposition 4.1.2, α ≤ 4.23 and x ≥ 85 give|Q1| < 6.5 · 14.217m1 (3√3x)m0+m1 .74It follows thatx <(1.707 · 14.217m1 (3√3)m0+m1) 1m0−m1 . (4.45)We may check that m0 ≥ 3.14m1, for all m0 ≥ 98 (and m1 ≥ 31) and hence, forthese m0, we have α ≥ 4.14 and sox < 1.7071/67 · 14.2171/2.14 · (3√3)4.14/2.14,which contradicts x ≥ 85.For m0 ≤ 97, we again find thatm1m2m1 +m2≤ m1 =⌈ m03.23⌉≤⌈ 973.23⌉= 31 < xand hence, from Proposition 4.2.4, (4.44) and x ≥ 85,x <(C(m2, 3)3.763(m1 +m2m1)) 1m0−m1,contradicting x ≥ 85, unless we have m0 = 4 and x ≤ 220, or m0 = 7 andx ≤ 138, or m0 = 10 and x ≤ 99, or m0 = 13 and x ≤ 110, or m0 = 20 andx ≤ 87. In each case, we may verify that there are no solutions to equation (4.2).By way of example, if m0 = 4, then m = 13 and a short computation reveals that,for 85 ≤ x ≤ 220, there are no corresponding solutions to (4.2).4.4.3 Case (3) : n = 5, d = 2, n0 = 2, x ≥ 720In this case, we will takem1 =⌈ m05.906⌉and m2 = m0 +⌈ m05.906⌉,so that α ≤ 6.906. From (4.41) and Proposition 4.1.2,|δ| < M(α, x)1/2pix(4αxM(α, x))m1.75Appealing to (4.30), since x ≥ 720 and α ≤ 6.906, it follows that4αxM(α, x)≤ 4α(1 + 719720α)α719 (α+ 1)< 1,whence, from (4.29),|δ| < M(α, x)1/2pix<(e (α+ 1))1/2pix< 0.003.We may therefore apply Proposition 4.2.2 to conclude that|Q1| > 2.552x 32m0 . (4.46)On the other hand, from (4.25), Proposition 4.1.2, α ≤ 6.906 and x ≥ 720 wehave|Q1| < 9.058 · 21.491m1 (4x)m0+m1 .It follows thatx <(3.550 · 21.491m1 4m0+m1) 2m0−2m1 .We may check that m0 > 5.809m1 (so that α > 6.809), for all m0 ≥ 332 andhence, for these m0, we havex < 3.5501/108 · 21.4912/3.809 · 42+6/3.809which contradicts x ≥ 720. For m0 ≤ 331,m1m2m1 +m2≤ m1 =⌈ m05.906⌉≤⌈ 3315.906⌉= 57 < xand hence Proposition 4.2.4, (4.46) and x ≥ 720 imply thatx <(C(m2, 2)2.548(m1 +m2m1)) 2m0−2m1,76contradicting x ≥ 720, unless we have m0 and 720 ≤ x ≤ x0 as follows :m0 x0 m0 x0 m0 x0 m0 x0 m0 x03 63090 12 2780 19 992 31 834 54 8366 578712 13 2531 20 909 36 859 55 7237 12601 14 1177 24 1101 37 777 65 7658 2605 15 755 25 847 42 849 71 7689 762 18 1667 30 1103 48 767 83 734Since we are assuming that m0 is odd, because gcd(m− 1, n− 1) = 2, this tablereduces to the following:m0 x0 m0 x0 m0 x0 m0 x0 m0 x03 63090 13 2531 25 847 55 723 83 7347 12601 15 755 31 834 65 7659 762 19 992 37 777 71 768For these remaining triples (x, n,m) = (x, 5, 2m0 + 1), with 720 ≤ x ≤ x0, justas in the cases n = 3 and n = 4, we reach a contradiction upon explicitly verifyingthat there are no integers y satisfying equation (4.2).4.4.4 Case (4) : n = 5, d = 4, n0 = 1, x ≥ 300In this case, we will takem1 =⌈ m02.93⌉and m2 = m0 +⌈ m02.93⌉,so that α ≤ 3.93. From (4.41) and Proposition 4.1.2,|δ| <√2M(α, x)1/42pix(8αxM(α, x))m1.77Appealing to (4.30), since x ≥ 300 and α ≤ 3.93, it follows that8αxM(α, x)≤ 8α(1 + 299300α)α299 (α+ 1)< 1,whence, from (4.29),|δ| <√2M(α, x)1/42pix<√2(e (α+ 1))1/42pix< 0.002.We may therefore apply Proposition 4.2.2 to conclude that|Q1| > 5.109x2m0 . (4.47)On the other hand, from (4.25), Proposition 4.1.2, α ≤ 3.93 and x ≥ 300 wehave|Q1| < 6.19 · 13.402m1 (8x)m0+m1 .It follows thatx <(1.212 · 13.402m1 8m0+m1) 1m0−m1 .We may check that m0 ≥ 2.87m1 (so that α ≥ 3.87) for all m0 ≥ 133 (and hencefor m1 ≥ 46) and hence, for these m0, we havex < 1.2121/87 · 13.4021/1.87 · 83.87/1.87which contradicts x ≥ 300.For m0 ≤ 132,m1m2m1 +m2≤ m1 =⌈ m02.93⌉≤⌈ 1322.93⌉= 46 < xand hence Proposition 4.2.4, (4.47) and x ≥ 300 imply thatx <(C(m2, 4)5.091(m1 +m2m1)) 1m0−m1.A short computation leads to the conclusion that x < 300 for all m0 ≤ 132, unless78we have m0 and x ≤ x0 as follows :m0 x0 m0 x0 m0 x03 33791 7 350 15 3434 600 9 502 18 3156 1131 12 434In the remaining cases, we again reach a contradiction upon explicitly verify-ing that there are no integers y satisfying equation (4.2) (assuming thereby x ≥300).4.4.5 Treating the remaining small values of x for n ∈ {3, 4}To deal with the remaining pairs (x, n) for n ∈ {3, 4, 5}, we can, in each case,reduce the problem to finding “integral points” on particular models of genus onecurves. Such a reduction is not apparently available for larger values of n. Incase n ∈ {3, 4}, this approach enables us to complete the proof of Theorem 4.0.2.When n = 5 (where we are left to treat values 2 ≤ x < 720), the resultingcomputations are much more involved. To complete them, we must work ratherharder; we postpone the details to the next section.Small values of x for n = 3To complete the proof of Theorem 4.0.2 for n = 3, it remains to solve equation(4.2) with 2 ≤ x ≤ 39. In this case, (4.2) becomesy2 + y + 1 =xm − 1x− 1 , (4.48)whereby(4(x− 1)2(2y + 1))2 = 64(x− 1)3xm − 16(3x+ 1)(x− 1)3.79Writing m = 3κ+ δ for κ ∈ Z and δ ∈ {0, 1, 2}, we thus haveY 2 = X3 − k, (4.49)forX = 4(x−1)xκ+δ, Y = 4(x−1)2(2y+ 1)xδ and k = 16(3x+ 1)(x−1)3x2δ.We solve equation (5.2.1) for the values of k arising from 2 ≤ x ≤ 39 and 0 ≤δ ≤ 2 rather quickly using Magma’s IntegralPoints routine (see [19]). The onlysolutions we find with the property that 4(x − 1)x2 | X are those coming fromtrivial solutions corresponding to m = 2, together with (x, δ,X, |Y |) equal to oneof(2, 1, 128, 1448), (2, 2, 32, 176), (5, 2, 800, 22400), (8, 2, 3584, 213248),(19, 2, 389880, 243441072), (26, 2, 11897600, 41038270000) or(27, 2, 227448, 108416880).Of these, only (x, δ,X, |Y |) = (2, 1, 128, 1448) and (2, 2, 32, 176) have the prop-erty thatX = 4(x−1)xt for t an integer, corresponding to the solutions (x, y,m) =(2, 90, 13) and (2, 5, 5) to equation (4.48), respectively.Small values of x for n = 4If n = 4 and we write m = 2κ + δ, for κ ∈ Z and δ ∈ {0, 1}, then (4.2)becomesxδ(xκ)2 = (x− 1)(y3 + y2 + y + 1) + 1,wherebyY 2 = X3 + xδ(x− 1)X2 + x2δ(x− 1)2X + x1+3δ (x− 1)2,forX = (x− 1)xδy and Y = (x− 1)xκ+2δ.80Once again applying Magma’s IntegralPoints routine, we find that the only pointsfor 2 ≤ x ≤ 84 and δ ∈ {0, 1}, and having (x − 1)x2 | Y correspond to eithertrivial solutions to (4.2) with either y = 0 or m = 4, or have δ = 1 and (x,X, |Y |)among(4, 48, 384), (9, 648, 17496), (16, 3840, 245760), (21, 1680, 79380),(21, 465360, 317599380), (25, 15000, 1875000), (36, 45360, 9797760),(41, 33620, 6320560), (49, 115248, 39530064), (64, 258048, 132120576),(65, 10400, 1352000), (81, 524880, 382637520).None of these triples lead to nontrivial solutions to (4.2) with n = 4.4.5 Small values of x for n = 5In case n = 5, solving equation (4.2) can, for a fixed choice of x, also be reducedto a question of finding integral points on a particular model of a genus 1 curve.Generally, for m odd, say m = 2κ+ 1, we can rewrite (4.2) asx (xκ)2 = (x− 1) (y4 + y3 + y2 + y + 1)+ 1,so that (xκ+1)2= (x2 − x) (y4 + y3 + y2 + y)+ x2.Applying Magma’s IntegralQuarticPoints routine, we may find solutions to themore general Diophantine equationY 2 = (x2 − x) (y4 + y3 + y2 + y)+ x2; (4.50)note that we always have, for each x, solutions (y, Y ) = (0,±x), (−1,±x) and(x,±x3).Unfortunately, it does not appear that this approach is computationally efficientenough to solve equation (4.50) in a reasonable time for all values of x with 2 ≤x < 720 (though it does work somewhat quickly for 2 ≤ x ≤ 59 and variousother x < 720). The elliptic curve defined by (4.50) has, in each case, rank at least812 (the solutions corresponding to (y, Y ) = (0, x) and (−1, x) are independentnon-torsion points). Magma’s IntegralQuarticPoints routine is based on boundsfor linear forms in elliptic logarithms and hence requires detailed knowledge of thegenerators of the Mordell-Weil group. Thus, when the rank is much larger than 2,Magma’s IntegralQuarticPoints routine can, in practice, work very slowly. This isthe case, for example, when x = 60 (where the corresponding elliptic curve hasrank 5 over Q).Instead, we will argue somewhat differently. We write (4.2) asFx(y, 1) = xm, (4.51)whereFx(y, z) = (x− 1)(y4 + y3z + y2z2 + yz3) + xz4.For the remainder of this section, we consider the homogeneous quartic form (4.51)for fixed x. Notably, we observe that this equation is a special case of the Thue-Mahler equation (3.1). In particular, if x = pα11 · · · pαvv is the prime factorizationof x with αi ≥ 0, then equation (4.51) becomesFx(y, 1) = pZ11 . . . pZvv (4.52)where Zi = mαi.To find all solutions to this equation, we will use linear forms in p-adic logarithmsto generate a very large upper bound on m. Then, applying several instances of theLLL lattice basis reduction algorithm, we will reduce the bound on m until it issufficiently small enough that we may perform a brute force search efficiently. Theremainder of this section is devoted to the details of this approach.4.5.1 First steps and small boundsFollowing arguments of Chapter 3 for solving Thue-Mahler equations, put S ={p1, . . . , pv}. This is the set of all distinct rational primes dividing x. As we82seek only those solutions (y, z, Z1, . . . , Zv) to (4.52) for which z = 1, here andhenceforth we write, for concision, F (y) = Fx(y, 1).Recall in Section 3.1 of Chapter 3 the setD. This set consists of all positive rationalintegers m dividing (x− 1) such that ordp(m) ≤ ordp(c) for all primes p /∈ S. Inour case, c = 1 so thatD = {1}. Thus the only possible values for ud, cd areud = (x− 1)3 and cd = (x− 1)3.Under the appropriate change of variables associated to ud, cd, this yieldsg(t) = (x−1)3F(tx− 1)= t4 +(x−1)t3 +(x−1)2t2 +(x−1)3t+x(x−1)3.Note that g(t) is irreducible in Z[t]. Writing K = Q(θ) with g(θ) = 0, it followsthat (4.52) is equivalent toNK/Q((x− 1)y − θ) = (x− 1)3pZ11 . . . pZvv . (4.53)Let(pi)OK =mi∏j=1pe(pij |pi)ijbe the factorization of pi into prime ideals in the ring of integers OK of K. In thisdecomposition, e(pij |pi) and f(pij |pi) denote the ramification index and residuedegree of pij respectively. Then, since N(pij) = pf(pij |pi)i , equation (4.53) leadsto finitely many ideal equations of the form((x− 1)y − θ)OK = am1∏j=1pz1j1j · · ·mv∏j=1pzvjvj (4.54)where a is an ideal of norm (x − 1)3 and the zij are unknown integers related tom by∑mij=1 f(pij |pi)zij = Zi = mαi. Applying Algorithms 3.3.3 and 3.3.6, wereduce the number of prime ideals appearing to a large power in this equation. In83doing so, we are reduced to solving finitely many equations of the form((x− 1)y − θ)OK = apu11 · · · puvv (4.55)in integer variables y, u1, . . . , uv with ui ≥ 0 for i = 1, . . . , v. Here• for i ∈ {1, . . . , v}, pi is a prime ideal of OK arising from Algorithm 3.3.3and Algorithm 3.3.6 applied to p ∈ {p1, . . . , pv}, such that (b, pi) ∈Mp forsome ideal b;• for any pi ∈ S such that Mpi = ∅, pi denotes the trivial ideal pi = (1)OK ;• a is an ideal ofOK of norm (x−1)3·pt11 · · · ptvv such that ui+ti = Zi = mαi.Remark 4.5.1. Unlike in [116] and [46], if, after applying Algorithm 3.3.3 andAlgorithm 3.3.6, we are in the situation that ui = 0 for some i in {1, . . . , v}, itfollows thatm =Ziαi=ui + tiαi=tiαi.We iterate this computation over all i ∈ {1, . . . , v} such that ui = 0 and take thesmallest m as our bound. For all of the values of x that we are interested in, thisbound on m is small enough that we may go directly to the final brute force searchfor solutions.Following Remark 4.5.1, for the remainder of this chapter, we assume that ui 6=0 for all i = 1, . . . , v. As in 3.4.3, we fix a complete set of fundamental units{ε1, . . . , εr} of OK . Recall r = s + t − 1, where s denotes the number of realembeddings of K into C and t denotes the number of complex conjugate pairs ofnon-real embeddings ofK intoC. A quick computation in Maple shows thatg(t) = t4 + (x− 1)t3 + (x− 1)2t2 + (x− 1)3t+ x(x− 1)3has only complex roots for x ≥ 2. It follows that we have no real embeddingsof K into R, two pairs of complex conjugate embeddings, and hence only onefundamental unit, ε1.Now, for each choice of a and prime ideals p1, . . . , pv, we reduce each equa-84tion (4.55) to a number of so-called “S-unit equations” via either procedure out-lined in Section 3.4.1 and Section 3.4.2 of Chapter 3. Regardless of which of theseprincipalization methods is used, we arrive at finitely many equations of the form(x− 1)y − θ = αζεa11 γn11 · · · γnvv (4.56)with unknowns a1 ∈ Z, ni ∈ Z≥0, and ζ in the set T of roots of unity in OK .Since T is also finite, we will treat ζ as another parameter. Moreover, we note thatthe ideal generated by α has norm(x− 1)3 · pt1+r11 · · · ptv+rvv , (4.57)and the ni are related to m viamαi = Zi = ui + ti =v∑j=1njaij + ri + ti.To summarize, our original problem of solving (4.52) is now reduced to the prob-lem of solving finitely many equations of the form (4.57) for the variablesy, a1, n1, . . . , nv.From here, we follow the arguments of Section 3.4.3 to deduce a so-called S-unitequation. In doing so, we eliminate the variable y and set ourselves the task ofbounding the exponents a1, n1, . . . , nv.In particular, let p ∈ {p1, . . . , pv,∞}. Denote the roots of g(t) in Qp (whereQ∞ = R = C) by θ(1), . . . , θ(4). Let i0, j, k ∈ {1, . . . , 4} be distinct indices andconsider the three embeddings of K into Qp defined by θ 7→ θ(i0), θ(j), θ(k). Weuse z(i) to denote the image of z under the embedding θ 7→ θ(i). Applying theseembeddings to β = (x− 1)y − θ yieldsλ = δ1(ε(k)1ε(j)1)a1 v∏i=1(γ(k)iγ(j)i)ni− 1 = δ2(ε(i0)1ε(j)1)a1 v∏i=1(γ(i0)iγ(j)i)ni, (4.58)85whereδ1 =θ(i0) − θ(j)θ(i0) − θ(k) ·α(k)ζ(k)α(j)ζ(j), δ2 =θ(j) − θ(k)θ(k) − θ(i0) ·α(i0)ζ(i0)α(j)ζ(j)are constants.Note that δ1 and δ2 are constants, in the sense that they do not depend upony, a1, n1, . . . , nv.Let l ∈ {1, . . . , v} and consider the prime p = pl. From now on we make thefollowing choice for the index i0. Let gl(t) be the irreducible factor of g(t) inQpl [t] corresponding to the prime ideal pl. Since pl has ramification index andresidue degree equal to 1, deg(gl[t]) = 1. We choose i0 ∈ {1, . . . , 4} so that θ(i0)is the root of gl(t). The indices of j, k are fixed, but arbitrary.By Lemma 3.5.2, if ordpl(δ1) 6= 0 for any l ∈ {1, . . . , v}, thenv∑i=1niali = min{ordpl(δ1), 0} − ordpl(δ2).For us, if this bound holds for any prime pl ∈ S, it follows thatm =∑vj=1 njalj + rl + tlαl=min{ordpl(δ1), 0} − ordpl(δ2) + rl + tlαl.In particular, we iterate this computation over all i ∈ {1, . . . , v} for which Lemma 3.5.2holds and take the smallest m as our bound on the solutions. We then compute allsolutions below this bound using a simple brute force search.For the remainder of this chapter, we may assume that ordpl(δ1) = 0, since other-wise a reasonable bound is afforded by Lemma 3.5.2.Following the notation of Section 3.5, we letb1 = 1, b1+i = ni for i ∈ {1, . . . , v},86andbv+2 = a1.Putα1 = logpl δ1, α1+i = logpl(γ(k)iγ(l)i)for i ∈ {1, . . . , v},andαv+2 = logpl(ε(k)1ε(l)1).DefineΛl =v+2∑i=1biαi.Let L be a finite extension of Qpl containing δ1,γ(k)iγ(l)i(for i = 1, . . . , v), and ε(k)1ε(l)1.Since finite p-adic fields are complete, αi ∈ L for i = 1, . . . , v+2 as well. Chooseφ ∈ Qpl such that L = Qpl(φ) and ordpl(φ) > 0. Let G(t) be the minimalpolynomial of φ over Qpl and let s be its degree. For i = 1, . . . , v+ 2 writeαi =s∑h=1αihφh−1, αih ∈ Qpl .ThenΛl =s∑h=1Λlhφh−1, (4.59)withΛlh =v+2∑i=1biαihfor h = 1, . . . , s.We recall several important lemmata from Section 3.5 which we restate here.Lemma 4.5.2. For every h ∈ {1, . . . , s}, we haveordpl(Λlh) > ordpl(Λl)−12ordpl(Disc(G(t))).87Lemma 4.5.3. Ifv∑i=1niali >1pl − 1 − ordpl(δ2),thenordpl(Λl) =v∑i=1niali + ordpl(δ2).Lemma 4.5.4. Letwl =⌊1pl − 1 − ordpl(δ2)⌋.(i) If ordpl(α1) < min2≤i≤v+2ordpl(αi), thenv∑i=1niali ≤ max{wl,⌈min2≤i≤v+2ordpl(αi)− ordpl(δ2)⌉− 1}(ii) For all h ∈ {1, . . . , s}, if ordpl(α1h) < min2≤i≤v+2ordpl(αih), thenv∑i=1niali ≤ max{wl,⌈min2≤i≤v+2ordpl(αih)− ordpl(δ2) + νl⌉− 1},whereνl =12ordpl(Disc(G(t))).Similar to Lemma 3.5.2, if Lemma 4.5.4 holds for pl givingv∑i=1niali ≤ Blfor some bound Bl as in the lemma, it follows thatm =∑vj=1 njalj + rl + tlαl≤ Bl + rl + tlαl.Again, we iterate this computation over all l ∈ {1, . . . , v} for which Lemma 4.5.4holds and take the smallest m as our bound on the solutions. We then compute all88solutions below this bound using a simple naive search.4.5.2 Bounding the∑vj=1 njaijAt this point, similar to [116], a very large upper bound for|a1|, v∑j=1nja1j , . . . ,v∑j=1njavjis derived using the theory of linear forms in logarithms. In practice, however,this requires that we compute the absolute logarithmic height of all terms of ourso-called S-unit equation, (4.58). More often than not, this proves to be a com-putational bottleneck, and is best avoided whenever possible. In particular, theapproach of Tzanakis and de Weger [116] requires the computation of the absolutelogarithmic height of each algebraic number in the product of (4.58). Unfortu-nately, in many such instances, the fundamental units may be very large, with eachcoefficient having over 105 digits in their representation. Similarly, the generatorsof our principal ideals may also be very large, making elementary operations onthem (such as division) a very time-consuming process. In the particular instanceof x = 60, by way of example, each coefficient of α has in excess of 20,000 digits.As a result of this, computing the absolute logarithmic height of these elements,a process which must be done for each choice of parameters ζ, a, p1, . . . , pv, iscomputationally painful. Instead of this approach, we appeal to results of Bugeaudand Gyo˝ry [23] to generate a (very large) upper bound for these quantities, which,while not sharp, will nevertheless prove adequate for our purposes. Following thenotation of [23], we now describe this bound.Arguing as in [23], put Zi = 4Ui+Vi with Ui, Vi ∈ Z, 0 ≤ Vi < 4 for i = 1, . . . , vand let RK and hK be the regulator and class number of K, respectively. Let Tbe the set of all extensions to K of the places of {p1, . . . , pv}. Let P denotemax{p1, . . . , pv}, and let RT denote the T -regulator of K. Further, let H be anupper bound for the maximum absolute value of the coefficients of F , namely89H = |x| = x. Let B = 3, let log∗ a denote max(log(a), 1), and letC8 = exp{c24PNRT (log∗RT )(log∗(PRT )log∗ P)(RK + vhK + log(HB′))},where N = 24, B′ ≤ BHP 4v = 2xP 4v, andc24 = 3v+1+25(v + 1)5(v+1)+12N3(v+1)+16= 3v+26(v + 1)5v+17N3v+19.Then, [23] shows that pUii ≤ C8. Now, log∗(PRT )/ log∗ P ≤ 2 log∗RT , sothatC8 ≤ exp{c24PNRT 2(log∗RT )2(RK + vhK + log(HB′))}.Lastly, we have, by [23] RT ≤ RKhK(4 log∗ P )4v. We note that the fundamentalunits of K may be very large, and so computing the regulator of K can be a verycostly computation. To avoid this, we simply appeal to the upper bound of [23],namelyRK <|Disc(K)|1/2(log |Disc(K)|)33!hK.Now we have all of the components necessary to explicitly compute an upperbound on C8, denoted C9 in [23], from which it follows thatUi ≤ log(C9)log piand hencemαi = Zi = 4Ui + Vi <4 log(C9)log(pi)+ Vi <4 log(C9)log(pi)+ 4.We thus obtain the inequalitym <4 log(C9)αi log(pi)+4αi= C10;we compute this for all pi ∈ {1, . . . , v} and select the smallest value of C10 as ourbound on m.90From (4.57), it follows that0 ≤v∑j=1njaij = mαi − ri − ti ≤ C10αi − ri − ti.At this point, converting this bound to a bound onm would yield far too large of anexponent to apply our brute force search. Instead, we must argue somewhat morecarefully. Note that||n||∞ = ||A−1(u− r)||∞ ≤ ||u− r||∞||A−1||∞,and somax1≤i≤v|ni| ≤ ||A−1||∞ max1≤i≤vv∑j=1njaij ≤ ||A−1||∞ max1≤i≤v(C10αi− ri− ti) = C11.4.5.3 A bound for |a1|In this subsection, we establish an upper bound for |a1| by considering two casesseparately. Our argument is based loosely on [116] but differs substantially in orderto accommodate our new S-unit equation, which, unlike in [116], may now havenegative exponents, ni. In this subsection, θ(1), . . . , θ(4) will denote the roots ofg(t) in C. We order the roots of g(t) in C so thatθ(1) = θ3 and θ(2) = θ4 ∈ C.PutC12 =∣∣∣∣∣∣log (x− 1)3min1≤i≤4|α(i)ζ(i)| + C10 log x∣∣∣∣∣∣andC13 =v∑j=1max1≤i≤4| log |γ(i)j ||91SetC14 = min(| log |ε(1)1 ||, | log |ε(2)1 ||)and let C15 be any number satisfying 0 < C15 < C143 . So we haveC14 − C15 > C14 − 3C15 > 0.Lemma 4.5.5. If min1≤i≤4|(x− 1)y − θ(i)| > e−C15|a1|, we have|a1| < C12 + C11C13C14 − 3C15 .Proof. Let k ∈ {1, 2} be an index such thatC14 = min(| log |ε(1)1 ||, | log |ε(2)1 ||)= | log |ε(k)1 ||.By (4.53),|β(k)| ·∏i 6=k|β(i)| = (x− 1)3 · pZ11 · · · pZvv ,therefore|(x− 1)y − θ(k)| = |β(k)| < (x− 1)3 · xC10 · e3C15|a1|.Now,|ε(k)a11 | =|(x− 1)y − θ(k)||α(k)ζ(k)||γ(k)1 |n1 · · · |γ(k)v |nv<(x− 1)3 · xC10 · e3C15|a1|min1≤i≤4|α(i)ζ(i)| · |γ(k)1 |n1 · · · |γ(k)v |nvfrom which it follows thatlog |ε(k)a11 | < log(x− 1)3min1≤i≤4|α(i)ζ(i)| + C10 log x+ 3C15|a1| −v∑j=1nj log |γ(k)j |.92Taking absolute values yields|a1|C14 = |a1|| log |ε(k)1 | < C12 + 3C15|a1|+v∑j=1|nj | log γ(k)j ||.Now|a1| <C12 +v∑j=1|nj || log |γ(k)j ||C14 − 3C15<C12 + C11v∑j=1| log |γ(k)j ||C14 − 3C15<C12 + C11C13C14 − 3C15 .Now, putC16 =⌊− 1C15log min1≤j≤t|Im(θ(j))|⌋.Lemma 4.5.6. If min1≤i≤n|(x− 1)y − θ(i)| ≤ e−C15|a1|, then|a1| ≤ C16.Proof.e−C15|a1| ≥ |(x− 1)y − θ(i)| ≥ |Im(θ(i))| ≥ min1≤j≤t|Im(θ(j))|,hence |a1| ≤ C16.It follows that|a1| ≤ max{C12 + C11C13C14 − 3C15 , C16}.934.5.4 The reduction strategyThe upper bounds on |a1|, v∑j=1nja1j , . . . ,v∑j=1njavjare expected to be very large. Enumeration of the solutions by a naive search atthis stage would be prohibitively expensive computationally. Instead, following themethods of [116], we reduce the above bound considerably by applying the LLL-algorithm to approximation lattices associated to the linear forms in logarithmsobtained from (4.58).In the standard algorithm for Thue-Mahler equations, this procedure is applied re-peatedly to the real/complex and p-adic linear forms in logarithms until no furtherimprovement on the bound is possible. The search space for solutions below thisreduced bound can then be narrowed further using the Fincke-Pohst algorithm ap-plied to the real/complex and p-adic linear forms in logarithms. Lastly, a sievingprocess and final enumeration of possibilities determines all solutions of the Thue-Mahler equation. In our situation however, after obtaining the above bounds, weapply the LLL algorithm for the p-adic linear forms in logarithms only.In each step, we let Nl denote the current best upper bound on∑vj=1 njalj , letA0 denote the current best upper bound on |a1|, and let M denote the current bestupper bound on m. We will use the notationb1 = 1, b1+i = ni for i ∈ {1, . . . , v},andbv+2 = a1of Section 4.5.1 frequently. It will therefore be convenient to let Bl denote thecurrent best upper bound for |bl| for l = 1, . . . , v + 2. ThenB1 = 1 and Bv+2 = A0.94For l = 1, . . . , v, using thatv∑j=1njalj < Nl, for l = 1, . . . , v,we compute|nl| ≤ max1≤i≤v|ni| ≤ ||A−1||∞ max1≤i≤vv∑j=1njaij ≤ ||A−1||∞ max1≤i≤v(Ni) = Bl+1.For each l ∈ {1, . . . , v}, our expectation is that the LLL algorithm will reduce theupper bound Nl to roughly logNl. Note that we expect the original upper boundsto be of size 10120 and hence a single application of our pl-adic reduction procedureshould yield a new bound Nl that is hopefully much smaller than 3000. Then wewould havem =∑vj=1 njalj + rl + tlαl<Nl + rl + tlαl= M < 3000at which point we could simply search naively (i.e. by brute force) for all solutionsarising from this S-unit equation. Of course, if this does not occur, we use ournew upper bound on m, M , to reduce the bounds N1, . . . , Nl−1, Nl+1, . . . , Nvviav∑j=1njaij = mαi − ri − ti ≤Mαi − ri − ti = Ni.We then repeat this procedure with pl+1 until M < 3000. We note that for allx with 2 ≤ x ≤ 719, the bound m < 3000 is, in each case, attained in 1 or 2iterations of LLL.Note also that if a bound on∑vj=1 njaij is obtained via Lemma 4.5.4, then wesimilarly compute the bound M on m and enter the final search. We may do sobecause this bound always furnishes a bound on m that is smaller than 3000 for xwith 2 ≤ x ≤ 719.Lastly, rather than testing each possible tuple (|a1|, |n1|, . . . , |nv|) as in [116], our95brute force search simply checks for solutions of (4.51) using the smallest boundobtained on m. Because of this, we may omit the reduction procedures on thereal/complex linear forms in logarithms, and furthermore, we need only to reducethe bounds on∑vj=1 njaij so that M < 3000.4.5.5 The pl-adic reduction procedureIn this section, we set some notation and give some preliminaries for the pl-adicreduction procedures. Consider a fixed index l ∈ {1, . . . , v}. Following Section4.5.1, we haveordpl(α1) ≥ min2≤i≤v+2ordpl(αi) and ordpl(α1h) ≥ min2≤i≤v+2(αih) h = (1, . . . , s).Let I be the set of all indices i′ ∈ {2, . . . , v + 2} for whichordpl(αi′) = min2≤i≤v+2ordpl(αi).We will identify two cases, the special case and the general case. The special caseoccurs when there is some index i′ ∈ I such that αi/αi′ ∈ Qpl for i = 1, . . . , v+2.The general case is when there is no such index.In the special case, let iˆ be an arbitrary index in I for which αi/αiˆ ∈ Qpl for everyi = 1, . . . , v + 2. We further defineβi = −αiαiˆi = 1, . . . , v + 2,andΛ′l =1αiˆΛl =v+2∑i=1bi(−βi).In the general case, we fix an h ∈ {1, . . . , s} arbitrarily. Then we let iˆ be an index96in {2, . . . , v + 2} such thatordpl(αiˆh) = min2≤i≤v+2(αih),and defineβi = −αihαiˆhi = 1, . . . , v + 2,andΛ′l =1αiˆhΛlh =v+2∑i=1bi(−βi).Now in both cases we have βi ∈ Zpl for i = 1, . . . , v + 2.Lemma 4.5.7. Supposev∑i=1niali >1pl − 1 − ordpl(δ2).In the special case, we haveordpl(Λ′l) =v∑i=1niali + dlwithdl = ordpl(δ2)− ordpl(αiˆ).In the general case we haveordpl(Λ′l) ≥v∑i=1niali + dlwithdl = ordpl(δ2)−12ordpl(Disc(G(t)))− ordpl(αiˆh).Proof. Immediate from Lemma 4.5.2 and Lemma 4.5.3.We now describe the pl-adic reduction procedure. Let µ,W2, . . . ,Wv+2 denotepositive integers. These are parameters that we will need to balance in order to97obtain a good reduction for the upper bound of∑vi=1 niali. We will discuss howto choose these parameters later in this section. For each x ∈ Zpl , let x{µ} denotethe unique rational integer in [0, pµl −1] such that ordpl(x−xµ) ≥ µ (ie. x ≡ x{µ}(mod pµl )). Let Γµ be the (v + 1)-dimensional lattice generated by the columnvectors of the matrixAµ =W2. . . 0Wiˆ−1Wiˆ+10. . .Wiˆβ{µ}2 · · · Wiˆβ{µ}iˆ−1 Wiˆβ{µ}iˆ+1· · · Wiˆβ{µ}v+2 Wiˆpµl.Putλ =1pµlv+2∑i=1bi(−β{µ}i)andy =0...0−Wiˆβµ1 ∈ Zv+1.Of course, we must compute the βi to pl-adic precision at least µ in order to avoiderrors here. We observe that y ∈ Γµ if and only if y = 0. To see that thisis true, note that y ∈ Γµ means there are integers z1, . . . , zv+1 such that y =Aµ[z1, . . . , zv+1]T . The last equation of this equivalence forces z1 = · · · = zv = 0and −β{µ}1 = zv+1pml . Since β{µ}1 ∈ [0, pml − 1], we must then have zv+1 = 0also. Hence y = 0.PutQ =v+2∑i=2W 2i B2i .98Lemma 4.5.8. If `(Γµ,y) > Q1/2 thenv∑i=1niali ≤ max{1pl − 1 − ordpl(δ2), µ− dl − 1, 0}Proof. We prove the contrapositive. Assumev∑i=1niali >1pl − 1 − ordpl(δ2),v∑i=1niali > µ− dl andv∑i=1niali > 0.Consider the vectorx = Aµb2...biˆ−1biˆ+1...bv+2λ=W2b2...Wiˆ−1biˆ−1Wiˆ+1biˆ+1...Wv+2bv+2−Wiˆbiˆ+ y.By Lemma 6.6.2,ordpl(v+2∑i=1bi(−βi))= ordpl(Λ′l) ≥v∑i=1niali + dl ≥ µ.Since ordpl(β{µ}i − βi) ≥ µ for i = 1, . . . , v + 2, it follows thatordpl(v+2∑i=1bi(−β{µ}i ))≥ µ,so that λ ∈ Z. Hence x ∈ Γµ. Now∑vi=1 niali > 0 so that there exists some isuch that niali 6= 0, and in particular, b1+i = ni 6= 0. Thus we cannot have x = y.99Therefore,`(Γµ,y)2 ≤ |x− y|2 =v+2∑i=2W 2i b2i ≤v+2∑i=2W 2i |bi|2 ≤v+2∑i=2W 2i B2i = Q.The reduction procedure works as follows. Taking Aµ as input, we first computean LLL-reduced basis for Γµ. Then, we find a lower bound for `(Γµ,y). If thelower bound is not greater than Q1/2 so that Lemma 4.5.8 does not give a newupper bound, we increase µ and try the procedure again. If we find that severalincreases of µ have failed to yield a new upper bound Nl and that the value of µhas become significantly larger than it was initially, we move onto the next l ∈{1, . . . , v}.If the lower bound is greater than Q1/2, Lemma 4.5.8 gives a new upper bound Nlfor∑vi=1 niali and hence for mm =∑vj=1 njalj + rl + tlαl<Nl + rl + tlαl= M.If M < 3000, we exit the algorithm and enter the brute force search. Otherwise,we update the bounds N1, . . . , Nl−1, Nl+1, . . . , Nv viav∑j=1njaij = mαi − ri − ti ≤Mαi − ri − ti = Ni.Then using|nl| ≤ max1≤i≤v|ni| ≤ ||A−1||∞ max1≤i≤vv∑j=1njaij ≤ ||A−1||∞ max1≤i≤v(Ni) = Bl+1.we update the Bi and repeat the above procedure until M < 3000 or until nofurther improvement can be made on the Bi, in which case we move onto the nextl ∈ {1, . . . , v}.1004.5.6 Computational conclusionsBottlenecks for this computation are generating the class group, generating the ringof integers of the splitting field ofK (this is entirely because of a Magma issue andcannot be avoided) and generating the unit group.An implementation of this algorithm is available athttp://www.nt.math.ubc.ca/BeGhKr/GESolverCode.As before, we have, for each x, solutions (x, y,m) = (x,−1, 1), (x, 0, 1), and(x, x, 5). For xwith 2 ≤ x ≤ 719, we find additional solutions (x, y,m) among(4, 1, 2), (5, 2, 3), (10,−2, 2), (10,−6, 4), (30, 2, 2), (60,−3, 2),(120, 3, 2), (204,−4, 2), (340, 4, 2), (520,−5, 2).Altogether, this computation took 3 weeks on a 16-core 2013 vintage MacPro, withthe case x = 710 being the most time-consuming, taking roughly 5 days and 16hours on a single core. This is the better timing attained for this value of x from ourtwo approaches, computed using the class group to generate the S-unit equations.The most time-consuming job when computing the class group was x = 719,which took 10 days and 8 hours. However, using our alternate code, the bettertiming for x = 719 was only 2 hours. Without computing the class group, the mosttime-consuming process was x = 654, which took 2 days and 7 hours. However,this is the faster timing that was attained for this value of x, as computing the classgroup took roughly 4 days and 8 hours.We list below some timings for our computation. These times are listed in seconds,with the second column indicating the algorithm requiring the computation of theclass group, and the third column indicating the time taken by the algorithm whichavoids the class group. In implementing these two algorithms, we terminated thelatter algorithm if the program ran longer than its class group counterpart took.From these timings, it is clear that it is not always easy to predict which algorithmwill prove faster.101x Timing with Cl(K) Timing without Cl(K) Solutions689 647.269 Terminated [−1, 1], [0, 1], [689, 5]690 215306.420 Terminated [−1, 1], [0, 1], [690, 5]691 456194.210 1821.049 [−1, 1], [0, 1], [691, 5]692 152385.640 Terminated [−1, 1], [0, 1], [692, 5]693 36922.540 1908.230 [−1, 1], [0, 1], [693, 5]694 8288.190 Terminated [−1, 1], [0, 1], [694, 5]695 362453.820 9786.649 [−1, 1], [0, 1], [695, 5]696 76273.470 Terminated [−1, 1], [0, 1], [696, 5]697 14537.219 725.340 [−1, 1], [0, 1], [697, 5]698 451700.650 2708.920 [−1, 1], [0, 1], [698, 5]Full computational details are available athttp://www.nt.math.ubc.ca/BeGhKr/GESolverData,including the timings obtained for each value of x, under both iterations of thealgorithm.This completes the proof of Theorem 4.0.2.4.6 Bounding C(k, d) : the proof of Proposition 4.1.2To complete the proof of Proposition 4.1.2, from (4.12), it remains to show that∏p|d p1/(p−1) < 2 log d, provided d > 2. We verify this by explicit calculation forall d ≤ d0 = 105.Since log p/(p− 1) is decreasing in p, if we denote by ω(d) the number of distinctprime divisors of d, we have∑p|dlog pp− 1 ≤∑p≤pω(d)log pp− 1 , (4.60)102where pk denotes the kth smallest prime. Since we have∑p≤p10log pp− 1 < log(2 log(d0)),we may thus suppose that ω(d) ≥ 11, wherebyd ≥ d1 = 2 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29 · 31 = 200560490130.The fact that ∑p≤p21log pp− 1 < log(2 log(d1))thus implies that ω(d) ≥ 22 andd ≥ d2 =∏1≤i≤22pi > 3 · 1030.We iterate this argument, finding that∑p≤pκ(j)log pp− 1 < log(2 log(dj)),so thatd ≥ dj+1 =∏1≤i≤κ(j)+1pi,for j = 0, 1, 2, 3, 4 and 5, whereκ(0) = 10, κ(1) = 21, κ(2) = 50, κ(3) = 130, κ(4) = 361 and κ(5) = 1055.We thus have that ω(d) ≥ 1056 andd ≥∏1≤i≤1056pi > e8316.103We may thus apply The´ore`me 12 of of Robin [95] to conclude thatω(d) ≤ log dlog log d+ 1.4573log d(log log d)2<7 log d6 log log d,while the Corollary to Theorem 3 of Rosser-Schoenfeld yieldspn < n(log n+ log log n) <109n log n.It follows thatpω(d) <3527log dlog log dlog(7 log d6 log log d)<3527log d.By Theorem 6 of Rosser-Schoenfeld, we have∑p<xlog pp< log x− 1.33258 + 12 log x, (4.61)for all x ≥ 319. Also, if j ≥ 2,∫ ∞klog uujdu =(j − 1) log(k) + 1(j − 1)2kj−1 . (4.62)For 2 ≤ j ≤ 10, we have∑p<xlog ppj<∑p<106log ppj+∑p>106log ppj<∑p<106log ppj+∫ ∞106log uujdu,whereby ∑p<xlog ppj<∑p<106log ppj+(j − 1) log(106) + 1(j − 1)2106(j−1) . (4.63)By explicit computation, from (4.63), we find that10∑j=2∑p<xlog ppj< 0.755, (4.64)104while, from (4.62),∑j≥11∑p<xlog ppj<∑j≥11(j − 1) log(2) + 1(j − 1)22j−1 <∑j≥111(j − 1)2j−1 . (4.65)Evaluating this last sum explicitly, it follows that∑j≥2∑p<xlog ppj< 0.755 + log(2)− 447047645120< 0.756,whereby, from (4.61), if x ≥ 319,∑p<xlog pp− 1 < log x− 0.489.Applying this last inequality with x = 3527 log d >3527 · 8316 = 10780, we concludefrom our earlier arguments that∑p|dlog pp− 1 < log log d.This completes the proof of Proposition 4.1.2.4.7 Concluding remarksThe techniques employed in this chapter may be used, with very minor modifica-tions, to treat equation (4.2), subject to condition (4.1), with the variables x and yintegers (rather than just positive integers). Since(−a− 1)3 − 1(−a− 1)− 1 =a3 − 1a− 1 ,in addition to the known solutions (x, y,m, n) = (2, 5, 5, 3) and (2, 90, 13, 3) to(4.2), we also find (x, y,m, n) = (2,−6, 5, 3) and (2,−91, 13, 3), where we haveassumed that |y| > |x| > 1. Beyond these, a short computer search uncovers only105three more integer tuples (x, y,m, n) satisfyingxm − 1x− 1 =yn − 1y − 1 , m > n ≥ 3, |y| > |x| > 1,namely(x, y,m, n) = (−2,−7, 7, 3), (−2, 6, 7, 3) and (−6, 10, 5, 4).Perhaps there are no others; we can prove this to be the case if, for example,n = 3, subject to (4.1). This result was obtained earlier as Corollary 4.1 ofYuan [121], though the statement there overlooks the solutions (x, y,m, n) =(−2, 6, 7, 3), (2,−6, 5, 3) and (2,−91, 13, 3).106Chapter 5Computing Elliptic Curves overQIn the chapter at hand, we outline an algorithm to compute elliptic curves overQ, based upon techniques of solving Thue-Mahler equations. Our aim is to givea straightforward demonstration of the link between the conductors of the ellipticcurves in question and the corresponding equations, and to make the Diophantineapproximation problem that follows as easy to tackle as possible. It is worth notinghere that these connections are quite straightforward for primes p > 3, but requirecareful analysis at the primes 2 and 3. We will demonstrate our approach for anumber of specific conductors and sets S, and then focus our main computationalefforts on curves with bad reduction at a single prime (i.e. curves of conductor p orp2 for p prime). In these cases, the computations simplify significantly and we areable to find all curves of prime conductor up to 2× 109 (1010 in the case of curvesof positive discriminant) and conductor p2 for p ≤ 5× 105. We then extend thesecomputations in the case of conductor p, for prime p ≤ 2× 1013, and conductor p2for prime p ≤ 1010. We are not, however, able to guarantee completeness for theseextended computations (we will discuss this further in what follows).1075.1 Elliptic curvesOur basic problem is to find a model for each isomorphism class of elliptic curvesover Q with a given conductor. Let S = {p1, . . . , pv} where the pi are distinctprimes, and fix a conductor N = pη11 · · · pηvv for ηi ∈ N. Any curve of conductorN has a minimal modelE : y2 + a1xy + a3y = x3 + a2x2 + a4x+ a6with the ai integral and discriminant∆E = (−1)δpγ11 · · · pγvv ,where the γi are positive integers satisfying γi ≥ ηi, for each i = 1, 2, . . . , v, andδ ∈ {0, 1}.Writingb2 = a21 + 4a2, b4 = a1a3 + 2a4, b6 = a23 + 4a6, c4 = b22 − 24b4andc6 = −b32 + 36b2b4 − 216b6,we have 1728∆E = c34 − c26 and jE = c34/∆E . It follows thatc26 = c34 + (−1)δ+126 · 33 · pγ11 · · · pγkk . (5.1)In fact, it is equation (5.1) that lies at the heart of our method (see also Cremonaand Lingham [35] for an approach to the problem that takes as its starting pointequation (5.1), but subsequently heads in a rather different direction).Let νp(x) = ordp(x) be the largest power of a prime p dividing a nonzero integerx. Since our model is minimal, we may suppose (via Tate’s algorithm; see, forexample, Papadopoulos [88]) thatmin{3νp(c4), 2νp(c6)} < 12 + 12νp(2) + 6νp(3),108for each prime p, whileνp(NE) ≤ 2 + νp(1728).For future use, it will be helpful to have a somewhat more precise determination ofthe possible values of νp(c4) and νp(c6) we encounter. We compile this data fromPapadopoulos [88] and summarize it in Tables 5.1, 5.2 and 5.3.ν2(c4) ν2(c6) ν2(∆E) ν2(N)0 0 ≥ 0 min{1, ν2(∆E)}≥ 4 3 0 0≥ 4 5 4 2, 3 or 4≥ 4 ≥ 6 6 5 or 64 6 7 74 6 8 2, 3 or 44 6 9 54 6 10 or 11 3 or 44 6 ≥ 12 45 7 8 7≥ 6 7 8 2, 3 or 4ν2(c4) ν2(c6) ν2(∆E) ν2(N)5 ≥ 8 9 8≥ 6 8 10 66 ≥ 9 12 5 or 66 9 ≥ 14 67 9 12 5≥ 8 9 12 46 9 13 77 10 14 77 ≥ 11 15 8≥ 8 10 14 6Table 5.1: The possible values of ν2(c4), ν2(c6), ν2(∆E) and ν2(N).ν3(c4) ν3(c6) ν3(∆E) ν3(N)0 0 ≥ 0 min{1, ν3(∆E)}1 ≥ 3 0 0≥ 2 3 3 2 or 32 4 3 32 ≥ 5 3 22 3 4 42 3 5 32 3 ≥ 6 2≥ 3 4 5 53 5 6 4ν3(c4) ν3(c6) ν3(∆E) ν3(N)3 ≥ 6 6 2≥ 4 5 7 5≥ 4 6 9 2 or 34 7 9 34 ≥ 8 9 24 6 10 44 6 11 3≥ 5 7 11 55 8 12 4≥ 6 8 13 5Table 5.2: The possible values of ν3(c4), ν3(c6), ν3(∆E) and ν3(N).109νp(c4) νp(c6) νp(∆E) νp(N)0 0 ≥ 1 1≥ 1 1 2 21 ≥ 2 3 2≥ 2 2 4 2≥ 2 ≥ 3 6 2νp(c4) νp(c6) νp(∆E) νp(N)2 3 ≥ 7 2≥ 3 4 8 23 ≥ 5 9 2≥ 4 5 10 2Table 5.3: The possible values of νp(c4), νp(c6), νp(∆E) and νp(N) whenp > 3 is prime and p | ∆E .5.2 Cubic forms : the main theorem and algorithmWe now turn our attention to cubic forms and our main result. Fix integers a, b, cand d, and consider the binary cubic formF (x, y) = ax3 + bx2y + cxy2 + dy3, (5.2)with discriminantDF = −27a2d2 + b2c2 + 18abcd− 4ac3 − 4b3d. (5.3)To any such form, we can associate a pair of covariants, the HessianH = HF :H = HF (x, y) = −14(∂2F∂x2∂2F∂y2−(∂2F∂x∂y)2)and the Jacobian determinant of F andH , a cubic formG = GF defined byG = GF (x, y) =∂F∂x∂H∂y− ∂F∂y∂H∂x.These satisfy the syzygy4H(x, y)3 = G(x, y)2 + 27DFF (x, y)2 (5.4)110as well as the resultant identities:Res(F,G) = −8D3F and Res(F,H) = D2F . (5.5)Note here that we could just as readily work with −G instead of G here (corre-sponding to taking the Jacobian determinant of H and F , rather than of F and H).Indeed, as we shall observe in Section 5.4.4, for our applications we will, in somesense, need to consider both possibilities.Notice that if we set (x, y) = (1, 0) and multiply through byD6/4 (for any rationalD), then this syzygy can be rewritten as(D2H(1, 0))3 −(D32G(1, 0))2= 1728 · D6DF256F (1, 0)2.Given an elliptic curve with corresponding invariants c4, c6 and ∆E , we will showthat it is always possible to construct a binary cubic form F , with correspondingDfor whichD2H(1, 0) = c4, −12D3G(1, 0) = c6 and ∆E = D6DFF (1, 0)2256(and hence equation (5.1) is satisfied). This is the basis of the proof of our main re-sult, which provides an algorithm for computing all isomorphism classes of ellipticcurves E/Q with conductor a fixed positive integer N . Though we state our resultfor curves with jE 6= 0, the case jE = 0 is easy to treat separately (see Section5.2.1).Theorem 5.2.1. Let E/Q be an elliptic curve of conductor N = 2α3βN0, whereN0 is coprime to 6 and 0 ≤ α ≤ 8, 0 ≤ β ≤ 5. Suppose further that jE 6= 0. Thenthere exists an integral binary cubic form F of discriminantDF = sign(∆E)2α03β0N1,and relatively prime integers u and v withF (u, v) = ω0u3 + ω1u2v + ω2uv2 + ω3v3 = 2α1 · 3β1 ·∏p|N0pκp , (5.6)111such that E is isomorphic over Q to ED, whereED : 3[β0/3]y2 = x3 − 27D2HF (u, v)x+ 27D3GF (u, v) (5.7)and, for [r] the greatest integer not exceeding a real number r,D =∏p|gcd(c4(E),c6(E))pmin{[νp(c4(E))/2],[νp(c6(E))/3]}. (5.8)The α0, α1, β0, β1 and N1 are nonnegative integers satisfying N1 | N0,(α0, α1) =(2, 0) or (2, 3) if α = 0,(3,≥ 3) or (2,≥ 4) if α = 1,(2, 1), (4, 0) or (4, 1) if α = 2,(2, 1), (2, 2), (3, 2), (4, 0) or (4, 1) if α = 3,(2,≥ 0), (3,≥ 2), (4, 0) or (4, 1) if α = 4,(2, 0) or (3, 1) if α = 5,(2,≥ 0), (3,≥ 1), (4, 0) or (4, 1) if α = 6,(3, 0) or (4, 0) if α = 7,(3, 1) if α = 8and(β0, β1) =(0, 0) if β = 0,(0,≥ 1) or (1,≥ 0) if β = 1,(3, 0), (0,≥ 0) or (1,≥ 0) if β = 2,(β, 0) or (β, 1) if β ≥ 3.The κp are nonnegative integers withνp(∆E) ={νp(DF ) + 2κp if p - D,νp(DF ) + 2κp + 6 if p | D(5.9)112andκp ∈ {0, 1} whenever p2 | N1. (5.10)Further, we haveif β0 ≥ 3, then 3 | ω1 and 3 | ω2, (5.11)andif νp(N) = 1, for p ≥ 3, then p | DFF (u, v). (5.12)Here, as we shall make explicit in the next subsection, the form F corresponding tothe curveE in Theorem 5.2.1 determines the 2-division field ofE. This connectionwas noted by Rubin and Silverberg [96] in a somewhat different context – theyproved that if K is a field of characteristic 6= 2, 3, F (u, v) is a binary cubic formdefined over K, E is an elliptic curve defined by y2 = F (x, 1), and E0 is anotherelliptic curve over K with the property that E[2] ∼= E0[2] (as Galois modules),then E0 is isomorphic to the curvey2 = x3 − 3HF (u, v)x+GF (u, v),for some u, v ∈ K.5.2.1 RemarksBefore we proceed, there are a number of observations we should make regardingTheorem 5.2.1.Historical commentsTheorem 5.2.1 is based upon a generalization of classical work of Mordell [77](see also Theorem 3 of Chapter 24 of Mordell [79]), in which the DiophantineequationX2 + kY 2 = Z3113is treated through reduction to binary cubic forms and their covariants, under theassumption that X and Z are coprime. That this last restriction can, with somecare, be eliminated, was noted by Sprindzuk (see Chapter VI of [107]). A similarapproach to this problem can be made through the invariant theory of binary quarticforms, where one is led to solve, instead, equations of the shapeX2 + kY 3 = Z3.We will not carry out the analogous analysis here.2-division fields and reducible formsIt might happen that the form F whose existence is guaranteed by Theorem 5.2.1is reducible over Z[x, y]. This occurs precisely when the elliptic curve E has anontrivial rational 2-torsion point. This follows from the more general fact thatthe cubic form F (u, v) = ω0u3 + ω1u2v + ω2uv2 + ω3v3 corresponding to anelliptic curve E has the property that the splitting field of F (u, 1) is isomorphic tothe 2-division field of E. This is almost immediate from the identity33 ω20 F(x−ω13ω0, 1)= x3 + (9ω0ω2 − 3ω21)x+ 27ω20ω3 − 9ω0ω1ω2 + 2ω31= x3 − 3HF (1, 0)x+GF (1, 0).Indeed, from (5.7), the elliptic curve defined by y2 = x3−3HF (1, 0)x+GF (1, 0)is a quadratic twist of that given by the model y2 = x3 − 27c4(E)x − 54c6(E),and hence also of E (whereby they have the same 2-division field).Imprimitive formsIt is also the case that the cubic forms arising need not be primitive (in the sensethat gcd(ω0, ω1, ω2, ω3) = 1). This situation can occur if each of the coefficientsof F is divisible by some integer g ∈ {2, 3, 6}. Since the discriminant is a quartic114form in the coefficients of F , for this to take place one requires thatDF ≡ 0 mod g4.This is a necessary but not sufficient condition for the form F to be imprimitive. Itfollows, if we wish to restrict attention to primitive forms in Theorem 5.2.1, thatthe possible values for νp(DF ) that can arise areν2(DF ) ∈ {0, 2, 3, 4}, ν3(DF ) ∈ {0, 1, 3, 4, 5} (5.13)and νp(DF ) ∈ {0, 1, 2}, for p > 3. (5.14)Possible twistsWe note that necessarilyD | 23 · 32 ·∏p|N0p, (5.15)so that, givenN , there is a finite set ofED to consider (we can restrict our attentionto quadratic twists of the curve defined via y2 = x3 − 3HF (1, 0)x+GF (1, 0), bysquarefree divisors of 6N ). In case we are dealing with squarefree conductor N(i.e. for semistable curves E), then, from Tables 5.1, 5.2 and 5.3, it follows thatD ∈ {1, 2}.Necessity, but not sufficiencyIf we search for elliptic curves of conductorN , say, there may exist a cubic form Ffor which the corresponding Thue-Mahler equation (5.6) has a solution, where allof the conditions of Theorem 5.2.1 are satisfied, but for which the correspondingED has conductor NED 6= N for all possible D. This can happen when certainlocal conditions at primes dividing 6N are not met; these local conditions are, inpractice, easy to check and only a minor issue when performing computations.Indeed, when producing tables of elliptic curves of conductor up to some given115bound, we will, in many cases, apply Theorem 5.2.1 to find all curves with goodreduction outside a fixed set of primes – in effect, working with multiple conduc-tors simultaneously. For such a computation, the conductor of every twist ED weencounter will be of interest to us.Special binary cubic formsIf, for a given binary form F (x, y) = ax3 + bx2y+ cxy2 + dy3, 3 divides both thecoefficients b and c (say b = 3b0 and c = 3c0), then 27 | DF and, consequently,we can write DF = 27D˜F , whereD˜F = −a2d2 + 6ab0c0d+ 3b20c20 − 4ac30 − 4b30d.One can show that the set of binary cubic forms with b ≡ c ≡ 0 mod 3 is closedwithin the larger set of all binary cubic forms in Z[x, y], under the action of eitherSL2(Z) or GL2(Z). Also note that for such forms we haveH˜F (x, y) =HF (x, y)9= (b20 − ac0)x2 + (b0c0 − ad)xy + (c20 − b0d)y2and G˜F (x, y) = GF (x, y)/27, so thatG˜F (x, y) = (−a2d+ 3ab0c0 − 2b30)x3 + 3(−b20c0 − ab0d+ 2ac20)x2y+3(b0c20 − 2b20d+ ac0d)xy2 + (−3b0c0d+ 2c30 + ad2)y3.The syzygy now becomes4H˜F (x, y)3 = G˜F (x, y)2 + D˜FF (x, y)2. (5.16)We note, from Theorem 5.2.1, that we will be working exclusively with formsof this shape whenever we wish to treat elliptic curves of conductor N ≡ 0mod 33.116The case jE = 0This case is treated over a general number field in Proposition 4.1 of Cremona andLingham [35]. The elliptic curves E/Q with jE = 0 and a given conductor Nare particularly easy to determine. Indeed, a curve with this property is necessarilyisomorphic over Q to a Mordell curve with a model Y 2 = X3 − 54c6 where c6 =c6(E). Such a model is minimal except possibly at 2 and 3 and has discriminant−26 · 39 · c26 (whereby any primes p > 2 which divide c6 necessarily also divideN ). Here, without loss of generality, we may suppose that c6 is sixth-power-free.Further, from Tables 5.1, 5.2, and 5.3, we have that ν2(N) ∈ {0, 2, 3, 4, 6}, thatν3(N) ∈ {2, 3, 5}, and that νp(N) = 2 whenever p | N for p > 3. Given apositive integer N satisfying these constraints, it is therefore a simple matter tocheck to see if there are elliptic curves E/Q with conductor N and j-invariant 0.One needs only to compute the conductors of the curves given by Y 2 = X3−54c6for each sixth-power-free integer (positive or negative) c6 dividing 64N3.5.2.2 The algorithmIt is straightforward to convert Theorem 5.2.1 into an algorithm for finding allE/Qof conductor N . We can proceed as follows.1. Begin by finding allE/Q of conductorN with jE = 0, as outlined in Section5.2.1.2. Next, compute GL2(Z)-representatives for every binary form F with dis-criminant∆F = ±2α03β0N1for each divisor N1 of N0, and each possible pair (α0, β0) given in the state-ment of Theorem 5.2.1 (see (5.13) for specifics). We describe an algorithmfor listing these forms in Section 5.4.3. Solve the corresponding Thue-Mahler equations, finding pairs of integers(u, v) such thatF (u, v) is an S-unit, where S = {p prime : p | N}∪{2} and117F (u, v) satisfies the additional conditions given in the statement of Theorem5.2.1.4. For each cubic form F and pair of integers (u, v), consider the elliptic curveE1 : y2 = x3 − 27HF (u, v)x+ 27GF (u, v)and all its quadratic twists by squarefree divisors of 6N . Output those curveswith conductor N (if any).The first, second and fourth steps here are straightforward; the first and second canbe done efficiently, while the fourth is essentially trivial. The main bottleneck isstep (3). While there is a deterministic procedure for carrying this out (see Tzanakisand de Weger [115], [116]), it is both involved and, often, computationally taxing.Instead, we apply the implementation outlined in Chapter 3 and Chapter 6 of thisthesis. The version used for this computation (which we will reference here andhenceforth as UBC-TM) is available athttp://www.nt.math.ubc.ca/BeGhRe/Code/UBC-TMCodeWe give a number of examples of this general procedure in Section 5.5. In Sec-tion 5.6, we show that in the special cases where the conductor is prime or thesquare of a prime, the Thue-Mahler equations (5.6) (happily) reduce to Thue equa-tions (i.e. the exponents on the right hand side of (5.6) are absolutely bounded).This situation occurs because, for such elliptic curves, a very strong form of Szpiro’sconjecture (bounding the minimal discriminant of an elliptic curve from above interms of its conductor) is known to hold. Thue equations can be solved by routinesthat are computationally much easier than is currently the case for Thue-Mahlerequations; such procedures have been implemented in Pari/GP [89] and Magma[19]. Further, in this situation, it is possible to apply a much more computation-ally efficient argument to find all such elliptic curves heuristically but not, perhaps,completely (see Section 5.7).1185.3 Proof of Theorem 5.2.1Proof. Given an elliptic curve E/Q of conductor N = 2α3βN0 and invariantsc4 = c4(E) 6= 0 and c6 = c6(E), we will construct a corresponding cubic formF explicitly. In fact, our form F will have the property that its leading coefficientwill be supported on the primes dividing 6N , i.e. thatF (1, 0) = 2α1 · 3β1 ·∏p|N0pκp .Define D as in (5.8), i.e. take D to be the largest integer whose square divides c4and whose cube divides c6. We then setX = c4/D2 and Y = c6/D3,whereby, from (5.1),Y 2 = X3 + (−1)δ+1M, (5.17)forM = D−6 · 26 · 33 · |∆E |.Note that the assumption that c4(E) 6= 0 ensures that both the j-invariant jE 6= 0and that X 6= 0.It will prove useful to us later to understand precisely the possible common factorsamong X,Y,D and M . For any p > 3, we have νp(N) ≤ 2. When νp(N) = 1,from Table 5.3 we find that(νp(D), νp(X), νp(Y ), νp(M)) = (0, 0, 0,≥ 1), (5.18)while, if νp(N) = 2, then eitherνp(D) = 1 and min{νp(X), νp(Y )} = 0, νp(M) = 0, (5.19)119orνp(D) ≤ 1, (νp(X), νp(Y ), νp(M)) = (0, 0,≥ 1), (≥ 1, 1, 2), (1,≥ 2, 3) (5.20)or (≥ 2, 2, 4). (5.21)Things are rather more complicated for the primes 2 and 3; we summarize this inTables 5.4 and 5.5 (which are, in turn, compiled from the data in Tables 5.1 and5.2).ν2(N) (ν2(X), ν2(Y ), ν2(M), ν2(D))0 (≥ 2, 0, 0, 1) or (0, 0, 6, 0)1 (0, 0,≥ 7, 0)2 (≥ 2, 2, 4, 1), (≥ 2, 1, 2, 2) or (0, 0, 2, 2)3 (≥ 2, 2, 4, 1), (≥ 2, 1, 2, 2) or (0, 0, t, 2), t = 2, 4 or 54 (≥ 2, 2, 4, 1), (≥ 2, 1, 2, 2), (≥ 2, 0, 0, 3) or (0, 0, t, 2), t = 2 or t ≥ 45 (≥ 0,≥ 0, 0, 2), (0,≥ 0, 0, 3), (0, 0, 3, 2) or (1, 0, 0, 3)6 (≥ 0,≥ 0, 0, 2), (0,≥ 0, 0, 3), (≥ 2, 2, 4, 2), (≥ 2, 1, 2, 3) or (0, 0,≥ 2, 3)7 (0, 0, 1, 2), (0, 0, 1, 3), (1, 1, 2, 2) or (1, 1, 2, 3)8 (1,≥ 2, 3, 2) or (1,≥ 2, 3, 3).Table 5.4: The possible values of ν2(N), ν2(X), ν2(Y ), ν2(M) and ν2(D)ν3(N) (ν3(X), ν3(Y ), ν3(M), ν3(D))0 (1,≥ 3, 3, 0) or (0, 0, 3, 0)1 (0, 0,≥ 4, 0)2 (≥ 0, 0, 0, 1), (0,≥ 2, 0, 1), (0, 0,≥ 3, 1), (1,≥ 3, 3, 1), (≥ 0, 0, 0, 2)or (0,≥ 2, 0, 2)3 (≥ 0, 0, 0, 1), (≥ 0, 0, 0, 2), (0, 1, 0, 1), (0, 1, 0, 2), (0, 0, 2, 1)or (0, 0, 2, 2)4 (0, 0, 1, 1), (0, 0, 1, 2), (1, 2, 3, 1) or (1, 2, 3, 2)5 (≥ 1, 1, 2, 1), (≥ 1, 1, 2, 2), (≥ 2, 2, 4, 1) or (≥ 2, 2, 4, 2).Table 5.5: The possible values of ν3(N), ν3(X), ν3(Y ), ν3(M) and ν3(D)120We will construct a cubic formF1(x, y) = ax3 + 3b0x2y + 3c0xy2 + dy3,one coefficient at a time; our main challenge will be to ensure that the a, b0, c0 andd we produce are actually integral rather than just rational. The form F whoseexistence is asserted in the statement of Theorem 5.2.1 will turn out to be either F1or F1/3.Let us writeM = M1 ·M2where M2 is the largest integer divisor of M that is coprime to X , so thatM1 =∏p |Xpνp(M) and M2 =∏p -Xpνp(M).We definea1 =∏p|M1p[νp(M)−12](5.22)and seta2 = 3−1∏p|M2 p[νp(M)2]if ν3(X) = 0, ν3(M) = 2t, t ∈ Z, t ≥ 2,∏p|M2 p[νp(M)2]otherwise.(5.23)Define a = a1 · a2. It follows that a21 | M1 and, from (5.18), (5.19), (5.20), andTables 5.4 and 5.5, that botha1 | X and a21 | Y.We write X = a1 · X1 and observe that a22 | M2. Note that a2 is coprime toX and hence to a1. Since a2 | M , we may thus define a positive integer K via121K = M/a2, so that (5.17) becomesY 2 −X3 = (−1)δ+1Ka2.From the fact that gcd(a2, X) = 1 and X 6= 0, we may choose B so thata2B ≡ −Y/a1 mod X3,wherebyaB + Y ≡ 0 mod a1X3. (5.24)Note that, since a21 | Y and a1 | X , it follows that a1 | B. Let us defineb0 =aB + YX, c0 =b20 −Xaand d =b0c0 − 2Ba. (5.25)We now demonstrate that these are all integers. That b0 ∈ Z is immediate from(5.24). Since b0X − Y = aB, we know that b0X ≡ Y mod a. Squaring bothsides thus givesb20X2 ≡ Y 2 ≡ X3 + (−1)δ+1Ka2 ≡ X3 mod a1 · a2,and, since gcd(a2, X) = 1,b20 ≡ X mod a2.From (5.24), we have b0 ≡ 0 mod a1X2, whereby, since a1 | X ,b20 ≡ X ≡ 0 mod a1.The fact that gcd(a1, a2) = 1 thus allows us to conclude that b20 ≡ X mod a andhence that c0 ∈ Z.It remains to show that d is an integer. Let us rewrite ad asad = b0c0 − 2B =(aB + YaX)((aB + YX)2−X)− 2B,122so thatad =(aB + YaX)((−1)δ+1Ka2 + 2aBY + a2B2X2)− 2B.Expanding, we find thatX3d = (−1)δ+1KY + 3Y B2 + aB3 + (−1)δ+13KaB. (5.26)We wish to show that(−1)δ+1KY + 3Y B2 + aB3 + (−1)δ+13KaB ≡ 0 mod X3.From (5.24), we have that(−1)δ+1KY+3Y B2+aB3+(−1)δ+13KaB ≡ 2Y(B2 + (−1)δK)mod a1X3.Multiplying congruence (5.24) by aB − Y (which, from our prior discussion, isdivisible by a21), we find thata2B2 ≡ Y 2 ≡ X3 + (−1)δ+1Ka2 mod a31X3and hence, dividing through by a21,a22B2 ≡ a1X31 + (−1)δ+1Ka22 mod a1X3.It follows thatB2 + (−1)δK ≡ a−22 a1X31 mod a1X3, (5.27)and so, since a21 | Y ,Y(B2 + (−1)δK)≡ 0 mod X3,whence we conclude that d is an integer, as desired.With these values of a, b0, c0 and d, we can then confirm (with a quick computa-123tion) that the cubic formF1(x, y) = ax3 + 3b0x2y + 3c0xy2 + dy3has discriminantDF1 =108a2(X3 − Y 2) = (−1)δ · 22 · 33 ·KWe also note thatF1(1, 0) = a, H˜F1(1, 0) = b20 − ac0 = Xand−12G˜F1(1, 0) =12(a2d− 3ab0c0 + 2b30) = Y,where G˜F and H˜F are as in Section 5.2.1.Summarizing Table 5.5, we find that we are in one of the following four cases :(i) ν3(X) = 1, ν3(Y ) = 2, ν3(M) = 3 and ν3(N) = 4,(ii) ν3(X) ≥ 2, ν3(Y ) = 2, ν3(M) = 4, ν3(N) = 5,(iii) ν3(M) ≤ 2 and ν3(N) ≥ 2, or(iv) ν3(M) ≥ 3 and either ν3(XY ) = 0 or ν3(X) = 1, ν3(Y ) ≥ 3.In cases (i), (ii), and (iii), we choose F = F1, i.e.(ω0, ω1, ω2, ω3) = (a, 3b0, 3c0, d),so thatF (1, 0) = a, DF = (−1)δ22 · 33 ·K, c4 = D2H˜F (1, 0)andc6 = −12D3G˜F (1, 0).124It follows that E is isomorphic over Q to the curvey2 = x3 − 27c4x− 54c6 = x3 − 3D2HF (1, 0)x+D3GF (1, 0).In case (iv), observe that, from definitions (5.22) and (5.23),ν3(a) =[ν3(M)− 12]and ν3(K) = ν3(M)− 2ν3(a), (5.28)so that 3 | a and 3 | K. From equation (5.26), 3 | X3d. If ν3(X) = 0 this impliesthat 3 | d. On the other hand, if ν3(X) = 1, then, from (5.27), we may concludethat 3 | B. Since each of a,B and K is divisible by 3, while ν3(X) = 1 andν3(Y ) ≥ 3, equation (5.26) once again implies that 3 | d. In this case, we cantherefore write a = 3a0 and d = 3d0, for integers a0 and d0 and set F = F1/3, i.e.take(ω0, ω1, ω2, ω3) = (a0, b0, c0, d0).We haveF (1, 0) = a/3, DF = (−1)δ22 ·K/3, c4 = D2HF (1, 0)andc6 = −12D3GF (1, 0).The curve E is now isomorphic over Q to the modely2 = x3 − 27c4x− 54c6 = x3 − 27D2HF (1, 0)x+ 27D3GF (1, 0).Since |DF |/DF = (−1)δ and a2K | 1728∆E , we may writeF (1, 0) = 2α1 · 3β1 ·∏p|N0pκp and DF = (|∆E |/∆E)2α03β0N1,for nonnegative integers α0, α1, β0, β1, κp and a positive integer N1, divisible only125by primes dividing N0. More explicitly, we haveα0 = ν2(K) + 2 and β0 = ν3(K) +{3 in case (i), (ii) or (iii), or−1 in case (iv),andα1 = ν2(a) and β1 = ν3(a) +{0 in case (i), (ii) or (iii), or−1 in case (iv).It remains for us to prove that these integers satisfy the conditions listed in thestatement of the theorem. It is straightforward to check this, considering in turneach possible triple (X,Y,M) from (5.18), (5.19), (5.20), and Tables 5.4 and 5.5,and using the fact that K = M/a2.In particular, if p > 3, we have νp(∆E) = 6νp(D) + νp(DF ) + 2κp. From Table5.3 and (5.8), we have νp(D) ≤ 1, whereby (5.9) follows. Further,νp(a) =[νp(M)−12]if p | X,[νp(M)2]if p - X,(5.29)and so, if p - X ,νp(M)− 2νp(a) ≤ 1.Since a2K = M , if p2 | DF , then νp(N) = 2 and it follows that we are in case(5.20), with p | X . We may thus conclude that νp(M) ∈ {2, 3, 4} and hence, from(5.29), that νp(a) ≤ 1. This proves (5.10).For (5.11), note that, in cases (i), (ii) and (iii), we clearly have that 3 | ω1 and3 | ω2. In case (iv), from (5.28),β0 = ν3(DF ) = ν3(K)− 1 = ν3(M)− 2[ν3(M)− 12]− 1 ∈ {0, 1}.Finally, to see (5.12), note that if νp(N) = 1, for p > 3, then we have (5.18) andhenceνp(DF ) + 2νp(F (u, v)) = νp(M) ≥ 1,126whereby p | DF or p | F (u, v). We may also readily check that the same conclu-sion obtains for p = 3 (since, equivalently, β0 +β1 ≥ 1). This completes the proofof Theorem 5.2.1.To illustrate this argument, suppose we consider the elliptic curve (denoted 109a1in Cremona’s database) defined viaE : y2 + xy = x3 − x2 − 8x− 7,with ∆E = −109. We havec4(E) = 393 and c6(E) = 7803,so that gcd(c4(E), c6(E)) = 3. It follows thatD = 1, X = 393, Y = 7803, δ = 1, M = 26 · 33 · 109,and hence we haveM1 = 33, M2 = 26 · 109, a1 = 3, a2 = 23, a = 23 · 3 and K = 3 · 109.We solve the congruence 8B ≡ −2601 mod 3933 to find that we may chooseB = 7586982, so thatb0 = 463347, c0 = 8945435084 and d = 172701687278841.We are in case (iv) and thus setF (x, y) = 8x3 + 463347x2y + 8945435084xy2 + 57567229092947y3,with discriminant DF = −4 · 109,GF (1, 0) = −15606 = −2c6(E) and HF (1, 0) = 393 = c4(E).127The curve E is thus isomorphic to the modelED : y2 = x3 − 27D2HF (1, 0)x+ 27D3GF (1, 0) = x3 − 10611x− 421362.(5.30)We observe that the form F is GL2(Z)-equivalent to a “reduced” form (see Section5.4 for details), given byF˜ (x, y) = x3 + 3x2y + 4xy2 + 6y3.In fact, this is the only form (up to GL2(Z)-equivalence) of discriminant ±4 · 109.We can check that the solutions to the Thue equation F˜ (u, v) = 8 are given by(u, v) = (2, 0) and (u, v) = (−7, 3). The minimal quadratic twist ofy2 = x3 − 27HF˜ (2, 0)x+ 27GF˜ (2, 0)has conductor 25 ·109 and hence cannot correspond toE. For the solution (u, v) =(−7, 3), we find that the curve given by the modely2 = x3 − 27HF˜ (−7, 3)x+ 27GF˜ (−7, 3) = x3 − 10611x+ 421362,is the quadratic twist by −1 of the curve (5.30). This situation arises from the factthat GF is an SL2(Z)-covariant, but not a GL2(Z)-covariant of F (we will discussthis more in the next section).5.4 Finding representative formsAs Theorem 5.2.1 illustrates, we are able to tabulate elliptic curves over Q withgood reduction outside a given set of primes, by finding a set of representatives forGL2(Z)-equivalence classes of binary cubic forms with certain discriminants, andthen solving a number of Thue-Mahler equations. In this section, we will providea brief description of techniques to find distinguished reduced representatives forequivalence classes of cubic forms over a given range of discriminants. For bothpositive and negative discriminants, the notion of reduction arises from associating128a particular definite quadratic form to a given cubic form.5.4.1 Irreducible FormsFor forms of positive discriminant, there is a well developed classical theory ofreduction dating back to work of Hermite [55], [56] and, later, Davenport (seee.g. [36], [37] and [39]). We can actually apply this method to both reducible andirreducible forms. Initially, though, we will assume the forms are irreducible, sincewe will treat the elliptic curves corresponding to reducible forms by a somewhatdifferent approach (see Section 5.4.2). Note that when one speaks of “irreducible,reduced forms”, as Davenport observes, “the terminology is unfortunate, but canhardly be avoided” ([38], page 184).In each of Belabas [7], Belabas and Cohen [8] and Cremona [33], we find veryefficient algorithms for computing cubic forms of both positive and negative dis-criminant, refining classical work of Hermite, Berwick and Mathews [13], and Julia[57]. These are readily translated into computer code to loop over valid (a, b, c, d)-values (with corresponding forms ax3 + bx2y + cxy2 + dy3). The running timein each case is linear in the upper bound X . Realistically, this step (finding rep-resentatives for our cubic forms) is highly unlikely to be the bottleneck in ourcomputations.5.4.2 Reducible formsOne can make similar definitions of reduction for reducible forms (see [9] for ex-ample). However, for our purposes, it is sufficient to note that a reducible form isequivalent toF (x, y) = bx2y + cxy2 + dy3 with 0 ≤ d ≤ c,which has discriminant∆F = b2(c2 − 4bd).129To find all elliptic curves with good reduction outside S = {p1, p2, . . . , pk}, cor-responding to reducible cubics in Theorem 5.2.1 (i.e. those E with at least onerational 2-torsion point), it is enough to find all such triples (b, c, d) for whichthere exist integers x and y so that bothb2(c2 − 4bd) and bx2y + cxy2 + dy3are S∗-units (with S∗ = S ∪ {2}). For this to be true, it is necessary that each ofthe integersb, c2 − 4bd, y and µ = bx2 + cxy + dy2is an S∗-unit. Taking the discriminant of µ as a function of x, we thus requirethat(c2 − 4bd)y2 + 4bµ = Z2, (5.31)for some integer Z. This is an equation of the shapeX + Y = Z2 (5.32)in S∗-units X and Y .An algorithm for solving such equations is described in detail in Chapter 7 ofde Weger [118] (see also [119]); it relies on bounds for linear forms in p-adicand complex logarithms and various reduction techniques from Diophantine ap-proximation. An implementation of this is available athttp://www.nt.math.ubc.ca/BeGhRe/Code/UBC-TMCode.While a priori equation (5.32) arises as only a necessary condition for the existenceof an elliptic curve of the desired form, given any solution to (5.32) in S∗-units Xand Y and integer Z, the curvesE1(X,Y ) : y2 = x3 + Zx2 +X4x130andE2(X,Y ) : y2 = x3 + Zx2 +Y4xhave nontrivial rational 2-torsion (i.e. the point corresponding to (x, y) = (0, 0))and discriminant X2Y and XY 2, respectively (and hence good reduction at allprimes outside S∗).Though a detailed analysis of running times for solving equations of the shape(5.32), or for solving more general cubic Thue-Mahler equations, has not to ourknowledge been carried out, our experience from carrying out such computationsfor several thousand sets S is that, typically, the former can be done significantlyfaster than the latter. By way of example, solving (5.32) for S = {2, 3, 5, 7, 11}takes only a few hours on a laptop, while treating the analogous problem of deter-mining all elliptic curves over Q with trivial rational 2-torsion and good reductionoutside S (see Section 5.5.4) requires many thousand machine-hours.5.4.3 Computing forms of fixed discriminantFor our purposes, we will typically compute and tabulate a large list of irreducibleforms of absolute discriminant bounded by a given positive number X (of size upto 1012 of so, beyond which storage becomes problematical). In certain situations,however, we will want to compute all forms of a given fixed, larger discriminant(perhaps up to size 1015). To carry this out and find desired forms of the shapeax3 + bx2y + cxy2 + dy3, we can argue as in, for example, Cremona [33], torestrict our attention to O(X3/4) triples (a, b, c). From (5.3), the definition of DF ,we have thatd =9abc− 2b3 ±√4(b2 − 3ac)3 − 27a2DF27a2and hence it remains to check that the quantity 4(b2−3ac)3−27a2DF is an integersquare, that the relevant conditions modulo 27a2 are satisfied, and that a variety offurther inequalities from [33] are satisfied. The running time for finding forms withdiscriminants of absolute value of sizeX via this approach is of orderX3/4.1315.4.4 GL2(Z) vs SL2(Z)One last observation which is very important to make before we proceed, is thatwhileG2F is GL2(Z)-covariant, the same is not actually true forGF (it is, however,an SL2(Z)-covariant). This may seem like a subtle point, but what it means forus in practice is that, having found our GL2(Z)-representative forms F and corre-sponding curves of the shape ED from Theorem 5.2.1, we need, in every case, toalso check to see ifE˜D : 3[β0/3]y2 = x3 − 27D2HF (u, v)x− 27D3GF (u, v),the quadratic twist of ED by −1, yields a curve of the desired conductor.5.5 ExamplesIn this section, we will describe a few applications of Theorem 5.2.1 to computingall elliptic curves of a fixed conductor N , or all curves with good reduction outsidea given set of primes S. We restrict our attention to examples with compositeconductors, since the case of conductors p and p2, for p prime, will be treated atlength in Section 5.6 (and subsequently). For the examples in Sections 5.5.1, 5.5.2,5.5.2 and 5.5.2, since the conductors under discussion are not “square-full”, thereare necessarily no curves E encountered with jE = 0.In our computations in this section, we executed all jobs in parallel via the shelltool [111]. We note that our Magma code lends itself easily to parallelization, andwe made full use of this fact throughout.We carried out a one-time computation of all irreducible cubic forms that can arisein Theorem 5.2.1, of absolute discriminant bounded by 1010. This computationtook slightly more than 3 hours on a cluster of 40 cores; roughly half this time wastaken up with sorting and organizing output files. There are 996198693 classesof irreducible cubic forms of positive discriminant and 3079102475 of negativediscriminant in the range in question; storing them requires roughly 120 gigabytes.We could also have tabulated and stored representatives for each class of reducible132form of absolute discriminant up to 1010, but chose not to since our approach tosolving equation (5.32) does not require them.5.5.1 Cases without irreducible formsWe begin by noting an obvious corollary to Theorem 5.2.1 that, in many cases,makes it a relatively routine matter to determine all elliptic curves of a given con-ductor, provided we can show the nonexistence of certain corresponding cubicforms.Corollary 5.5.1. Let N be a square-free positive integer with gcd(N, 6) = 1 andsuppose that there do not exist irreducible binary cubic forms in Z[x, y] of discrim-inant ±4N1, for each positive integer N1 | N . Then every elliptic curve over Q ofconductor N1, for each N1 | N , has nontrivial rational 2-torsion.We will apply this result to a pair of examples (chosen somewhat arbitrarily). Cur-rently, such an approach is feasible for forms of absolute discriminant (and hencepotentially conductors) up to roughly 1015. We observe that, among the positiveintegers N < 108 satisfyingν2(N) ≤ 8, ν3(N) ≤ 5 and νp(N) ≤ 2 for p > 3,i.e. those for which there might actually exist elliptic curves E/Q of conductor N ,we find that 708639 satisfy the hypotheses of Corollary 5.5.1.It is somewhat harder to modify the statement of Corollary 5.5.1 to include re-ducible forms (with corresponding elliptic curves having nontrivial rational 2-torsion). One of the difficulties one encounters is that there actually do exist re-ducible forms of, by way of example, discriminant 4p for every p ≡ 1 mod 8;writing p = 8k + 1, for instance, the formF (x, y) = 2x2y + xy2 − ky3has this property.133Conductor 2655632887 = 31 · 9007 · 9511In the notation of Theorem 5.2.1, we have α = β = 0 and hence α0 = 2 andβ0 = 0, so that, in order for there to be an elliptic curve with trivial rational 2-torsion and this conductor, we require the existence of an irreducible cubic formof discriminant 4N1 where N1 | 31 · 9007 · 9511, i.e. discriminant ±4 · 31δ1 ·9007δ2 · 9511δ3 , for δi ∈ {0, 1}. We check that there are no such forms, directlyfrom our table of forms, except for the possibility of DF = ±4 · 31 · 9007 · 9511,which exceeds 1010 in absolute value. For these latter possibilities, we argue as inSection 5.4.3 to show that no such forms exist. We may thus appeal to Corollary5.5.1.For the possible cases with rational 2-torsion, we solveX+Y = Z2 withX and YS-units for S = {2, 31, 9007, 9511}. The solutions to this equation with X ≥ Y ,Z > 0 and gcd(X,Y ) squarefree are precisely those with(X,Y ) = (2,−1), (2, 2), (8, 1), (32,−31), (62, 2), (256,−31), (961, 128),(992,−31), (3968, 1), (76088,−9007), (294841, 8)and (492032,−9007).A short calculation confirms that each elliptic curve arising from these solutionsvia quadratic twist has bad reduction at the prime 2 (and, in particular, cannot haveconductor 2655632887). There are thus no elliptic curves over Q with conductor2655632887. Observe that these calculations in fact ensure that there do not existelliptic curves over Q with conductor dividing 2655632887.Full computational details are available athttp://www.nt.math.ubc.ca/BeGhRe/Examples/2655632887-data.We observe that it is much more challenging computationally to try to extend thisargument to tabulate curves E with good reduction outside S = {31, 9007, 9511}.To do this, we would have to first determine whether or not there exist irreduciblecubic forms of discriminant, say, DF = ±4 · 312 · 90072 · 95112 > 2.8 × 1019.This appears to be at or beyond current computational limits.134Conductor 3305354359 = 41 · 409 · 439 · 449For there to exist an elliptic curve with trivial rational 2-torsion and conductor3305354359, we require the existence of an irreducible cubic form of discriminant±4 · 41δ1 · 409δ2 · 439δ3 · 449δ4 , with δi ∈ {0, 1}. We check that, again, thereare no such forms (once more employing a short auxiliary computation in the caseDF = ±4 ·41 ·409 ·439 ·449). If we solveX+Y = Z2 withX and Y S-units forS = {2, 41, 409, 439, 449}, we find that the solutions to this equation withX ≥ Y ,Z > 0 and gcd(X,Y ) squarefree are precisely(X,Y ) = (2,−1), (2, 2), (8, 1), (41,−16), (41,−32), (41, 8), (82,−1),(128, 41), (409,−328), (409, 32), (439, 2), (449,−328), (449,−8),(512, 449), (818, 82), (898, 2), (3272, 449), (3362, 2), (7184, 41),(16769,−128), (16769,−14368), (18409,−16384),(33538,−18409), (36818, 818), (41984, 41), (68921,−57472),(183641,−1312), (183641,−56192), (183641, 41984),(359102, 898), (403202,−33538), (403202,−359102),(403202, 17999), (737959, 183641), (754769,−6544),(6858521,−919552), (8265641,−16)and (7095601778,−5610270178).Once again, a short calculation confirms that each elliptic curve arising from thesesolutions via twists has even conductor. There are thus no elliptic curves over Qwith conductor 3305354359.Full computational details are available athttp://www.nt.math.ubc.ca/BeGhRe/Examples/3305354359-data.1355.5.2 Cases with fixed conductor (and corresponding irreducibleforms)Conductor 399993 = 3 · 11 · 17 · 23 · 31We next choose an example where full data is already available for comparison inthe LMFDB [68]. In particular, there are precisely 10 isogeny classes of curves ofthis conductor (labelled 399993a to 399993j in the LMFDB), containing a total of21 isomorphism classes. Of these, 7 isogeny classes (and 18 isomorphism classes)have nontrivial rational 2-torsion.According to Theorem 5.2.1, the curves arise from consideration of cubic formsof discriminant discriminant ±4K, where K | 3 · 11 · 17 · 23 · 31. The (re-duced) irreducible cubic forms F (u, v) of these discriminants are as follows, whereF (u, v) = ω0u3 + ω1u2v + ω2uv2 + ω3v3.(ω0, ω1, ω2, ω3) DF (ω0, ω1, ω2, ω3) DF(1, 1, 1, 3) −4 · 3 · 17 (2, 4,−6,−3) 4 · 3 · 17 · 23(1, 2, 2, 2) −4 · 11 (2, 5, 2, 6) −4 · 3 · 17 · 23(1, 2, 2, 6) −4 · 11 · 17 (3, 3,−8,−2) 4 · 3 · 23 · 31(1, 4,−16,−2) 4 · 11 · 17 · 31 (3, 3, 44, 66) −4 · 3 · 11 · 17 · 23 · 31(1, 8,−2, 42) −4 · 3 · 17 · 23 · 31 (3, 4, 10, 14) −4 · 11 · 23 · 31(1, 11,−12,−6) 4 · 3 · 11 · 17 · 31 (3, 7, 5, 7) −4 · 3 · 23 · 31(2, 0, 7, 1) −4 · 23 · 31 (4, 17, 10, 28) −4 · 11 · 17 · 23 · 31(2, 1, 14,−2) −4 · 11 · 17 · 31In each case, we are thus led to solve the Thue-Mahler equationF (u, v) = 23δ3β111κ1117κ1723κ2331κ31 , (5.33)where gcd(u, v) = 1, δ ∈ {0, 1} and β1, κ11, κ17, κ23 and κ31 are arbitrary non-negative integers. Applying (5.12), in order to find a curve of conductor 399993,136we require additionally that, for a corresponding solution to (5.33),F (u, v)DF ≡ 0 mod 3 · 11 · 17 · 23 · 31. (5.34)We readily check that the congruence F (u, v) ≡ 0 mod p has only the solutionu ≡ v ≡ 0 mod p for the following forms F and primes p (whereby (5.34) cannotbe satisfied by coprime integers u and v for these forms) :(ω0, ω1, ω2, ω3) p (ω0, ω1, ω2, ω3) p(1, 1, 1, 3) 11, 23 (2, 0, 7, 1) 3, 17(1, 2, 2, 2) 3, 23, 31 (2, 5, 2, 6) 11, 31(1, 4,−16,−2) 3, 23 (3, 3,−8,−2) 11(1, 8,−2, 42) 11 (4, 17, 10, 28) 3(1, 11,−12,−6) 23For the remaining 6 forms under consideration, we appeal to UBC-TM. The onlysolutions we find satisfying (5.34) are as follows.(ω0, ω1, ω2, ω3) (u, v)(1, 2, 2, 6) (−1851, 892), (14133,−3790)(2, 1, 14,−2) (13,−5), (−29,−923)(2, 4,−6,−3) (10,−3), (64, 49), (−95, 199), (−3395, 1189),(3677,−1069), (5158, 4045), (−23546, 57259),(−77755, 30999)(3, 3, 44, 66) (1, 0), (1, 2), (−3, 4), (3,−2), (−11, 9), (25,−3),(231, 2), (−317, 240), (489, 61), (1263,−878), (6853,−4119)(3, 7, 5, 7) (1, 12), (−29, 26), (78, 1), (423,−160)(3, 4, 10, 14) (−41, 84), (95,−69), (307, 90)From these, we compute the conductors ofED in (5.7), whereD ∈ {1, 2}, togetherwith their twists by −1. The only curves with conductor 399993 arise from theform F with (ω0, ω1, ω2, ω3) = (2, 4,−6,−3) and the solutions(u, v) ∈ {(10,−3), (5158, 4045), (−23546, 57259)} .137In each case, D = 2. The solution (u, v) = (10,−3) corresponds to, in the no-tation of the LMFBD, curve 399993.j1, (u, v) = (5158, 4045) to 399993.i1, and(u, v) = (−23546, 57259) to 399993.h1. Note that every form and solution weconsider leads to elliptic curves with good reduction outside {2, 3, 11, 17, 23, 31},just not necessarily of conductor 399993. By way of example, if (ω0, ω1, ω2, ω3) =(2, 4,−6,−3) and (u, v) = (−77755, 30999), we find curves with minimal quadratictwists of conductor25 · 3 · 11 · 17 · 23 · 31 = 25 · 399993.To determine the curves of conductor 399993 with nontrivial rational 2-torsion, weare led to solve the equation X + Y = Z2 in S-units X and Y , and integers Z,where S = {2, 3, 11, 17, 23, 31}. We employ Magma code available athttp://nt.math.ubc.ca/BeGhRe/Code/UBC-TMCodeto find precisely 2858 solutions with X ≥ |Y | and gcd(X,Y ) squarefree (thiscomputation took slightly less than 2 hours). Of these, 1397 have Z > 0, with Zlargest for the solution corresponding to the identity48539191572432− 40649300451407 = 24 · 34 · 11 · 237− 175 · 315 = 28088952.As in subsection 5.4.2, we attach to each solution a pair of elliptic curvesE1(X,Y )and E2(X,Y ). Of these, the only twists we find to have conductor 399993 are thequadratic twists by t of Ei(X,Y ) given in the following table. Note that thereis some duplication – the curve labelled 399993.f2 in the LMFDB, for example,arises from three distinct solutions to X + Y = Z2.138X Y Ei t LMFDB16192 −4743 E1 −1 399993.g216192 −4743 E2 2 399993.g123529 18496 E1 −2 399993.f223529 18496 E2 1 399993.f3116281 −75072 E1 2 399993.f4116281 −75072 E2 −1 399993.f2371008 4761 E1 1 399993.f2371008 4761 E2 −2 399993.f1519777 −131648 E1 2 399993.d2519777 −131648 E2 −1 399993.d1534336 −506447 E1 −1 399993.e2534336 −506447 E2 2 399993.e11311552 −527 E1 1 399993.a21311552 −527 E2 −2 399993.a11414017 −1045568 E1 2 399993.b21414017 −1045568 E2 −1 399993.b16305121 3027904 E1 2 399993.c16305121 3027904 E2 −1 399993.c26988113 18496 E1 2 399993.c26988113 18496 E2 −1 399993.c37745089 −2731968 E1 2 399993.c47745089 −2731968 E2 −1 399993.c2Full computational details are available athttp://www.nt.math.ubc.ca/BeGhRe/Examples/399993-data.139Conductor 106 − 1We next treat a slightly larger conductor, which is not available in the LMFDBcurrently (but probably within computational range). We have106 − 1 = 33 · 7 · 11 · 13 · 37.From Theorem 5.2.1, we thus need to consider binary cubic forms F (u, v) =ω0u3 + ω1u2v + ω2uv2 + ω3v3 of discriminant DF = ±108N1, where N1 |7 · 11 · 13 · 37 and ω1 ≡ ω2 ≡ 0 mod 3. The irreducible forms of this shape areas follows.140(ω0, ω1, ω2, ω3) DF p(1, 0,−6,−2) 108 · 7 37(1, 0, 21, 16) −108 · 11 · 37 7, 13(1, 0, 30, 2) −108 · 7 · 11 · 13 none(1, 3, 3, 3) −108 7, 13, 37(1, 3, 6, 16) −108 · 37 7(1, 3, 12, 26) −108 · 7 · 13 none(1, 3, 33, 117) −108 · 7 · 11 · 37 none(1, 6,−36,−34) 108 · 7 · 13 · 37 11(1, 6, 3, 6) −108 · 37 7(1, 6, 9, 26) −108 · 11 · 13 none(1, 9, 0, 74) −108 · 7 · 13 · 37 none(1, 12, 12, 14) −108 · 13 · 37 11(2, 0,−18,−5) 108 · 11 · 37 13(2, 0, 3, 3) −108 · 11 7, 37(2, 0, 15, 3) −108 · 7 · 37 11, 13(2, 0, 18, 7) −108 · 13 · 37 11(2, 3,−78,−26) 108 · 7 · 11 · 13 · 37 none(2, 3, 6, 3) −108 · 7 11, 37(2, 3, 6, 8) −108 · 37 7(2, 6,−12, 1) 108 · 11 · 13 7(2, 6, 21, 88) −108 · 11 · 13 · 37 none(2, 12, 0, 13) −108 · 7 · 11 · 13 none(2, 21,−6, 80) −108 · 7 · 11 · 13 · 37 none(3, 3, 18, 20) −108 · 7 · 11 · 13 none(4, 6, 15, 14) −108 · 13 · 37 11(5, 6, 27, 14) −108 · 7 · 11 · 37 none(5, 9, 3, 21) −108 · 7 · 11 · 37 none(7, 0, 12, 14) −108 · 7 · 11 · 37 none(10, 3, 42,−16) −108 · 7 · 11 · 13 · 37 none(10, 6, 12, 3) −108 · 13 · 37 none(11, 6, 12, 6) −108 · 7 · 11 · 13 none(21, 12, 27, 20) −108 · 7 · 11 · 13 · 37 none141Here, we list primes p for which a local obstruction exists modulo p to findingcoprime integers u and v satisfying (5.12). It is worth noting at this point that therestriction to forms with ω1 ≡ ω2 ≡ 0 mod 3 that follows from the fact that weare considering a conductor divisible by 33 is a helpful one. There certainly canand do exist irreducible forms F with 108 | DF that fail to satisfy ω1 ≡ ω2 ≡ 0mod 3.We are thus left to treat 17 Thue-Mahler equations which we solve using UBC-TM;seehttp://www.nt.math.ubc.ca/BeGhRe/Examples/999999-datafor computational details. From (5.12), we require that DFF (u, v) ≡ 0 mod 7 ·11·13·37; the only solutions we find satisfying this constraint are as follows.142(ω0, ω1, ω2, ω3) (u, v)(1, 0, 30, 2) (−1, 21), (1, 16), (27, 25)(1, 3, 33, 117) (26,−7)(1, 9, 0, 74) (−19, 2)(2, 3,−78,−26) (−1, 3), (−3, 2), (−5,−1), (9,−1), (13, 2), (−17,−58),(−39,−61), (−57,−10), (−59, 9), (65,−6), (79,−330),(159,−23)(2, 6, 21, 88) (3, 1), (165,−43)(2, 12, 0, 13) (−1, 9), (18, 23)(2, 21,−6, 80) (1,−10), (2, 1), (4,−3), (4,−1), (17, 1),(19,−5), (21,−2), (138,−11), (1356,−127)(3, 3, 18, 20) (9, 13), (97,−12)(5, 6, 27, 14) (14, 1), (19, 6), (−21, 44)(5, 9, 3, 21) (−1, 2), (6, 1), (8,−3), (−649, 284), (1077,−464)(7, 0, 12, 14) (−1, 5), (−7, 9), (301,−62), (−459, 553)(10, 3, 42,−16) (1, 1), (1, 2), (2,−1), (3, 1), (4,−17), (20, 19), (−22,−69),(127, 339)(10, 6, 12, 3) (2,−1), (5,−13), (−12, 83), (−24, 89), (81,−107),(125,−437)(11, 6, 12, 6) (−1, 22), (47,−72), (223,−429)(21, 12, 27, 20) (1,−3), (1, 0), (1, 5), (4,−9), (4, 3), (9,−29),(19,−15), (29,−40), (316,−455), (551,−805)The only ones of these for which we find an ED in (5.7) of conductor 999999 areas follows, where ED is isomorphic over Q to a curve with modely2 + a1xy + a3y = x3 + a2x2 + a4x+ a6.143(ω0, ω1, ω2, ω3) (u, v) D a1 a2 a3 a4 a6(1, 0, 30, 2) (27, 25) 6 0 0 1 −40395 5402579(1, 0, 30, 2) (27, 25) −2 0 0 1 −363555 −145869640(5, 6, 27, 14) (14, 1) 1 1 −1 0 14700 55223(5, 6, 27, 14) (14, 1) −3 1 −1 1 1633 −2590(5, 9, 3, 21) (−1, 2) 6 0 0 1 30 2254(5, 9, 3, 21) (−1, 2) −2 0 0 1 270 −60865(10, 6, 12, 3) (125,−437) 2 0 0 1 −17205345 −27554570341(10, 6, 12, 3) (125,−437) −6 0 0 1 −1911705 1020539642(21, 12, 27, 20) (4, 3) −1 1 −1 0 12432 −164125(21, 12, 27, 20) (4, 3) 3 1 −1 1 1381 5618Each of these listed curves has trivial rational 2-torsion. To search for curves ofconductor 999999 with nontrivial rational 2-torsion, we solve the equation X +Y = Z2 in S-units X and Y , and integers Z, where S = {2, 3, 7, 11, 13, 37}. Wefind that there are precisely 4336 solutions with X ≥ |Y | and gcd(X,Y ) square-free. Of these, 2136 have Z > 0, with Z largest for the solution corresponding tothe identity103934571636753−68209863326528 = 315 ·11·13·373−26 ·713 ·11 = 59770152.Once again, we attach to each solution a pair of elliptic curves E1(X,Y ) andE2(X,Y ). We find 505270 isomorphism classes of E/Q with good reductionoutside of {2, 3, 7, 11, 13, 37} and nontrivial rational 2-torsion. None of them haveconductor 999999, whereby we conclude that there are precisely 10 isomorphismclasses of elliptic curves overQwith conductor 106−1. Checking that these curveseach have distinct traces of Frobenius a47 shows that they are nonisogenous.144Conductor 109 − 1This example is chosen to be somewhat beyond the current scope of the LMFDB.We have109 − 1 = 34 · 37 · 333667and so, applying Theorem 5.2.1, we are led to consider binary cubic forms ofdiscriminant±4 ·34 ·37δ1 ·333667δ2 , where δi ∈ {0, 1}. These include imprimitiveforms with the property that each of its coefficients ωi is divisible by 3. For suchforms, from Theorem 5.2.1, we necessarily have β1 ∈ {0, 1} and hence β1 = 1.Dividing through by 3, we may thus restrict our attention to primitive forms ofdiscriminant ±4 · 3κ · 37δ1 · 333667δ2 , where δi ∈ {0, 1} and κ ∈ {0, 4}. For theirreducible forms, we have, by slight abuse of notation (since, for the F listed herewith DF 6≡ 0 mod 3, the form whose existence is guaranteed by Theorem 5.2.1is actually 3F ), the following.145(ω0, ω1, ω2, ω3) DF p(1, 1,−3,−1) 4 · 37 333667(1, 4, 52, 250) −4 · 333667 37(1, 9, 37, 279) −4 · 333667 none(1, 21, 117, 2135) −4 · 34 · 333667 37(2, 0, 3, 1) −4 · 34 37(2, 17,−26,−31) 4 · 333667 37(4, 30, 117, 665) −4 · 34 · 333667 37(4, 35, 14, 216) −4 · 37 · 333667 none(5, 6, 9, 6) −4 · 34 · 37 none(5, 7, 19, 51) −4 · 333667 37(5, 14, 19, 54) −4 · 333667 37(6, 18, 168, 323) −4 · 34 · 333667 37(6, 27, 42, 356) −4 · 34 · 333667 37(6, 54,−48, 115) −4 · 34 · 333667 37(10, 18, 96, 229) −4 · 34 · 333667 37(26, 9, 102, 4) −4 · 34 · 333667 none(27, 7, 70, 32) −4 · 37 · 333667 none(31, 9, 87,−25) −4 · 34 · 333667 none(49, 51, 63, 55) −4 · 34 · 333667 none(52, 55, 72, 37) −4 · 37 · 333667 noneOnce again, we list primes p for which a local obstruction exists modulo p tofinding coprime integers u and v satisfying (5.12). There are thus 8 Thue-Mahlerequations left to solve. In the (four) cases where DF 6≡ 0 mod 3, these take theshapeF (u, v) = 23δ1 · 37γ1 · 333667γ2 ,where δ1 ∈ {0, 1}, γ1 and γ2 are nonnegative integers, and u and v are coprimeintegers. For the remaining F , the analogous equation isF (u, v) = 23δ1 · 3δ2 · 37γ1 · 333667γ2 ,146where δi ∈ {0, 1}, γ1, γ2 ∈ Z+ and u, v ∈ Z with gcd(u, v) = 1. We solve theseequations using the UBC-TM Thue-Mahler solver. The only cases where we findthatDFF (u, v) ≡ 0 mod 37 · 333667occur for (ω0, ω1, ω2, ω3) = (4, 35, 14, 216) and (u, v) = (−8, 1) or (u, v) =(−2, 1), for (ω0, ω1, ω2, ω3) = (27, 7, 70, 32) and (u, v) = (1,−2) or (2,−1),and for (ω0, ω1, ω2, ω3) = (52, 55, 72, 37) and (u, v) = (0, 1) or (−3, 5). In eachcase, all resulting twists have bad reduction at 2 (and hence cannot have conductor109 − 1).To search for curves with nontrivial rational 2-torsion and conductor 109 − 1, wesolve the equation X + Y = Z2 in S-units X and Y , and integers Z, where S ={2, 3, 37, 333667}. There are precisely 98 solutions with X ≥ |Y | and gcd(X,Y )squarefree. Of these, 41 have Z > 0, with Z largest for the solution coming fromthe identity27027027− 101306 = 34 · 333667− 2 · 373 = 51892.These correspond via twists to elliptic curves of conductor as large as 28 · 32 ·372 · 3336672, but none of conductor 109 − 1. There thus exist no curves E/Q ofconductor 109 − 1.Full computational details are available athttp://www.nt.math.ubc.ca/BeGhRe/Examples/999999999-data.5.5.3 Curves with good reduction outside {2, 3, 23} : an example ofKoutsianis and of von Kanel and MatchkeThis case was worked out by Koutsianis [61] (and also by von Kanel and Matschke[58], who actually computed E/Q with good reduction outside {2, 3, p} for allprime p ≤ 163), by rather different methods from those employed here. We includeit here to provide an example where we determine all E/Q with good reductionoutside a specific set S, which is of somewhat manageable size in terms of the set147of cubic forms encountered. Our data agrees with that of [58] and [61].To begin, we observe that the elliptic curves with good reduction outside {2, 3, 23}and j-invariant 0 are precisely those with models of the shapeE : Y 2 = X3 ± 2a3b23c, where 0 ≤ a, b, c ≤ 5.Appealing to (5.13), we next look through our precomputed list to find all theirreducible primitive cubic forms of discriminant ±2α3β23γ , whereα ∈ {0, 2, 3, 4}, β ∈ {0, 1, 3, 4, 5} and γ ∈ {0, 1, 2}.The imprimitive forms we need consider correspond to primitive forms F witheither ν2(DF ) = 0 or ν3(DF ) ∈ {0, 1}. We find precisely 95 irreducible, primitivecubic forms of the desired discriminants.148(ω0, ω1, ω2, ω3) DF (ω0, ω1, ω2, ω3) DF (ω0, ω1, ω2, ω3) DF(1, 0,−18,−6) 22 · 35 · 23 (2, 0, 3, 4) −23 · 35 (4, 9, 24, 29) −22 · 34 · 232(1, 0,−3,−1) 34 (2, 3, 6, 4) −22 · 35 (4, 12, 12, 27) −24 · 33 · 232(1, 0, 3, 2) −23 · 33 (2, 3, 12, 8) −24 · 33 · 23 (4, 12, 12, 73) −24 · 35 · 232(1, 0, 6, 2) −22 · 35 (2, 3, 36, 29) −23 · 34 · 232 (4, 18, 9, 24) −22 · 35 · 232(1, 0, 6, 4) −24 · 34 (2, 3, 36, 98) −23 · 35 · 232 (4, 18, 27, 48) −22 · 35 · 232(1, 0, 9, 6) −24 · 35 (2, 5, 8, 15) −23 · 3 · 232 (5, 6, 7, 4) −23 · 232(1, 0, 33, 32) −22 · 34 · 232 (2, 6,−12,−1) 22 · 35 · 23 (5, 6, 15, 8) −23 · 35 · 23(1, 1, 2, 1) −23 (2, 6, 6, 5) −22 · 35 (5, 9, 12, 10) −22 · 35 · 23(1, 1, 8, 6) −22 · 232 (2, 6, 6, 25) −22 · 33 · 232 (5, 12, 18, 20) −24 · 35 · 23(1, 3,−9,−4) 35 · 23 (2, 6, 27, 117) −23 · 35 · 232 (5, 18, 30, 46) −22 · 35 · 232(1, 3,−6,−4) 22 · 33 · 23 (2, 9,−6,−4) 22 · 35 · 23 (5, 24,−3, 26) −24 · 35 · 232(1, 3,−3,−2) 33 · 23 (2, 9, 0,−4) 24 · 33 · 23 (6, 3, 12,−7) −23 · 33 · 232(1, 3,−6,−2) 23 · 35 (2, 9, 48, 185) −24 · 35 · 232 (6, 3, 12, 16) −24 · 33 · 232(1, 3, 3, 3) −22 · 33 (2, 12, 24, 85) −22 · 35 · 232 (6, 6, 9, 13) −23 · 33 · 232(1, 3, 3, 5) −24 · 33 (2, 18,−15, 31) −23 · 35 · 232 (6, 9, 12, 23) −23 · 34 · 232149(ω0, ω1, ω2, ω3) DF (ω0, ω1, ω2, ω3) DF (ω0, ω1, ω2, ω3) DF(1, 3, 3, 7) −22 · 35 (3, 0, 3, 2) −24 · 34 (6, 18, 18, 29) −22 · 35 · 232(1, 3, 3, 13) −24 · 35 (3, 4, 12, 12) −24 · 3 · 232 (7, 6, 9, 4) −23 · 34 · 23(1, 3, 18, 50) −23 · 35 · 23 (3, 6, 4, 6) −22 · 3 · 232 (7, 15, 3, 17) −22 · 35 · 232(1, 6,−24,−4) 24 · 35 · 23 (3, 6, 9, 8) −23 · 33 · 23 (8, 9, 12, 13) −22 · 34 · 232(1, 6, 3, 32) −23 · 35 · 23 (3, 9, 9, 7) −24 · 35 (8, 15, 18, 21) −23 · 34 · 232(1, 6, 6, 16) −24 · 33 · 23 (3, 9, 9, 49) −22 · 35 · 232 (9, 9, 3, 31) −24 · 35 · 232(1, 6, 12, 54) −22 · 33 · 232 (3, 18, 36, 116) −24 · 35 · 232 (10, 6, 15, 1) −23 · 33 · 232(1, 6, 12, 100) −24 · 33 · 232 (3, 27, 9, 29) −24 · 35 · 232 (11, 6, 12, 2) −22 · 33 · 232(1, 9,−12,−16) 24 · 35 · 23 (4, 0,−18,−3) 24 · 35 · 23 (11, 15, 15, 17) −22 · 35 · 232(1, 9,−9,−3) 22 · 35 · 23 (4, 0, 6, 1) −24 · 35 (12, 9, 36, 16) −24 · 35 · 232(1, 9, 27, 165) −22 · 35 · 232 (4, 2, 8, 3) −24 · 232 (12, 36, 36, 35) −24 · 35 · 232(1, 9, 27, 303) −24 · 35 · 232 (4, 3, 6, 2) −22 · 33 · 23 (13, 9, 18, 12) −22 · 35 · 232(1, 12, 9, 18) −24 · 35 · 23 (4, 3, 12, 10) −23 · 35 · 23 (13, 15, 27, 7) −22 · 35 · 232(1, 12, 12, 44) −24 · 33 · 232 (4, 3, 18, 13) −23 · 33 · 232 (21, 9, 27, 11) −24 · 35 · 232(1, 15, 3,−7) 24 · 35 · 23 (4, 3, 18, 36) −22 · 35 · 232 (23, 30, 36, 20) −24 · 35 · 232(2, 0, 3, 1) −22 · 34 (4, 4, 9, 1) −24 · 232 (24, 27, 36, 16) −24 · 35 · 232(2, 0, 3, 2) −23 · 34 (4, 6, 3, 12) −22 · 33 · 232150In each case, we solve the corresponding Thue-Mahler equation specified by The-orem 5.2.1. For example, if DF = ±24 ·3t ·232, with t ≥ 3, then we actually needonly solve the (eight) Thue equations of the shapeF (u, v) = 2δ13δ223δ3 , where δi ∈ {0, 1}.For all other discriminants, we must treat “genuine” Thue-Mahler equations (whereat least one of the exponents on the right-hand-side of equation (5.6) is, a priori,unconstrained). Details of this computation are available athttp://www.nt.math.ubc.ca/BeGhRe/Examples/2-3-23-data.In total, we found precisely 730 solutions to these equations, leading, after twisting,to 3856 isomorphism classes of E/Q with good reduction outside {2, 3, 23} andtrivial rational 2-torsion.Once again, to find the curves with nontrivial rational 2-torsion, we solvedX+Y =Z2 in S-units X and Y , and integers Z, where S = {2, 3, 23}. There are precisely118 solutions with X ≥ |Y | and gcd(X,Y ) squarefree (this computation took lessthan 1 hour). Of these, 55 have Z > 0, with Z largest for the solution coming fromthe identity89424− 23 = 24 · 35 · 23− 23 = 2992.These correspond via twists to elliptic curves of conductor as large as 28 · 32 · 232,a total of 1664 isomorphism classes. There thus exist a total of 5520 isomorphismclasses (in 3968 isogeny classes) of elliptic curves E/Q with good reduction out-side {2, 3, 23}. Note that 432 = 2× 63 of these have jE = 0.5.5.4 Curves with good reduction outside {2, 3, 5, 7, 11} : an exampleof von Kanel and MatschkeThis is the largest computation carried out along these lines by von Kanel andMatschke [58] (and also a very substantial computation via our approach, takingmany thousand machine hours on 80 cores).151As in the preceding example, note that the curves with models of the shapeE : Y 2 = X3 ± 2a3b5c7d11e, 0 ≤ a, b, c, d, e ≤ 5are precisely the E/Q with good reduction outside {2, 3, 5, 7, 11} and j-invariant0. We next proceed by searching our precomputed list for all irreducible primitivecubic forms of discriminant 2α3βM , whereα ∈ {0, 2, 3, 4}, β ∈ {0, 1, 3, 4, 5} and M | 52 · 72 · 112.The imprimitive forms we need consider again correspond to primitive forms Fwith either ν2(DF ) = 0 or ν3(DF ) ∈ {0, 1}. We encounter 1796 irreducible cubicforms, which we tabulate athttp://www.nt.math.ubc.ca/BeGhRe/Examples/2-3-5-7-11-datawhere details on the resulting Thue-Mahler computation may also be found. Con-firming the results of von Kanel and Matschke [58], we find that there exist atotal of 592192 isomorphism classes (in 453632 isogeny classes) of elliptic curvesE/Q with good reduction outside {2, 3, 5, 7, 11}, including 15552 = 2 × 65 withjE = 0.5.6 Good reduction outside a single primeFor the remainder of this chapter, we will focus our attention on the case of ellipticcurves with bad reduction at a single prime, i.e. curves of conductor p or p2,for p prime. In this case, our approach simplifies considerably and rather thanbeing required to solve Thue-Mahler equations, the problem reduces to one ofsolving Thue equations, i.e. equations of the shape F (x, y) = m, where F isa form and m is a fixed integer. While, once again, we do not have a detailedcomputational complexity analysis either of algorithms for solving Thue equationsor of more general algorithms for solving Thue-Mahler equations, computations todate strongly support the contention that the former is, usually, much, much fasterthan the latter, particularly if the set of primes S considered for the Thue-Mahler152equations is anything other than tiny. Since none of these conductors are divisibleby 9, we may always suppose that jE 6= 0. We note that the data we have producedin these cases totals several terabytes. As a result, we have not yet determined howbest to make it publicly available; interested readers should contact the authors of[11] further details.5.6.1 Conductor N = pSuppose that E is a curve with conductor N = p prime with invariants c4 and c6.From Tables 5.1, 5.2 and 5.3, we necessarily have one of(ν2(c4), ν2(c6)) = (0, 0) or (≥ 4, 3), and ν2(∆E) = 0, or(ν3(c4), ν3(c6)) = (0, 0) or (1,≥ 3), and ν3(∆E) = 0, or(νp(c4), νp(c6)) = (0, 0) and νp(∆E) ≥ 1.From this we see that D = 1 or 2. Theorem 5.2.1 then implies that there is a cubicform of discriminant ±4 or ±4p, and integers u, v, withF (u, v) = pκp or 8pκp , c4 = D2HF (u, v) and c6 = −12D3GF (u, v),forD ∈ {1, 2} and κp a nonnegative integer. Note that, while the smallest absolutediscriminant for an irreducible cubic form in Z[x, y] is 23, there do exist reduciblecubic forms of discriminants 4 and −4 which we must consider.Appealing to The´ore`me 2 of Mestre and Oesterle´ [74] (and using [20]), we canactually restrict the choices for m dramatically. In fact, we have 3 possibilities –either p ∈ {11, 17, 19, 37}, or p = t2 + 64 for some integer t, or, in all other cases,∆E = ±p. There are precisely 14 isomorphism classes of E/Q with conductorin {11, 17, 19, 37}; one may consult Cremona [31] for details. If we can writep = t2 + 64 for an integer t (which we may, without loss of generality, assume tosatisfy t ≡ 1 mod 4), then the (2-isogenous) curves defined byy2 + xy = x3 +t− 14· x2 − x153andy2 + xy = x3 +t− 14· x2 + 4x+ thave rational points of order 2 given by (x, y) = (0, 0) and (x, y) = (−t/4, t/8),respectively, and discriminants t2 + 64 and −(t2 + 64)2, respectively. In the finalcase (in which ∆E = ±p), we have (using the notation of Section 5.2 and, inparticular, appealing to (5.9) which, in this case yields the equation 1 = νp(∆E) =νp(DF ) + 2κp)α0 = 2, α1 ∈ {0, 3}, β0 = β1 = 0, κp = 0 and N1 ∈ {1, p}.Theorem 5.2.1 thus tells us that to determine the elliptic curves of conductor p, weare led to to find all binary cubic forms (reducible and irreducible) F of discrimi-nant ±4 and ±4p and then solve the Thue equationsF (x, y) = 1 and F (x, y) = 8.Since for any solution (x, y) to the equation F (x, y) = 1, we have F (2x, 2y) = 8,we may thus restrict our attention to the equation F (x, y) = 8 (where we assumethat gcd(x, y) | 2).5.6.2 Conductor N = p2In case E has conductor N = p2, we have that either E is a either a quadratictwist of a curve of conductor p, or we have νp(∆E) ∈ {2, 3, 4}. To see this,note that, via Table 5.3, p | c4, p | c6 and D | 2p, and we may suppose that(νp(c4(E)), νp(c6(E)), νp(∆E)) is one of(≥ 1, 1, 2), (1,≥ 2, 3), (≥ 2, 2, 4), (≥ 2,≥ 3, 6), (2, 3,≥ 7), (≥ 3, 4, 8), (3,≥ 5, 9),or (≥ 4, 5, 10). In each case with νp(c6(E)) ≥ 3, denote by E1 the quadratic twistof E by (−1)(p−1)/2p. For curves E with(νp(c4(E)), νp(c6(E)), νp(∆E)) = (≥ 2,≥ 3, 6),154one may verify that E1 has good reduction at p and hence conductor 1, a contra-diction. If we have(νp(c4(E)), νp(c6(E)), νp(∆E)) = (2, 3,≥ 7),then(νp(c4(E1)), νp(c6(E1)), νp(∆E1)) = (0, 0, νp(∆E)− 6)and so E1 has conductor p. In the remaining cases, where(νp(c4(E)), νp(c6(E)), νp(∆E)) ∈ {(≥ 3, 4, 8), (3,≥ 5, 9), (≥ 4, 5, 10)},we check that(νp(c4(E1)), νp(c6(E1)), νp(∆E1)) ∈ {(≥ 1, 1, 2), (1,≥ 2, 3), (≥ 2, 2, 4)}.It follows that, in order to determine all isomorphism classes of E/Q of conductorp2, it suffices to carry out the following program.• Find all curves of conductor p.• Find E/Q with minimal discriminant ∆E ∈ {±p2,±p3,±p4}, and then• consider all appropriate quadratic twists of these curves.The fact that we can essentially restrict attention toE/Qwith minimal discriminant∆E ∈ {±p2,±p3,±p4} (5.35)(once we have all curves of conductor p) was noted by Edixhoven, de Groot andTop in Lemma 1 of [41]. To find the E satisfying (5.35), Theorem 5.2.1 (withspecific appeal to (5.9)) leads us to consider Thue equations of the shapeF (x, y) = 8 for F a form of discriminant ± 4p2, (5.36)F (x, y) = 8p for F a form of discriminant ± 4p (5.37)155c4 c6 p ∆E NE4353 287199 17 17 1733 −81 17 17 17t2 + 48 −t(t2 + 72) t2 + 64 t2 + 64 t2 + 64273 4455 17 172 17t2 − 192 −t(t2 + 576) t2 + 64 −(t2 + 64)2 t2 + 641785 75411 7 73 72105 1323 7 −73 7233 12015 17 −174 17Table 5.6: All curves of conductor p and p2, for p prime, corresponding toreducible forms (i.e. with nontrivial rational 2-torsion). Note that t isany integer so that t2 + 64 is prime. For the sake of brevity, we haveomitted curves that are quadratic twists by ±p of curves of conductor p.andF (x, y) = 8p for F a form of discriminant ± 4p2, (5.38)corresponding to ∆E = ±p2, ±p3 and ±p4, respectively.5.6.3 Reducible formsTo find all elliptic curvesE/Qwith conductor p or p2 arising from reducible forms,via Theorem 5.2.1 we are led to solve equationsF (x, y) = 8pn with n ∈ Z and gcd(x, y) | 2, (5.39)where F is a reducible binary cubic form of discriminant±4, ±4p and±4p2. Thisis an essentially elementary, though rather painful, exercise. Alternatively, we mayobserve that curves of conductor p or p2 arising from reducible cubic forms areexactly those with at least one rational 2-torsion point. We can then use Theorem Iof Hadano [48] to show that the only such p are p = 7, 17 and p = t2 + 64 forinteger t. In any case, after some rather tedious but straightforward work, we canshow that the elliptic curves of conductor p or p2 corresponding to reducible forms,are precisely those given in Table 5.6 (up to quadratic twists by ±p).1565.6.4 Irreducible forms : conductor pA quick search demonstrates that there are no irreducible cubic forms of discrim-inant ±4. Consequently if we wish to find elliptic curves of conductor p comingfrom irreducible cubics in Theorem 5.2.1, we need to solve equations of the shapeF (x, y) = 8 for all cubic forms of discriminant ±4p. An almost immediate conse-quence of this is the following.Proposition 5.6.1. Let p > 17 be prime. If there exists an elliptic curve E/Qof conductor p, then either p = t2 + 64 for some integer t, or there exists anirreducible binary cubic form of discriminant ±4p.On the other hand, if we denote by h(K) the class number of a number field K,classical results of Hasse [51] imply the following.Proposition 5.6.2. Let p ≡ ±1 mod 8 be prime and δ ∈ {0, 1}. If there exists anirreducible cubic form of discriminant (−1)δ4p, thenh(Q(√(−1)δp))≡ 0 mod 3.Combining Propositions 5.6.1 and 5.6.2, we thus haveCorollary 5.6.3 (Theorem 1 of Setzer [98]). Let p ≡ ±1 mod 8 be prime. Ifthere exists an elliptic curve E/Q of conductor p, then either p = t2 + 64 for someinteger t, or we haveh (Q(√p)) · h (Q(√−p)) ≡ 0 mod 3.We remark that Proposition 5.6.1 is actually a rather stronger criterion for guar-anteeing the non-existence of elliptic curves of conductor p than Corollary 5.6.3.Indeed, by way of example, we may readily check that there are no irreduciblecubic forms of discriminant ±4p forp ∈ {23, 31, 199, 239, 257, 367, 439},(and hence no elliptic curves of conductor p for these primes) while, in each case,157we have that 3 | h (Q(√p)) · h (Q(√−p)).5.6.5 Irreducible forms : conductor p2As noted earlier, to determine all elliptic curves of conductor p2 for p prime, arisingvia Theorem 5.2.1 from irreducible cubics, it suffices to find those of conductor pand those of conductor p2 with ∆F = ±p2,±p3 and ±p4 (and subsequently twistthem). We explore these cases below.Elliptic curves of discriminant ±p3To find elliptic curves of discriminant ±p3, we need to solve Thue equations ofthe shape F (x, y) = 8p, where F runs over all cubic forms of discriminant ∆F =±4p. These forms are already required to compute curves of conductor p. Now,we can either proceed directly to solve F (x, y) = 8p or transform the probleminto one of solving a pair of new Thue equations of the shape Gi(x, y) = 8. Inpractice, we used the former when solving rigorously and the latter when solvingheuristically (see Section 5.7.3).We now describe this transformation. Let F (x, y) = ax3 + bx2y + cxy2 + dy3 bea reduced form of discriminant ±4p. Since p | ∆F , we haveF (x, y) ≡ a(x− r0y)2(x− r1y) mod p,where we must have that p - a, since F is a reduced form (which implies that1 ≤ a < p). Comparing coefficients of x shows that2r0 + r1 ≡ −b/a mod p, r20 + 2r0r1 ≡ c/a mod pandr20r1 ≡ −d/a mod p.158Multiply the first two of these by a and add them to get3ar20 + 2br0 + c ≡ 0 mod p.We can solve this for r0 and hence r1:(r0, r1) ≡ (3a)−1 (−b± t,−b∓ 2t) mod p,where we require that t satisfies t2 ≡ b2−3ac mod p. Finding square roots mod-ulo p can be done efficiently via the Tonelli-Shanks algorithm, for example (seee.g. Shanks [100]), and almost trivially if, say, p ≡ 3 mod 4. Once we have these(r0, r1), we can readily check which pair satisfies r20r1 ≡ −d/a mod p.Now if F (x, y) = 8p then we must have eitherx ≡ r0y mod p or x ≡ r1y mod p.In either case, write x = riy+pu, which maps the equation F (x, y) = 8p to a pairof equations of the shapeGi(u, y) = 8,whereGi(u, y) = ap2u3+(3apri+bp)u2y+(3ar2i +2bri+c)uy2+1p(ar3i +br2i +cri+d)y3.Notice that ∆Gi = p2∆F . In practice, for our deterministic approach, we willactually solve the equation F (x, y) = 8p directly. For our heuristic approach(where a substantial increase in the size of the form’s discriminant is not especiallyproblematic), we will reduce to consideration of the equationsGi(x, y) = 8.A (conjecturally infinite) family of forms and solutionsWe note that there are families of primes for which we can guarantee that theequation F (x, y) = 8p has solutions. By way of example, define a quartic form159pr,s viapr,s = r4 + 9r2s2 + 27s4.Then for a given r, s and p = pr,s the cubic formF (x, y) = sx3 + rx2y − 3sxy2 − ry3has discriminant 4p. Additionally one can readily verify the polynomial identi-tiesF (2r2/s+ 6s,−2r) = 8p and F (6s,−18s2/r − 2r) = 8p.If we set s ∈ {1, 2} in the first of these, or r ∈ {1, 2} in the second, then we arriveat four one-parameter families of forms of discriminant 4p for which the equationF (x, y) = 8p has a solution, namely:(p, x, y) = (r4 + 9r2 + 27, 2r2 + 6,−2r), (r4 + 36r2 + 432, r2 + 12,−2r),(27s4 + 9s2 + 1, 6s,−18s2 − 2), (27s4 + 36s2 + 16, 6s,−9s2 − 4).Similarly, if we definepr,s = r4 − 9r2s2 + 27s4then the formF (x, y) = sx3 + rx2y + 3sxy2 + ry3has discriminant −4p, and the equation F (x, y) = 8p has solutions(x, y) = (−2r2/s+ 6s, 2r) and (6s,−18s2/r + 2r)160and hence we again find (one parameter) families of primes corresponding to eitherr ∈ {1, 2} or s ∈ {1, 2}:(p, x, y) =(r4 − 9r2 + 27,−2r2 + 6, 2r), (r4 − 36r2 + 432,−r2 + 12, 2r),(27s4 − 9s2 + 1, 6s,−18s2 + 2), (27s4 − 36s2 + 16, 6s,−9s2 + 4).We expect that each of the quartic families described here attains infinitely manyprime values, but proving this is beyond current technology.Elliptic curves of discriminant p2 and p4To find elliptic curves of discriminant p2 and p4 via Theorem 5.2.1, we need tosolve Thue equations F (x, y) = 8 and F (x, y) = 8p, respectively, for cubic formsF of discriminant 4p2. Such forms are quite special and it turns out that they forma 2-parameter family.Indeed, in order for there to exist a cubic form of discriminant 4p2, it is necessaryand sufficient that we are able to write p = r2 + 27s2 for positive integers r and s,whereby F is equivalent to the formFr,s(x, y) = sx3 + rx2y − 9sxy2 − ry3.To see this, note that the existence of an irreducible cubic form of discriminant4p2 for prime p necessarily implies that of a (cyclic) cubic field of discriminantp2 and field index 2. From Silvester, Spearman and Williams [104], there is aunique such field up to isomorphism, which exists precisely when the prime p canbe represented by the quadratic form r2 + 27s2. We conclude as desired uponobserving thatDFr,s = 4(r2 + 27s2)2.It remains, then, to solve the Thue equationsFr,s(x, y) = 8 and Fr,s(x, y) = 8p.161We can transform the problem of solving the second of these equations to one ofsolving a related Thue equation of the form Gr,s(x, y) = 8. This transformation isquite similar to the one described in the previous subsection.First note that we may assume that p - y, since otherwise, we would require thatp | sx, contradicting the facts that s < √p and p2 - F . Since p2 | ∆F , it followsthat the congruencesu3 + ru2 − 9su− r ≡ 0 mod phas a unique solution modulo p; one may readily check that this satisfies u ≡ 9s/rmod p:su3 + ru2 − 9su− r ≡ −r−3 · (r2 − 27s2)(r2 + 27s2) ≡ 0 mod p.Consequently, we know that x ≡ uy mod p. Substituting x = uy + vp into FgivesFr,s(uy + vp, y) = a0v3 + b0v2y + c0vy2 + d0y3so, with a quick renaming of variables, we obtainGr,s(x, y) = a0x3 + b0x2y + c0xy2 + d0y3 = 8,wherea0 = sp2, b0 = (3us+ r)p, c0 = 3u2s+ 2ru− 9sand d0 = (u3s+ ru2 − 9us− r)/p. A little algebra confirms that∆Gr,s = 4p4.As noted in the previous subsection, we have solved Fr,s(x, y) = 8p directly in ourdeterministic approach, while we solved equation Gr,s(x, y) = 8 for our heuristicmethod.162Elliptic curves of discriminant −p2 and −p4Elliptic curves of discriminant −p2 and −p4 can be found through Theorem 5.2.1by solving the Thue equations F (x, y) = 8 and F (x, y) = 8p, respectively,this time for cubic forms F of discriminant −4p2. As in the cases treated inthe preceding subsection, these forms can be described as a 2-parameter family.Specifically, such forms arise precisely when there exist integers r and s such thatp = |r2 − 27s2|, in which case the form F is equivalent toFr,s(x, y) = sx3 + rx2y + 9sxy2 + ry3.The primes p for which we can write p = |r2 − 27s2| are those with p ≡ ±1mod 12. To see this, note first that if p ≡ 1 mod 3 and p = |r2 − 27s2|, thennecessarily p = r2 − 27s2, so that p ≡ 1 mod 4, while, if p ≡ −1 mod 3and p = |r2 − 27s2|, then p = 27s2 − r2 so that p ≡ −1 mod 4. It follows thatnecessarily p ≡ ±1 mod 12. To show that this is sufficient to have p = |r2−27s2|for integers r and s, we appeal to the following.Proposition 5.6.4. If p ≡ 1 mod 12 is prime, there exist positive integers r and ssuch thatr2 − 27s2 = p, with r < 32√6p and s <518√6p.If p ≡ −1 mod 12 is prime, there exist positive integers r and s such thatr2 − 27s2 = −p, with r < 52√2p and s <12√2p.This result is, in fact, an almost direct consequence of the following.Theorem 5.6.5 (Theorem 112 from Nagell [81]). If p ≡ 1 mod 12 is prime, thereexist positive integers u and v such thatp = u2 − 3v2, u <√3p/2 and v <√p/6.163If p ≡ −1 mod 12 is prime, there exist positive integers u and v such that−p = u2 − 3v2, u <√p/2 and v <√p/2.Proof of Proposition 5.6.4. If p ≡ ±1 mod 12, Theorem 5.6.5 guarantees theexistence of integers u and v such that p = |u2 − 3v2|. If 3 | v then we setr = u, s = v/3. Clearly 3 - u, so if 3 - v then we have (replacing v by −v isnecessary) that u ≡ v mod 3. If we now set r = 2u + 3v and s = (2v + u)/3,then it follows that|r2 − 27v2| = |(2u+ 3v)2 − 3(2v + u)2| = |u2 − 3v2| = pand hence either|r| ≤ 2√3p/2 + 3√p/6 =32√6p and |s| ≤ 13(2√p/6 +√3p/2) =518√6p,or|r| ≤ 2√p/2 + 3√p/2 =52√2p and |s| ≤ 13(2√p/2 +√p/2) =12√2p.Again, we are able to reduce the problem of solving Fr,s(x, y) = 8p to that oftreating a related equation Gr,s(x, y) = 8. As before, note that if u ≡ −9s/rmod p, thensu3 + ru2 + 9su+ r ≡ r−3(r2 − 27s2)(r2 + 27s2) ≡ 0 mod p.Again, write x = r0y + vp so that, after renaming v, we haveGr,s(x, y) = a0x3 + b0x2y + c0xy2 + d0y3 = 8,wherea0 = sp2, b0 = (3us+ r)p, c0 = 3u2s+ 2ru+ 9sand d0 = (u3s+ ru2 + 9us+ r)/p.164Note that, in contrast to the case where p = r2 + 27s2, here p is represented byan indefinite quadratic form and so the presence of infinitely many units in Q(√3)implies that a given representation is not unique. If, however, we have two solutionsto the equation |r2 − 27s2| = p, say (r1, s1) and (r2, s2), then the correspondingformss1x3 + r1x2y + 9s1xy2 + r1y3 and s2x3 + r2x2y + 9s2xy2 + r2y3may be shown to be GL2(Z)-equivalent.5.7 Computational detailsAs noted earlier, the computations required to generate curves of prime conductorp (and subsequently conductor p2) fall into a small number of distinct parts.5.7.1 Generating the required formsTo find the irreducible forms potentially corresponding to elliptic curves of primeconductor p ≤ X for some fixed positive real X , arguing as in Section 5.4, wetabulated all reduced forms F (x, y) = ax3 + bx2y + cxy2 + d with discriminantsin (0, 4X] and [−4X, 0), separately. As each form was generated, we checked tosee if it actually satisfied the desired definition of reduction. Of course, this doesnot only produce forms with discriminant ±4p – as each form was produced, wekept only those whose discriminant was in the appropriate range, and equal to±4pfor some prime p. Checking primality was done using the Miller-Rabin primalitytest (see [76], [93]; to make this deterministic for the range we require, we appeal to[106]). While it is straightforward to code the above in computer algebra packagessuch as sage [97], maple [72] or Magma [19], we instead implemented it in c++for speed. To avoid possible numerical overflows, we used the CLN library [49] forc++.We computed forms of discriminant ±4p in two separate runs — first to p ≤ 1012and then a second run to p ≤ 2 × 1013. In the first of these, we constructed all165the forms and explicitly saved them to files. Constructing all the required positivediscriminant forms took approximately 40 days of CPU time on a modern server,and about 300 gigabytes of disc space. Thankfully, the computation is easily par-allelised and it only took about 1 day of real time. We split the jobs by runninga manager which distributed a-values to the other cores. The output from eacha-value was stored as a tab-delimited text file with one tuple of p, a, b, c, d on eachline. Generating all forms of negative discriminant took about 3 times longer andrequired about 900 gigabytes of disc space. The distribution of forms is heavilyweighted to small values of a. To allow us to spread the load across many CPUswe actually split the task into 2 parts. We first ran a ≥ 3, with the master nodedistributing a-values to the other cores. We then ran a = 1 and 2 with the masternode distributing b-values to the other cores. Generating all forms took less than 1week of real time but required about 1.2 terabytes of disc space.These forms were then sorted by discriminant while keeping positive and nega-tive discriminant forms separated. Sorting a terabyte of data is a non-trivial task,and in practice we did this by first sorting1 the forms for each a-value and thensplitting them into files of discriminants in the ranges [n × 109, (n + 1) × 109)for n ∈ [0, 999]. Finally, all the files of each discriminant range were sorted to-gether. This process for positive and negative discriminant forms took around twodays of real time. We found 9247369050 forms of positive discriminant 4p and27938060315 of negative discriminant −4p, with p bounded by 1012. Of these,475831852 and 828238359, respectively had F (x, y) = 8 solvable (by the heuris-tic method described below), leading to 159552514 and 276339267 elliptic curvesof positive and negative discriminant, respectively, with prime conductor up to1012.The second run to p ≤ 2 × 1013 required a different workflow due to space con-straints. Saving all forms to disc was simply impractical — we estimated it to re-quire over 20 terabytes of space! Because of this we combined the form-generationcode with the heuristic solution method (see below) and kept only those formsF (x, y) for which solutions to F (x, y) = 8 existed. Since only a small fraction of1Using the standard unix sort command and taking advantage of multiple cores.166forms (asymptotically likely 0) have solutions, the disc space required was consid-erably less. Indeed to store all the required forms took about 250 and 400 gigabytesfor positive and negative forms respectively. This then translated into about 65and 115 gigabytes of positive and negative discriminant curves, respectively, withprime conductor up to 2 × 1013. This second computation took roughly 20 timeslonger than the first, requiring about 4 months of real-time. This led to a final countof 1738595275 and 3011354026 (isomorphism classes of) curves of positive andnegative discriminant, respectively, with prime conductor up to 2× 1013.5.7.2 Complete solution of Thue equations : conductor pFor each form encountered, we needed to solve the Thue equationax3 + bx2y + cxy2 + dy3 = 8in integers x and y with gcd(x, y) ∈ {1, 2}. We approached this in two distinctways.To solve the Thue equation rigorously, we appealed to by now well-known argu-ments of Tzanakis and de Weger [114], based upon lower bounds for linear forms incomplex logarithms, together with lattice basis reduction; these are implemented inseveral computer algebra packages, including Magma [19] and Pari/GP [89]. Themain computational bottleneck in this approach is typically that of computing thefundamental units in the corresponding cubic fields; for computations p of size upto 109 or so, we encountered no difficulties with any of the Thue equations aris-ing (in particular, the fundamental units occurring can be certified without relianceupon the Generalized Riemann Hypothesis).We ran this computation in Magma [19], using its built-in Thue equation solver.Due to memory consumption issues, we fed the forms into Magma in small batches,restarting Magma after each set. We saved the output as a tuplep, a, b, c, d, n, {(x1, y1), . . . , (xn, yn)},167where p, a, b, c, d came from the form, n counts the number of solutions of theThue equation and (xi, yi) the solutions. These solutions can then be converted intocorresponding elliptic curves in minimal form using Theorem 5.2.1 and standardtechniques.For positive discriminant, this approach works without issue for p < 1010. Forforms of negative discriminant −4p, however, the fundamental unit p in the asso-ciated cubic field can be extremely large (i.e. log |p| can be roughly of size √p).For this reason, finding all negative discriminant curves with prime conductor ex-ceeding 2 · 109 or so proves to be extremely time-consuming. Consequently, forlarge p, we turned to a non-exhaustive method, which, though it finds solutions tothe Thue equation, is not actually guaranteed to find them all.5.7.3 Non-exhaustive, heuristic solution of Thue equationsIf we wish to find all “small” solutions to a Thue equation (which, subject to variouswell-accepted conjectures, might actually prove to be all solutions), there is anobvious and very computationally efficient approach we can take, based upon theidea that, given any solution to the equation F (x, y) = m for fixed integer m, wenecessarily either have that x and y are (very) small, relative to m, or that x/y isa convergent in the infinite simple continued fraction expansion to a root of theequation F (x, 1) = 0.Such techniques were developed in detail by Petho˝ [91], [92]; in particular, heprovides a precise and computationally efficient distinction between “large” and“small” solutions. Following this, for each form F under consideration, we ex-panded the roots of F (x, 1) = 0 to high precision, again using the CLN libraryfor c++. We then computed the continued fraction expansion for each real root,along with its associated convergents. Each convergent x/y was then substitutedinto F (x, y) and checked to see if F (x, y) = ±1,±8. Replacing (x, y) by one of(−x,−y), (2x, 2y) or (−2x,−2y), if necessary, then provided the required solu-tions of F (x, y) = 8. The precision was chosen so that we could compute conver-gents x/y with |x|, |y| ≤ 2128 ≈ 3.4× 1038. We then looked for solutions of small168height using a brute force search over a relatively small range of values.To “solve” F (x, y) = 8 by this method, for all forms with discriminant ±4p withp ≤ 1012, took about 1 week of real time using 80 cores. The resulting solutionsfiles (in which we stored also forms with no corresponding solutions) requiredabout 1.5 terabytes of disc space. Again, the files were split into files of absolutediscriminant (or more precisely absolute discriminant divided by 4) in the ranges[n× 109, (n+ 1)× 109) for n ∈ [0, 999]. For the second computation run to p ≤2 × 1013, we combined the form-generation and heuristic-solutions steps, storingonly forms which had solutions. This produced about 235 and 405 gigabytes ofdata for positive and negative discriminants, respectively.5.7.4 Conversion to curvesOnce one has a tuple (a, b, c, d, x, y), one then computes GF (x, y) and HF (x, y),appeals to Theorem 5.2.1 and checks twists. This leaves us with a list of pairs(c4, c6) corresponding to elliptic curves. It is now straightforward to derive a1, a2,a3, a4 and a6 for a corresponding elliptic curve in minimal form (see e.g. Cremona[32]). For each curve, we saved a tuple p, a1, a2, a3, a4, a6,±1 with the last entrybeing the sign of the discriminant of the form used to generate the curve (whichcoincides with the sign of the discriminant of the curve). We then merged thecurves with positive and negative discriminants and added the curves with primeconductor arising from reducible forms (i.e. of small conductor or for primes ofthe form t2 + 64). After sorting by conductor, this formed a single file of about 17gigabytes for all curves with prime conductor p < 1012 and about 180 gigabytesfor all curves with conductor p < 2× 1013.5.7.5 Conductor p2The conductor p2 computation was quite similar, but was split further into parts.169Twisting conductor pThe vast majority of curves of conductor p2 that we encountered arose as quadratictwists of curves of conductor p. To compute these, we took all curves with conduc-tor p ≤ 1010 and calculated the invariants c4 and c6. The twisted curve then hascorresponding c-invariantsc′4 = p2c4 and c′6 = (−1)(p−1)/2p3c6.The minimal a-invariants were then computed as for curves of conductor p.We wrote a simple c++ program to read curves of conductor p and then twist them,recompute the a-invariants and output them as a tuple p2, a1, a2, a3, a4, a6,±1.The resulting code only took a few minutes to process the approximately 1.1×107curves.Solving F (x, y) = 8p with F of discriminant ±4pThere was no need to retabulate forms for this computation; we reused the posi-tive and negative forms of discriminant ±4p with p ≤ 1010 from the conductor-p computations. We subsequently rigorously solved the corresponding equationsF (x, y) = 8p for p ≤ 108. To solve the Thue equation F (x, y) = 8p for108 < p ≤ 1010, using the non-exhaustive, heuristic method, we first convertedthe equation to a pair of new Thue equations of the form Gi(u, y) = 8 as de-scribed in Section 5.6.5 and then applied Petho˝’s solution search method (wherewe searched for solutions to these new equations with |y| bounded by 2128 and|u| = |(x − riy)/p| bounded in such way as to guarantee that our original |x| isalso bounded by 2128).The solutions were then processed into curves as for the conductor p case above,and the resulting curves were twisted by ±p in order to obtain more curves ofconductor p2.170Solving F (x, y) ∈ {8, 8p} with F of discriminant ±4p2To find forms of discriminant 4p2 with p ≤ 1010 we need only check to see whichprimes are of the form p = r2 + 27s2 in the desired range. To do so, we simplylooped over r and s values and then again checked primality using Miller-Rabin.As each prime was found, the corresponding p, r, s tuple was converted to a formas in Section 5.6.5, and the Thue equations F (x, y) = 8 and F (x, y) = 8p weresolved, using the rigorous approach for p < 106 and the non-exhaustive method de-scribed previously for 106 < p ≤ 1010. Again, in the latter situation, the equationF (x, y) = 8pwas converted to a new equationG(x, y) = 8 as described in Section5.6.5. The process for forms of discriminant−4p2 was very similar, excepting thatmore care is required with the range of r and s (appealing to Proposition 5.6.4). Thenon-exhaustive method solving both F (x, y) = 8 and F (x, y) = 8p for positiveand negative forms took a total of approximately 5 days of real time on a smallerserver of 20 cores. The rigorous approach, even restricted to prime p < 106 wasmuch, much slower.The solutions were then converted to curves as with the previous cases and eachresulting curve was twisted by ±p to find other curves of conductor p2.5.8 Data5.8.1 Previous workThe principal prior work on computing table of elliptic curves of prime conductorwas carried out in two lengthy computations, by Brumer and McGuinness [21] inthe late 1980s and by Stein and Watkins [109] slightly more than ten years later.For the first of these computations, the authors fixed the a1, a2 and a3 invariants(12 possibilities) and looped over a4 and a6 chosen to make the correspondingdiscriminant small. By this approach, they were able to find 311243 curves ofprime conductor p < 108 (representing approximately 99.6% of such curves). Inthe latter case, the authors looped instead over c4 and c6, subject to (necessary)171local conditions. They obtained a large collection of elliptic curves of generalconductor to 108, and 11378912 of those with prime conductor to 1010 (which weestimate to be slightly in excess of 99.8% of such curves).5.8.2 Counts : conductor pBy way of comparison, we found the following numbers of isomorphism classesof elliptic curves over Q with prime conductor p ≤ X:172X ∆E > 0 ∆E < 0 Ratio2 Total Expected Total / Expected103 33 51 2.3884 84 68 1.2353104 129 228 3.1239 357 321 1.1122105 624 1116 3.1986 1740 1669 1.0425106 3388 5912 3.0450 9300 9223 1.0084107 19605 34006 3.0087 53611 52916 1.0131108 114452 198041 2.9941 312493 311587 1.0029109 685278 1187686 3.0038 1872964 1869757 1.00172× 109 1178204 2040736 3.0001 3218940 3216245 1.00081010 4171055 7226982 3.0021 11398037 11383665 1.00131011 25661634 44466339 3.0026 70127973 70107401 1.00031012 159552514 276341397 2.9997 435893911 435810488 1.00021013 999385394 1731017588 3.0001 2730402982 2730189484 1.000082× 1013 1738595275 3011354026 3.0000 4749949301 4749609116 1.00007173The data above the line is rigorous; for positive discriminant, we actually have arigorous result to 1010. For the positive forms this took about one week of realtime using 80 cores. Unfortunately, the negative discriminant forms took signifi-cantly longer, roughly 2 months of real time using 80 cores. Heuristics given byBrumer and McGuinness [21] suggest that the number of elliptic curves of nega-tive discriminant of absolute discriminant up to X should be asymptotically√3times as many as those of positive discriminant in the same range – here we reportthe square of this ratio in the given ranges. The aforementioned heuristic countof Brumer and McGuinness suggests that the expected number of E with primeNE ≤ X should be√312(∫ ∞11√u3 − 1du+∫ ∞−11√u3 + 1du)Li(X5/6),which we list (after rounding) in the table above. It should not be surprising thatthis “expected” number of curves appears to slightly undercount the actual number,since it does not take into account the roughly√X/ logX curves of conductor p =n2 + 64 and discriminant −p2 (counting only curves of discriminant ±p).5.8.3 Counts : conductor p2To compile the final list of curves of conductor p2, we combined the five lists ofcurves: twists of curves of conductor p, curves from forms of discriminant +4pand −4p, and curves from discriminant +4p2 and −4p2. The list was then sortedand any duplicates removed. The resulting list is approximately one gigabyte insize. The counts of curves are as follows; here we list numbers of isomorphismclasses of curves of conductor p2 for p prime with p ≤ X .174X ∆E > 0 ∆E < 0 Total Ratio2103 53 93 146 3.0790104 191 322 513 2.8421105 764 1304 2068 2.9132106 3764 6356 10120 2.8515107 20539 35096 55635 2.9198108 116894 200799 317693 2.9508109 691806 1195262 1887068 2.98511010 4189445 7247980 11437425 2.9931Subsequently we decided that we should recompute the discriminants of thesecurves as a sanity check, by reading the curves into sage and using its built-inelliptic curve routines to compute and then factor the discriminant. This took aboutone day on a single core.The only curves of genuine interest are those that do not arise from twisting, i.e.those of discriminant ±p2, ±p3 and ±p4. In the last of these categories, we foundonly 5 curves, of conductors 112, 432, 4312, 4332 and 330132. The first four ofthese were noted by Edixhoven, de Groot and Top [41] (and are of small enoughconductor to now appear in Cremona’s tables). The fifth, satisfying(a1, a2, a3, a4, a6) = (1,−1, 1,−1294206576, 17920963598714),has discriminant 330134. For discriminants ±p2 and ±p3, we found the followingnumbers of curves, for conductors p2 with p ≤ X :175X ∆E = −p2 ∆E = p2 ∆E = −p3 ∆E = p3103 12 4 7 4104 36 24 9 5105 80 58 12 9106 203 170 17 15107 519 441 24 23108 1345 1182 32 36109 3738 3203 48 581010 10437 9106 60 86It is perhaps worth observing that the majority of these curves arise from, in thecase of discriminant ±p2, forms with, in the notation of Sections 5.6.5 and 5.6.5,either r or s in {1, 8}. Similarly, for ∆E = ±p3, most of the curves we foundcome from forms in the eight one-parameter families described in Section 5.6.5.We are unaware of a heuristic predicting the number of curves of conductor p2 upto X that do not arise from twisting curves of conductor p.5.8.4 Thue equationsIt is noteworthy that all solutions we encountered to the Thue equations F (x, y) =8 and F (x, y) = 8p under consideration satisfied |x|, |y| < 230. The “largest” suchsolution corresponded to the equation355x3 + 293x2y − 1310xy2 − 292y3 = 8,where we have(x, y) = (188455233,−82526573).This leads to the elliptic curve of conductor 948762329069,E : y2 + xy + y = x2 − 2x2 + a4x+ a6,witha4 = −1197791024934480813341176anda6 = 15955840837175565243579564368641.Note that this curve does not actually correspond to a particularly impressive abcor Hall conjecture (see Section 5.9 for the definition of this term) example.In the following table, we collect data on the number of GL2(Z)-equivalence classesof irreducible binary cubic forms of discriminant 4p or−4p for p in [0, X], denotedP3(0, X) and P3(−X, 0), respectively. We also provide counts for those formswhere the corresponding equation F (x, y) = 8 has at least one integer solution,denoted P ∗3 (0, X) and P ∗3 (−X, 0) for positive and negative discriminant forms,respectively.X P3(0, X) P∗3 (0, X) P3(−X, 0) P ∗3 (−X, 0)103 23 22 78 61104 204 163 740 453105 1851 1159 6104 2641106 16333 7668 53202 16079107 147653 49866 466601 97074108 1330934 314722 4126541 582792109 12050910 1966105 36979557 35308202× 109 23418535 3408656 71676647 60802451010 109730653 12229663 334260481 215765851011 1004607003 76122366 3045402451 1331156511012 9247369050 475831852 27938060315 828238359Due to space limitations we did not compute these statistics in the second largecomputational run.Our expectation is that the number of forms for which the equation F (x, y) = 8 hassolutions with absolute discriminant up toX is o(X) (i.e. this occurs for essentially“zero” percent of forms; a first step in proving something is this direction can befound in recent work of Akhtari and Bhargava [2]).1775.8.5 Elliptic curves with the same prime conductorOne might ask how many isomorphism classes of curves of a given prime conduc-tor can occur. If one accepts recent heuristics that predict that the Mordell-Weilrank of E/Q is absolutely bounded (see e.g. [90] and [117]), then this numbershould also be so bounded. As noted by Brumer and Silverman [22], there are13 curves of conductor 61263451. Up to p < 1012, the largest number we en-countered was for p = 530956036043, with 20 isogeny classes, corresponding to(a1, a2, a3, a4, a6) as follows :(0,−1, 1,−1003, 37465) , (0,−1, 1,−1775, 45957) ,(0,−1, 1,−38939, 2970729) , (0,−1, 1,−659,−35439) ,(0,−1, 1, 2011, 4311) , (0,−2, 1,−27597,−1746656) ,(0,−2, 1, 57, 35020) , (1,−1, 0,−13337473, 18751485796) ,(0, 0, 1,−13921, 633170) , (0, 0, 1,−30292,−2029574) ,(0, 0, 1,−6721,−214958) , (0, 0, 1,−845710,−299350726) ,(0, 0, 1,−86411851, 309177638530) , (0, 0, 1,−10717, 428466) ,(1,−1, 0,−5632177, 5146137924) , (1,−1, 0, 878, 33379) ,(1,−1, 1, 1080, 32014) , (1,−2, 1,−8117,−278943) ,(1,−3, 0,−2879, 71732) , (1,−3, 0,−30415,−2014316) .All have discriminant −p. Elkies [42] found examples of rather larger conductorwith more curves, including 21 classes for p = 14425386253757 and discrim-inant p, and 24 classes for p = 998820191314747 and discriminant −p. Ourcomputations confirm, with high likelihood, that, for p < 2 × 1013, the numberof isomorphism classes of elliptic curves of conductor a fixed prime p is at most21.5.8.6 Rank and discriminant recordsIn the following table, we list the smallest prime conductor with a given Mordell-Weil rank. These were computed by running through our data, using Rubinstein’supper bounds for analytic ranks (as implemented in Sage) to search for candidate178curves of “large” rank which were then checked using mwrank [34]. The last entrycorresponds to a curve of rank 6 with minimal positive prime discriminant; we havenot yet ruled out the existence of a rank 6 curve with smaller absolute (negative)discriminant.N (a1, a2, a3, a4, a6) sign(∆E) rk(E(Q)37 (0, 0, 1,−1, 0) + 1389 (0, 1, 1,−2, 0) + 25077 (0, 0, 1,−7, 6) + 3501029 (0, 1, 1,−72, 210) + 419047851 (0, 0, 1,−79, 342) − 56756532597 (0, 0, 1,−547,−2934) + 6It is perhaps noteworthy that the curve listed here of rank 6 has the smallest knownminimal discriminant for such a curve (see Table 4 of Elkies and Watkins [44]).If we are interested in similar records over all curves, including composite conduc-tors, we haveN (a1, a2, a3, a4, a6) sign(∆E) rk(E(Q)37 (0, 0, 1,−1, 0) + 1389 (0, 1, 1,−2, 0] + 25077 (0, 0, 1,−7, 6) + 3234446 (1,−1, 0,−79, 289) + 419047851 (0, 0, 1,−79, 342) − 55187563742 (1, 1, 0,−2582, 48720) + 6382623908456 (0, 0, 0,−10012, 346900) + 7Here, the curves listed above the line are proven to be those of smallest conductorwith the given rank. Those listed below the line have the smallest known conductorfor the corresponding rank. It is our belief that the techniques of this chapter shouldenable one to determine whether the curve listed here of rank 5 has the smallest179conductor of any curve with this property.5.9 Completeness of our dataAs a final result, we will present some information that might, optimistically, beviewed as evidence that our “heuristic” approach, in practice, enables us to actuallyfind all elliptic curves of prime conductor p < 2× 1013.A conjecture of Hall, admittedly one that without modification is widely disbe-lieved at present, is that if x and y are integers for which x3 − y2 is nonzero, thenthe Hall ratioHx,y = |x|1/2|x3 − y2|is absolutely bounded. The pair (x, y) corresponding to the largest known Hallratio comes from the identity58538865167812233 − 4478849284284020423079182 = 1641843,noted by Elkies [43], with Hx,y > 46.6. All other examples known currently haveHx,y < 7. We prove the following.Proposition 5.9.1. If there is an elliptic curve E with conductor p < 2 × 1013,corresponding via Theorem 5.2.1 to a cubic form F and u, v ∈ Z, such thatF (u, v) = 8 and max{|u|, |v|} ≥ 2128,thenHc4(E),c6(E) > 1.5× 106. (5.40)In other words, if there is an elliptic curve E with conductor p < 2 × 1013 thatwe have missed in our heuristic search, then we necessarily have inequality (5.40)(and hence a record-setting Hall ratio).Proof. The main idea behind our proof is that the roots of the Hessian HF (x, 1)have no particularly good reason to be close to those of the polynomial F (x, 1).180It follows that, if we have relatively large integers u and v satisfying the Thueequation F (u, v) = 8 (so that u/v is close to a root of F (x, 1) = 0), our expec-tation is that not only does HF (u, v) fail to be small, but, in fact, we should haveinequalities of the order ofHF (u, v) (max{|u|, |v|})2 and GF (u, v) (max{|u|, |v|})3(where the Vinogradov symbol hides a possible dependence on p). Sincec4(E) = D2HF (u, v) and c6(E) = −12D3GF (u, v),where D ∈ {1, 2}, these would imply thatHc4(E),c6(E) p1pmax{|u|, |v|}.In fact, for forms (and curves) of positive discriminant, we can deduce inequalitiesof the shapeHc4(E),c6(E) p p−3/4 min{|u|, |v|}  p−5/4 max{|u|, |v|},where the implicit constants are absolute. For curves of negative discriminant, wehave a slightly weaker result :Hc4(E),c6(E) p p−1 min{|u|, |v|}  p−3/2 max{|u|, |v|}.To make this argument precise, let us write, for concision, c4 = c4(E) and c6 =c6(E). From the identity |c34 − c26| = 1728p, we have a Hall ratioHc4,c6 =|c4|1/21728p>|c4|1/23.456× 1016 ≥|HF (u, v)|1/23.456× 1016 .Our goal will thus be to obtain a lower bound upon |HF (u, v)| – we claim that, infact, |HF (u, v)| > 3× 1045, whereby this Hall ratio exceeds 1.5× 106, as stated.Suppose that we have a cubic form F and integers u and v with DF = ±4p for p181prime,max{|u|, |v|} ≥ 2128 and 2× 109 < p < 2× 1013. (5.41)Notice that F (u, 0) = ω0u3 = 8 and hence (5.41) implies that v 6= 0.WriteF (u, v) = ω0(u− α1v)(u− α2v)(u− α3v)and suppose that|u− α1v| = min{|u− αiv|, i = 1, 2, 3}.We may further assume, without loss of generality, that the form F is reduced.From (5.5), we haveω20 |HF (α1, 1)HF (α2, 1)HF (α3, 1)| = 16 p2. (5.42)For future use, we note that the main result of Mahler [70] implies the inequality|ω0|3∏i=1max{1, |αi|} ≤ |ω0|+ |ω1|+ |ω2|+ |ω3|. (5.43)Let us assume first that DF > 0, whereby HF has negative discriminant (DHF =−3DF ). Since F is reduced, we have|ω1ω2 − 9ω0ω3| ≤ ω21 − 3ω0ω2 ≤ ω22 − 3ω1ω3,and hence the identity(ω1ω2 − 9ω0ω3)2 − 4(ω21 − 3ω0ω2)(ω22 − 3ω1ω3) = −3DF (5.44)yields the inequalitiesDF ≥ (ω21 − 3ω0ω2)(ω22 − 3ω1ω3) ≥ (ω21 − 3ω0ω2)2. (5.45)182Since (5.44) and DF > 0 imply that ω21 − 3ω0ω2 6= 0, we may writeHF (α1, 1)ω21 − 3ω0ω2=(α1 − 9ω0ω3 − ω1ω2 +√−3DF2(ω21 − 3ω0ω2))(α1 − 9ω0ω3 − ω1ω2 −√−3DF2(ω21 − 3ω0ω2)).DefiningΓ1 = α1 − 9ω0ω3 − ω1ω22(ω21 − 3ω0ω2)and Γ2 =√3DF2(ω21 − 3ω0ω2),we haveHF (α1, 1) =(ω21 − 3ω0ω2) (Γ21 + Γ22)and so|HF (α1, 1)| > 3DF4(ω21 − 3ω0ω2). (5.46)Since α1 is “close” to u/v, it follows that the same is true for HF (α1, 1) andHF (u/v, 1) = v−2HF (u, v). To make this precise, note that, via the Mean ValueTheorem,|HF (α1, 1)−HF (u/v, 1)| =∣∣2(ω21 − 3ω0ω2)y + ω1ω2 − 9ω0ω3∣∣ ∣∣∣α1 − uv ∣∣∣ ,(5.47)for some y lying between α1 and u/v. We thus have|HF (α1, 1)−HF (u/v, 1)| ≤ (ω21−3ω0ω2)(2(|α1|+∣∣∣α1 − uv∣∣∣)+ 1) ∣∣∣α1 − uv∣∣∣ .(5.48)To derive an upper bound upon∣∣α1 − uv ∣∣, we can argue as in the proof of Theorem2 of Petho˝ [92] to obtain the inequality∣∣∣α1 − uv∣∣∣ ≤ 27/3D−1/6F v−2. (5.49)183Since |v| ≥ 1 and DF = 4p > 8× 109, we thus have that∣∣∣α1 − uv∣∣∣ < 0.12. (5.50)We may suppose that F is reduced,|ω0| ≤ 2D1/4F3√3and |ω1| ≤ 3ω02+(√DF − 27ω204)1/2<(1 +1√3)D1/4F .From Proposition 5.5 of Belabas and Cohen [8],|ω2| ≤(35 + 13√13216)1/3D1/3F and |ω3| ≤427D1/2F ,whence, after a little computation, we find that|ω0|+ |ω1|+ |ω2|+ |ω3| < D1/2F = 2p1/2.From (5.43), it follows that|α1| ≤ |ω0|+ |ω1|+ |ω2|+ |ω3| < 2p1/2,whereby inequalities (5.50) and (5.41) thus yield|u/v| < 2p1/2 + 0.12 < 223.1,and so, again appealing to (5.41), min{|u|, |v|} > 2104. Returning to inequality(5.48), we find that, after applying (5.45),|HF (α1, 1)−HF (u/v, 1)| ≤ 2p1/2(4p1/2 + 1.24)27/3(2p)−1/6v−2.From p < 2× 1013 and |v| > 2104, it follows that|HF (α1, 1)−HF (u/v, 1)| < 10−50.184Combining this with (5.45) and (5.46) yields the inequality|HF (u/v, 1)| > 2p|ω21 − 3ω0ω2|,whence|HF (u, v)| = v2 |HF (u/v, 1)| > 2v2p|ω21 − 3ω0ω2|≥ v2√p,where the last inequality follows from (5.45). From (5.41) and the fact that |v| >2104, we conclude that|HF (u, v)| > 1067.Next, suppose that F has negative discriminant, so that HF has positive discrimi-nant DHF = −3DF . If ω21 − 3ω0ω2 = 0, then, from (5.44), we have that3p = −(ω21 − 3ω0ω2)(ω22 − 3ω1ω3),which implies thatmax{|ω21 − 3ω0ω2|, |ω22 − 3ω1ω3|} ≥ p.On the other hand, from Lemma 6.4 of Belabas and Cohen [8], we have|ω0| ≤ 23/2p1/433/4, |ω1| ≤ 23/2p1/431/4, |ω1ω2| ≤ 8p1/231/2, (5.51)max{|ω0ω32|, |ω31ω3|} ≤(11 + 5√5)p2, and |ω0ω3| ≤ 2p1/231/2, (5.52)whereby a short calculation, together with the fact that p > 2 × 109, yields acontradiction. We may thus suppose that ω21 − 3ω0ω2 6= 0. We haveHF (αi, 1) = (ω21 − 3ω0ω2) (αi − β1) (αi − β2) ,185whereβj =9ω0ω3 − ω1ω2 + (−1)j√12p2(ω21 − 3ω0ω2)for j ∈ {1, 2}.It follows that|βj | ≤ |ω21 − 3ω0ω2|−144 · 3−1/2p1/2and, again from (5.43),|ω0αi| ≤ |ω0|+ |ω1|+ |ω2|+ |ω3|,whereby|ω0αi| ≤ 23/2p1/433/4+23/2p1/431/4+22/3(11 + 5√5)1/3p1/231/2|ω0|+2p1/231/2|ω0|,whence we find that|αi| ≤ 3.4 p1/4|ω0| +2.1 p1/2|ω0|2 <6.4 p1/2|ω0|2 .From (5.42), we thus have|HF (α1, 1)| ≥ ω−20 (ω21 − 3ω0ω2)−2 min{ω203.2,|ω21 − 3ω0ω2|12.8}2.If |ω21 − 3ω0ω2| > 4ω20 , it follows that|HF (α1, 1)| ≥ ω2010.24 (ω21 − 3ω0ω2)2and so|HF (α1, 1)| ≥ 110.24 (233−1/2p1/2 + 22/331/2(11 + 5√5)1/3p1/2)2which implies that|HF (α1, 1)| > 11561 p. (5.53)186If, conversely, |ω21 − 3ω0ω2| ≤ 4ω20 , then|HF (α1, 1)| ≥ 1163.84ω20>1253√pand hence (5.53) holds in either case.Now if α1 6∈ R, then, via Mahler [71],|Im(α1)| ≥ 118(|ω0|+ |ω1|+ |ω2|+ |ω3|)−2 > ω20738 p,so that ∣∣∣α1 − uv∣∣∣ > ω20738 pand hence8 = |ω0||v|3∣∣∣α1 − uv∣∣∣ ∣∣∣α2 − uv∣∣∣ ∣∣∣α3 − uv∣∣∣ > |ω0||v|3( ω20738 p)3.It follows that|v| < 1476p < 2.952× 1016,via (5.41). Since max{|u|, |v|} > 2128, we thus have|u/v| > 1.15× 1022.From|α1| < 6.4p1/2 < 6.4(2× 1013)1/2 < 3× 107,we may thus conclude that ∣∣∣α1 − uv∣∣∣ > 1.14× 1022and so8 ≥ (1.14× 1022)3 ,an immediate contradiction.We may thus suppose that α1 ∈ R (so that α2, α3 6∈ R). It follows from Mahler187[71] that ∣∣∣αi − uv∣∣∣ > ω20738 p, for i ∈ {2, 3},and so ∣∣∣α1 − uv∣∣∣ < 8|ω0||v|3(738pω20)2. (5.54)Appealing to (5.41) and the inequalities |α1| < 3× 107 and |v| ≥ 1, we thus havethat|u/v| < 1.75× 1033 + 3× 107 < 1.76× 1033,and so, from max{|u|, |v|} > 2128, |v| > 1.9 × 105. Inequality (5.54) thus nowimplies|u/v| < 2.6× 1017,whence |v| > 1.3× 1021. Substituting this a third time into (5.54),∣∣∣α1 − uv∣∣∣ < 10−30,so that |u/v| < 3.1 × 107 and |v| > 1031. One final use of (5.54) thus yields theinequality ∣∣∣α1 − uv∣∣∣ < 10−59.Appealing to (5.41), (5.47), (5.51), (5.52), and the fact that |α1| < 3 × 107, wethus have, after a little work,|HF (α1, 1)−HF (u/v, 1)| < 3.4× 10−44.With (5.53), this implies that|HF (u/v, 1)| > 11562 pand so|HF (u, v)| = v2 |HF (u/v, 1)| > v21562p>10623124× 1013 > 3× 1045,as claimed.1885.10 Concluding remarksMany of the techniques of this chapter can be generalized to potentially treat theproblem of determining elliptic curves of a given conductor over a number fieldK.In case K is an imaginary quadratic field of class number 1, then, in fact, such anapproach works without any especially new ingredients.189Chapter 6Towards Efficient Resolution ofThue-Mahler EquationsLet c denote a nonzero integer and let S = {p1, . . . , pv} be a set of rational primes.In this section, we specialize the results of Chapter 3 to the degree 3 Thue–MahlerequationF (X,Y ) = c0X3 + c1X2Y + c2XY2 + c3Y3 = cpZ11 · · · pZvv , (6.1)where (X,Y ) ∈ Z2, gcd(X,Y ) = 1, and Zi ≥ 0 for i = 1, . . . , v. In particular,to enumerate the set of solutions {X,Y, Z1, . . . , Zv} to this equation, we followSection 3.4 to reduce the problem of solving (6.1) to solving finitely many so-called “S-unit” equationsλ = δ1r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni− 1 = δ2r∏i=1(ε(i0)iε(j)i)ai ν∏i=1(γ(i0)iγ(j)i)ni,(6.2)whereδ1 =θ(i0) − θ(j)θ(i0) − θ(k) ·α(k)ζ(k)α(j)ζ(j), δ2 =θ(j) − θ(k)θ(k) − θ(i0) ·α(i0)ζ(i0)α(j)ζ(j)190are constants. Here, we adopt the notation of Chapter 3 and recall that we reduce(6.1) to a homogenous equation of the formf(x, y) = x3 + C1x2y + C2xy2 + C3y3 = cpz11 · · · pzvv , (6.3)where gcd(x, y) = 1 and gcd(c, pi) = 1 for i = 1, . . . , pv. Moreover, we setg(t) = f(t, 1) = t3 + C1t2 + C2t+ C3 (6.4)so that K = Q(θ) with g(θ) = 0. Recall that ζ in (6.2) denotes a root of unity inK, while {ε1, . . . , εr} is a set of fundamental units of OK . In this case, as K is adegree 3 extension ofQ, we either have 3 real embeddings of K into C, or one realembedding of K into C and a pair of complex conjugate embeddings of K into C.Thus either r = 1 or r = 2.In this section, we describe new techniques to solve equation (6.2) via a globalWeil height. This work is part of the on-going collaborative project [46]. Notably,the ideas presented in this chapter do not yet yield a full degree 3 Thue-Mahlersolver. Indeed, for the time being, only those Thue-Mahler equations with r = 2are considered. However, when r = 1, the general setup established in this chapterremains the same.6.1 Decomposition of the Weil heightThe sieves of [116] involve logarithms which are of local nature. To obtain aglobal sieve, we work instead with the global logarithmic Weil height. This heightis invariant under conjugation and admits a decomposition into local heights whichcan be related to complex and p-adic logarithms.Let n1, . . . , nν , a1, . . . , ar be a solution to (6.2) and set z = δ2λ , wherez =r∏i=1(ε(j)iε(i0)i)ai ν∏i=1(γ(j)iγ(i0)i)ni.Given the global Weil height of z, or all the local heights of z, we will construct191several ellipsoids ‘containing’ n1, . . . , nν , a1, . . . , ar such that the volume of theellipsoids are as small as possible. We begin by computing the height of z.Let L be the splitting field of K. Recall that for cubic extensions K, the Galoisgroup Gal(L/Q) is isomorphic to either the alternating group A3 or the symmetricgroup S3.Lemma 6.1.1. Let p be a prime ideal ofOK and let P denote an ideal ofOL lyingabove it. Suppose σi0 : L → L, θ 7→ θ(i0) and σj : L → L, θ 7→ θ(j) are twoautomorphisms of L such that (i0, j, k) forms a subgroup of Gal(L/Q) of order 3.Let P(i0) = σi0(P) and P(j) = σj(P) be the prime ideals lying over p(i0), p(j)respectively. For i = 1, . . . , ν,(γ(j)iγ(i0)i)OL =∏P|p1P(j) e(P(j)|p(j)1 )P(i0) e(P(i0)|p(i0)1 )a1i · · ·∏P|pνP(j) e(P(j)|p(j)ν )P(i0) e(P(i0)|p(i0)ν )aνiwhere P(j) 6= P(i0) for all P lying above p in K.Proof. Since(γi)OK = pa1i1 · · · paνiν ,for i = 1, . . . , ν, wherepiOL =∏P|piPe(P|pi),it holds that(γi)OL =∏P|p1Pe(P|p1)a1i · · ·∏P|pνPe(P|pν)aνi .Let P(i0),P(j) denote the ideal P under the automorphisms of Lσi0 : L→ L, θ 7→ θ(i0) and σj : L→ L, θ 7→ θ(j),192respectively. That is, P(i0) = σi0(P) and P(j) = σj(P). Then(γ(j)iγ(i0)i)OL =∏P|p1P(j) e(P(j)|p(j)1 )P(i0) e(P(i0)|p(i0)1 )a1i · · ·∏P|pνP(j) e(P(j)|p(j)ν )P(i0) e(P(i0)|p(i0)ν )aνi .To show that P(j) 6= P(i0) for all P lying above p in K, we consider the decom-position group of P,D(P|p) = {σ ∈ G : σ(P) = P}.Iterating through all possible decompositions of p in L, we observe that P(i0) 6=P(j) wheneverD(Pi|p) does not have cardinality 2. Since (i0, j, k) forms an order3 subgroup of Gal(L/Q), it cannot coincide with D(P|p) and therefore cannotlead to P(i0) = P(j).For the remainder of this paper, we assume that (i0, j, k) are automorphisms of Lselected as in Lemma 6.1.1.Lemma 6.1.2. Let p be a prime ideal ofOK and let P denote an ideal ofOL lyingabove it. Let P(i0) = σi0(P) and P(j) = σj(P) be the prime ideals lying overp(i0), p(j) respectively. We haveordP(δ2λ)=(ul − rl)e(P(j)|p(j)l ) if P(j) | pl, pl ∈ {p1, . . . , pν}(rl − ul)e(P(i0)|p(i0)l ) if P(i0) | pl, pl ∈ {p1, . . . , pν}0 otherwise.Proof. By Lemma 6.1.1, we have(δ2λ)OL =(γ(j)1γ(i0)1)n1· · ·(γ(j)νγ(i0)ν)nνOL=∏P|p1P(j) e(P(j)|p(j)1 )P(i0) e(P(i0)|p(i0)1 )u1−r1 · · ·∏P|pνP(j) e(P(j)|p(j)ν )P(i0) e(P(i0)|p(i0)ν )uν−rν .193It follows thatordP(δ2λ)=(ul − rl)e(P(j)|p(j)l ) if P(j) | pl, pl ∈ {p1, . . . , pν}(rl − ul)e(P(i0)|p(i0)l ) if P(i0) | pl, pl ∈ {p1, . . . , pν}0 otherwise.Let log+(·) denote the real valued function max(log(·), 0) on R≥0.Proposition 6.1.3. The height h(z) admits a decompositionh(z) =1[K : Q]ν∑l=1log(pl)|ul − rl|+ 1[L : Q]∑w:L→Clog+ |w(z)|. (6.5)In particular, when deg(g(t)) = 3,∑w:L→Clog+ |w(z)| = b maxw:L→Clog+ |w(z)|whereb =2 if Gal(L/Q) ∼= S31 if Gal(L/Q) ∼= A3.For ease of notation, let S∗ = S ∪ {w : L→ C} and writeh(z) =1[K : Q]∑v∈S∗hv(z).By Proposition 6.1.3, when v = pl is a finite place,hv(z) = log(pl)|ul − rl|,whereas we writehv(z) =1[L : K]log+ |w(z)|for all infinite places v = w : L→ C.194Proof of Proposition 6.1.3. Since z ∈ L, the definition of the absolute logarithmicWeil height givesh(z) =1[L : Q]∑w∈MLlog+ ‖z‖wwhere ||z||w and ML are the usual norms and set of inequivalent absolute valueson L, respectively. In particular, if w : L→ C is an infinite place,log+ ‖z‖w = log+ |w(z)|.If w = P is a finite place, we havelog+ ‖z‖w = log+(1N(P)ordP(z)).Let pl ∈ S and P(j) | pl. Applying Lemma 6.1.2, we obtainlog+ ‖z‖w = log+(1N(P)(ul−rl)e(P(j)|p(j)l ))= max{−(ul − rl)f(P(j) | pl)e(P(j) | p(j)l ) log(pl), 0}.For each pl ∈ S, there is only one unique prime ideal pl ∈ OK in the ideal equation(3.8) lying above pl. Hence, each P lying over pl must also lie over pl. Now, forw = P(j) lying over pl,∑w|p(j)llog+ ‖z‖w = max {(rl − ul) log(pl), 0} f(p(j)l | pl)[L : Q(θ(j))]= max {(rl − ul) log(pl), 0} f(p(j)l | pl)[L : K],where the last inequality follows from K = Q(θ) ∼= Q(θ(j)).Similarly, applying Lemma 6.1.2 to all w = P(i0) lying over pl ∈ S, we obtain∑w|p(i0)llog+ ‖z‖w = max {(ul − rl) log(pl), 0} f(p(i0)l | pl)[L : K].195Lastly, if w = P such that P 6= P(i0),P(j), we have log+ ‖z‖w = 0. Putting thisall together yields the first result (6.5).To prove the second statement, write z as the quotient z = d(j)/d(i0) ∈ L. Theorbit of z is{d(j)d(i0), d(k)d(j), d(i0)d(k), d(j)d(k), d(k)d(i0), d(i0)d(j)}if Gal(L/Q) ∼= S3{d(j)d(i0), d(k)d(j), d(i0)d(k)}if Gal(L/Q) ∼= A3.Choose a, b, c ∈ {i0, j, k} such that|d(a)| ≥ |d(b)| ≥ |d(c)|.If Gal(L/Q) ∼= S3,∑w:L→Clog+ |w(z)| = log+∣∣∣∣∣d(a)d(b)∣∣∣∣∣+ log+∣∣∣∣∣d(b)d(c)∣∣∣∣∣+ log+∣∣∣∣∣d(c)d(a)∣∣∣∣∣+ log+∣∣∣∣∣d(a)d(c)∣∣∣∣∣+ log+∣∣∣∣∣d(a)d(c)∣∣∣∣∣+ log+∣∣∣∣∣d(c)d(b)∣∣∣∣∣= log∣∣∣∣∣d(a)d(b)∣∣∣∣∣+ log∣∣∣∣∣d(b)d(c)∣∣∣∣∣+ log∣∣∣∣∣d(a)d(c)∣∣∣∣∣= 2 log∣∣∣∣∣d(a)d(c)∣∣∣∣∣ .Alternatively, if Gal(L/Q) ∼= A3,∑w:L→Clog+ |w(z)| = log+∣∣∣∣∣d(a)d(b)∣∣∣∣∣+ log+∣∣∣∣∣d(b)d(c)∣∣∣∣∣+ log+∣∣∣∣∣d(c)d(a)∣∣∣∣∣= log∣∣∣∣∣d(a)d(b)∣∣∣∣∣+ log∣∣∣∣∣d(b)d(c)∣∣∣∣∣= log∣∣∣∣∣d(a)d(c)∣∣∣∣∣ .1966.2 Initial height boundsLety˜ =r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni, x˜ =r∏i=1(ε(i0)iε(j)i)ai ν∏i=1(γ(i0)iγ(j)i)niand recall that we seek to compute all solutions to equaition (6.2). In our presentnotation, this equation isδ1y˜ − δ2x˜ = 1. (6.6)Let Σ denote the set of all pairs (x˜, y˜) satisfying (6.6). That is, Σ denotes the set ofall tuples (n1, . . . , nν , a1, . . . , ar) corresponding to (x˜, y˜) which satisfy (6.6).Let l,h ∈ Rν+m with 0 ≤ l ≤ h. By this we mean that if l = (l1, . . . , lν+m) andh = (h1, . . . , hν+m), then 0 ≤ li ≤ hi for all i = 1, . . . , ν + m. Define Σ(l,h)to be the set of all (x˜, y˜) ∈ Σ such that (hv(z)) ≤ h and such that (hv(z))  l,and write Σ(h) = Σ(l,h) if l = 0. Additionally, for each place w, we denote byΣw(l,h) the set of all (x˜, y˜) ∈ Σ(h) such that hw(z) > lw.Recall the minimal polynomial g(t) = f(t, 1) of K, wheref(x, y) = x3 + C1x2y + C2xy2 + C3y3 = cpz11 · · · pzvv .For S = {p1, . . . , pv}, let NS =∏p∈S p and setbS = 1728N2S∏p/∈Spmin(2,ordp(b))for any integer b. In particular, we take b = 432∆c2 with ∆ the discriminant of f .Denote by h(f − c) the maximum logarithmic Weil heights of the coefficients ofthe polynomial f − c,h(f − c) = max(log |C1|, log |C2|, log |C3|, log |c|).197Now, settingΩ = 2bS log(bS) + 172h(f − c),we obtain, by Corollary J (ii) of [58], the following height bound on any solution(x, y) of (6.3)max(h(x), h(y)) ≤ Ω.To translate this result for use with our logarithmic Weil height (6.5), we have thefollowing lemma.Lemma 6.2.1. Let m = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+ν be any solution of (6.2)and letΩ′ = [K : Q](2h(α) + 4Ω + 2h(θ) + 2 log(2)). (6.7)If h ∈ Rν+m with h = (Ω′), then m ∈ Σ(h).Proof. Let (x˜, y˜) ∈ Σ. We show that the corresponding value z arising from thischoice of x˜, y˜ satisfies0 < (hv(z)) ≤ h.As stated earlier, any solution x, y of f(x, y) = cpz11 · · · pzvv satisfiesmax(h(x), h(y)) ≤ Ω.Taking the height ofβ = x− yθ = αζεa11 · · · εarr · γn11 · · · γnνν ,we obtainh(β) = h(x) + h(θ) + h(y) + log 2 ≤ 2Ω + h(θ) + log 2.In particular, as h(β) = h(β(i)),h(β(i)) ≤ 2Ω + h(θ) + log 2.198Now,δ2x˜ =θ(j) − θ(k)θ(k) − θ(i0) ·α(i0)ζ(i0)α(j)ζ(j)r∏i=1(ε(i0)iε(j)i)ai ν∏i=1(γ(i0)iγ(j)i)ni=θ(j) − θ(k)θ(k) − θ(i0) ·β(i0)β(j),meaning that x˜ may be written asx˜ =β(i0)β(j)· α(j)ζ(j)α(i0)ζ(i0).Hence,h(x˜) = 2h(β) + 2h(α) ≤ 4Ω + 2h(θ) + 2 log 2 + 2h(α).Finally, we observe thath(z) = h(1/x˜) ≤ 4Ω + 2h(θ) + 2 log 2 + 2h(α).Together with1[K : Q]hv(z) ≤ h(z), this implieshv(z) ≤ [K : Q] (4Ω + 2h(θ) + 2 log 2 + 2h(α)) = Ω′.Of course, by definition, we have hv(z) ≥ 0, so that (x˜, y˜) ∈ Σ(h) as required.6.3 Coverings of ΣFrom Section 6.2, we now know that all solutions (x˜, y˜) ∈ Σ satisfy m ∈ Σ(h) ifh = (Ω′). In the notation of Section 6.2, we have the following result.Lemma 6.3.1. Let l,h ∈ Rν+m with 0 ≤ l ≤ h. It holds that Σ(h) = Σ(l,h) ∪Σ(l) and Σ(l,h) = ∪v∈S∗Σv(l,h).Proof. Suppose (x˜, y˜) ∈ Σ(h). By definition this means, (hv(z)) ≤ h and thathv(z) > 0 for at least one coordinate v. Since 0 ≤ l ≤ h, it follows that either(hv(z)) ≤ l or (hv(z))  l. That is, either all coordinates satisfy hv(z) ≤ lv,199or there is at least one coordinate for which hv(z) > lv. This means that either(x˜, y˜) ∈ Σ(l) or (x˜, y˜) ∈ Σ(l,h), and so Σ(h) ⊆ Σ(l,h) ∪ Σ(l).Conversely, suppose (x˜, y˜) ∈ Σ(l,h) ∪ Σ(l). It follows that either (hv(z)) ≤h and (hv(z))  l or (hv(z)) ≤ l and (hv(z))  0. In either case, this means that(hv(z)) ≤ h and (hv(z))  0. Hence (x˜, y˜) ∈ Σ(h) and Σ(h) ⊇ Σ(l,h) ∪ Σ(l).To prove the second equality, let (x˜, y˜) ∈ Σ(l,h). Then there exists w ∈ S∗with hw(z) > lw so that (x˜, y˜) lies in Σw(l,h). Hence Σ(l,h) ⊆ ∪v∈S∗Σv(l,h).Lastly, since each set Σv(l,h) is contained in Σ(l,h) it follows that Σ(l,h) =∪v∈S∗Σv(l,h) as required.Let h0 = (Ω′, . . . ,Ω′) denote the vector consisting of the initial bound Ω′. ByProposition 6.2.1, every solution of (6.2) is contained in h0. Therefore, we writeΣ = Σ(h0). Consider the pairs (ln,hn) ∈ Rν+m × Rν+m with 0 ≤ ln ≤ hn andhn+1 = ln for n = 0, . . . , N . Then we can cover Σ:Σ = Σ(lN ) ∪(∪Nn=0 ∪v∈S∗ Σv(ln,hn)).Indeed this follows directly by applying Lemma 6.3.1 N times. In particular,Lemma 6.3.1 givesΣ = Σ(h0), Σ(h) = Σ(l,h) ∪ Σ(l) and Σ(l,h) = ∪v∈S∗Σv(l,h).After choosing a good sequence of lower and upper bounds ln,hn covering thewhole space Σ, we are reduced to computing Σv(l,h) for each v ∈ S∗. In the fol-lowing section, we construct the ellipsoids associated to each Σv(l,h), after whichwe describe the sieve allowing us to compute the solutions of each Σv(l,h).6.4 Construction of the ellipsoidsIn Section 6.3, we establish that for a suitable pair of vectors l,h, solving (6.2)reduces to computing Σv(l,h) for each v ∈ S∗. In this section, we construct theellipsoids associated to each Σv(l,h), which will subsequently allow us to compute200all solutions of Σv(l,h).We begin with the quadratic form qf = ATD2A on Zν , where D2 is a ν × νdiagonal matrix with diagonal entries b log(pi)2log(2)2c for pi ∈ S. Recall that A is thematrix generated in either Section 3.4.1 or Section 3.4.2. As A is invertible, ourchoice of entries in D guarantees that this quadratic form is positive definite. Thiswill become very important later in the sieve when we will need to apply manyinstances of the Fincke-Pohst algorithm.Lemma 6.4.1. Consider any solution (n1, . . . , nν , a1, . . . , ar) of (6.2). Settingn = (n1, . . . , nν), we havelog(2)2qf (n) <ν∑l=1log(pl)2|ul − rl|2.Proof. Recall from Section 3.4.1 and Section 3.4.2 thatAn = u− r.Assume first that 2 /∈ S so thatqf (n) = (An)TD2An = (u− r)TD2(u− r) =ν∑l=1⌊log(pl)2log(2)2⌋|ul − rl|2.Multiplication by log(2)2 then giveslog(2)2qf (n) = log(2)2ν∑l=1⌊log(pl)2log(2)2⌋|ul − rl|2≤ν∑l=1log(pl)2|ul − rl|2,where all terms in the summand are clearly positive.If 2 ∈ S, we haveqf (n) = (An)TD2An = |u1 − r1|2 +ν∑l=2⌊log(pl)2log(2)2⌋|ul − rl|2.201It follows thatlog(2)2qf (n) ≤ log(2)2(|u1 − r1|2 +ν∑l=2log(pl)2log(2)2|ul − rl|2)=ν∑l=1log(pl)2|ul − rl|2.We now briefly re-examine the decomposition of h(z) into local heights,h(z) =1[K : Q]ν∑l=1log(pl)|ul − rl|+ 1[L : Q]∑w:L→Clog+ |w(z)|.For every finite place v, Lemma 6.4.1 tells us that any set of bounds {hv}v∈Son the set {hv(z)}v∈S yields a bound on log(2)2qf (n). In the remainder of thissection, we build analogous bounds on the exponents a1, . . . , ar of the fundamentalunits.Recall r = 1 or r = 2 for the degree 3 Thue-Mahler equation (6.3) in question.Choose a set I of embeddings L → C of cardinality r. For r = 1, consider thematrixR =(log∣∣∣∣( ε(j)1ε(i0)1)ι1∣∣∣∣) ,where I = {ι1}. Clearly, as long as we choose ι1 such that log∣∣∣∣( ε(j)1ε(i0)1)ι1∣∣∣∣ 6= 0,this matrix is invertible.When r = 2, we let I be the set of embeddings L → C of cardinality 2 such thatfor any α ∈ K, it holds that Iα(i0) ∪ Iα(j) = Gal(L/Q)α. For I = {ι1, ι2}, let Rbe the 2× 2 matrixR =log∣∣∣∣( ε(j)1ε(i0)1)ι1∣∣∣∣ log ∣∣∣∣( ε(j)2ε(i0)2)ι1∣∣∣∣log∣∣∣∣( ε(j)1ε(i0)1)ι2∣∣∣∣ log ∣∣∣∣( ε(j)2ε(i0)2)ι2∣∣∣∣ .202Lemma 6.4.2. When r = 2, the matrix R has an inverse,R−1 =(r11 r12r21 r22).Proof. Suppose that m ∈ Z2 satisfies Rm = 0. Then for each ι ∈ I it holds thatm1 log∣∣∣∣∣(ε(j)1ε(i0)1)ι∣∣∣∣∣+m2 log∣∣∣∣∣(ε(j)2ε(i0)2)ι∣∣∣∣∣ = 0,and hence ∣∣∣∣∣(ε(j)1ε(i0)1)ι∣∣∣∣∣m1·∣∣∣∣∣(ε(j)2ε(i0)2)ι∣∣∣∣∣m2= 1.This together with I(i) ∪ I(j) = Gal(L/Q) implies that all conjugates of α =εm11 εm22 have the same absolute value. Since all εi are units of OK , it follows that|α|[L:Q] = N(α) = 1 and hence α is a root of unity in K. On using that theelements εi are multiplicatively independent, we obtain that m = 0. Then linearalgebra gives R−1 ∈ R2×2, completing the proof.For the remainder of this chapter, we specialize to the real case, r = 2. The setupfor r = 1 follows closely the work described here, yet poses other difficulties whendefining the corresponding sieves. This case is treated in the on-going results of[46].Now, for any solution (x, y, n1, . . . , nν , a1, a2) of (6.2), setε =(a1 a2)T.We haveRε =log∣∣∣∣( ε(j)1ε(i0)1)ι1 a1·(ε(j)2ε(i0)2)ι1 a2∣∣∣∣log∣∣∣∣( ε(j)1ε(i0)1)ι2 a2·(ε(j)2ε(i0)2)ι2 a2∣∣∣∣ .203Since R is invertible with R−1 = (rnm), we findε =r11 log∣∣∣∣( ε(j)1ε(i0)1)ι1 a1·(ε(j)2ε(i0)2)ι1 a2∣∣∣∣+ r12 log ∣∣∣∣( ε(j)1ε(i0)1)ι2 a1·(ε(j)2ε(i0)2)ι2 a2∣∣∣∣r21 log∣∣∣∣( ε(j)1ε(i0)1)ι1 a1·(ε(j)2ε(i0)2)ι1 a2∣∣∣∣+ r22 log ∣∣∣∣( ε(j)1ε(i0)1)ι2 a1·(ε(j)2ε(i0)2)ι2 a2∣∣∣∣ ,givingal = rl1 log∣∣∣∣∣(ε(j)1ε(i0)1)ι1 a1·(ε(j)2ε(i0)2)ι1 a2∣∣∣∣∣+ rl2 log∣∣∣∣∣(ε(j)1ε(i0)1)ι2 a1·(ε(j)2ε(i0)2)ι2 a2∣∣∣∣∣for l = 1, 2.To estimate |al|, we begin to estimate the sum on the right hand side. For this, weconsiderz =(ε(j)1ε(i0)1)a1 (ε(j)2ε(i0)2)a2 ν∏i=1(γ(j)iγ(i0)i)ni.For any embedding ι : L→ C, we have(z)ιν∏i=1(γ(i0)iγ(j)i)ι ni=(ε(j)1ε(i0)1)ι a1 (ε(j)2ε(i0)2)ι a2.In particular ∣∣∣∣∣(z)ιν∏i=1(γ(i0)iγ(j)i)ι ni∣∣∣∣∣ =∣∣∣∣∣(ε(j)1ε(i0)1)ι a1 (ε(j)2ε(i0)2)ι a2∣∣∣∣∣ ,so thatlog∣∣∣∣∣(ε(j)1ε(i0)1)ι a1 (ε(j)2ε(i0)2)ι a2∣∣∣∣∣ = log |ι(z)| − log∣∣∣∣∣ν∏i=1(γ(j)iγ(i0)i)ι ni∣∣∣∣∣ .204Hence, for l = 1, 2,al = rl1 log∣∣∣∣∣(ε(j)1ε(i0)1)ι1 a1·(ε(j)2ε(i0)2)ι1 a2∣∣∣∣∣+ rl2 log∣∣∣∣∣(ε(j)1ε(i0)1)ι2 a1·(ε(j)2ε(i0)2)ι2 a2∣∣∣∣∣= rl1(log |ι1(z)| − log∣∣∣∣∣ν∏i=1(γ(j)iγ(i0)i)ι1 ni∣∣∣∣∣)++ rl2(log |ι2(z)| − log∣∣∣∣∣ν∏i=1(γ(j)iγ(i0)i)ι2 ni∣∣∣∣∣)= rl1 log |ι1(z)|+ rl2 log |ι2(z)| − n1βγ1l − · · · − nνβγν l,whereβγkl =(rl1 log∣∣∣∣∣ι1(γ(j)kγ(i0)k)∣∣∣∣∣+ rl2 log∣∣∣∣∣ι2(γ(j)kγ(i0)k)∣∣∣∣∣)for k = 1, . . . , ν. Recall that n = A−1(u − r) and suppose A−1 = (anm). Wehaven = A−1(u− r) =∑νk=1 a1k(uk − rk)...∑νk=1 aνk(uk − rk) ,so that we may rewrite each al asal = rl1 log |ι1(z)|+ rl2 log |ι2(z)| −ν∑k=1(uk − rk)αγlk,whereαγlk = a1kβγ1l + · · ·+ aνkβγν l.Taking absolute values, we obtain|al| ≤ |rl1|| log |ι1(z)||+ |rl2|| log |ι2(z)||+ν∑k=1|uk − rk||αγlk|.205Suppose log |ι1(z)| ≥ 0 and log |ι2(z)| ≥ 0. Then|al| ≤ |rl1| log |ι1(z)|+ |rl2| log |ι2(z)|+ν∑k=1|uk − rk||αγlk|≤ max{|rl1|, |rl2|}∑w:L→Clog+ |w(z)|+ν∑k=1|uk − rk||αγlk|.Applying Proposition 6.1.3 yields|al| ≤ bmax{|rl1|, |rl2|} maxw:L→Clog+ |w(z)|+ν∑k=1|uk − rk||αγlk|whereb =2 if Gal(L/Q) ∼= S31 if Gal(L/Q) ∼= A3.Alternatively, suppose that both log |ι1(z)| < 0 and log |ι2(z)| < 0. We recall thatz is a quotient of elements which are conjugate to one another. By taking the normof z in L, we obtain N(z) = 1. On the other hand, by definition, we have1 = N(z) =∏w:L→Cw(z).Taking absolute values and logarithms,0 =∑w:L→Clog |w(z)|so that− log |ι(z)| =∑w : L→ Cw 6= ιlog |w(z)|.In our present case, we use this equivalence to obtain a bound on |al| as fol-206lows.|al| ≤ |rl1|∑w : L→ Cw 6= ι1log |w(z)| − |rl2| log |ι2(z)|+ν∑k=1|uk − rk||αγlk|≤ bmax{|rl1|, |rl2|} maxw:L→Clog+ |w(z)|+ν∑k=1|uk − rk||αγlk|.Here, the second inequality follows again by Proposition 6.1.3. Lastly, if, withoutloss of generality, we have log |ι1(z)| < 0 and log |ι2(z)| ≥ 0, then|al| ≤ |rl1|∑w : L→ Cw 6= ι1log |w(z)|+ |rl2| log |ι2(z)|+ν∑k=1|uk − rk||αγlk|≤ (b+ 1) max{|rl1|, |rl2|} maxw:L→Clog+ |w(z)|+ν∑k=1|uk − rk||αγlk|.Now, letwεl = (b+ 1) max{|rl1|, |rl2|}, (6.8)andwγlk =|αγlk|log(pk), (6.9)whereαγlk = a1kβγ1l + · · ·+ aνkβγν l,andβγkl =(rl1 log∣∣∣∣∣(γ(j)kγ(i0)k)ι1∣∣∣∣∣+ rl2 log∣∣∣∣∣(γ(j)kγ(i0)k)ι2∣∣∣∣∣)for k = 1, . . . , ν. We have proven the following lemma.Lemma 6.4.3. For any solution (n1, . . . , nν , a1, . . . , ar) of (6.2), for l = 1, 2, wehave|al| ≤ wεl maxw:L→Clog+ |w(z)|+ν∑k=1wγlk log(pk)|uk − rk|.2076.4.1 The Archimedean ellipsoid: the real caseLet τ : L → R ⊂ C be an embedding and let lτ ≥ cτ and c > 0 be given realnumbers for cτ = log+(2|τ(δ2)|). We defineα0 = [c log |τ(δ1)|], αε1 =[c log∣∣∣∣∣τ(ε(k)1ε(j)1)∣∣∣∣∣], αε2 =[c log∣∣∣∣∣τ(ε(k)2ε(j)2)∣∣∣∣∣].(6.10)For i = 1, . . . , ν, defineαγi =[c log∣∣∣∣∣τ(γ(k)iγ(j)i)∣∣∣∣∣]. (6.11)Here, [ · ] denotes the nearest integer function.Letwε =wε1 + wε22, wγk =wγ1k + wγ2k2+12 log(pk)ν∑i=1|aik| (6.12)for k = 1, . . . , ν. Herewε1, wε2 andwγ1k, wγ2k are the coefficients (6.8) and (6.9),respectively. Let κτ = 3/2 and recall thathτ (z) =1[L : K]log+ |τ(z)|denotes the local height of z at τ in the decomposition of h(z).Lemma 6.4.4. Let (n1, . . . , nν , a1, . . . , ar) be any solution of (6.2). If hτ (z) > cτ ,then∣∣∣∣∣α0 +r∑i=1aiαεi +ν∑i=1niαγi∣∣∣∣∣≤ 12+ wε maxw:L→Clog+ |w(z)|+ν∑l=1wγl log(pl)|ul − rl|+ cκτe−hτ (z)208Proof. Letατ = α0 +r∑i=1aiαεi +ν∑i=1niαγiandΛτ = log∣∣∣∣∣τ(δ1r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni)∣∣∣∣∣ .We claim thatτ(δ1r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni)> 0.Indeed, hτ (z) > cτ by assumption, hencemax {|τ(z)|, 1} > max{2|τ(δ2)|, 1}.From this inequality, we must have that max{|τ(z)|, 1} = |τ(z)| and so2|τ(δ2)| < |τ(z)| = |τ(δ2)||τ(λ)| =⇒ |τ(λ)| <12.Recall that δ1y˜ − δ2x˜ = 1. This is the equation (6.6) defined earlier. In particular,observe that λ = δ2x˜ so that applying τ givesτ(λ) = τ(δ2x˜) = τ(δ1y˜)− 1.Thus|τ(λ)| < 12=⇒ τ(δ1y˜) = τ(λ) + 1 > 0.This proves our claimτ(δ1y˜) = τ(δ1r∏i=1(ε(k)iε(j)i)ai ν∏i=1(γ(k)iγ(j)i)ni)> 0.Having established this, we may now writeΛτ = log (τ (δ1)) +r∑i=1ai log(τ(ε(k)iε(j)i))+ν∑i=1ni log(τ(γ(k)iγ(j)i)).209By the triangle inequality,|ατ | ≤ |ατ − cΛτ |+ c|Λτ |,where|ατ − cΛτ | ≤ |[c log(τ(δ1))]− c log (τ (δ1))|+r∑i=1|ai|∣∣∣∣∣[c log(τ(ε(k)iε(j)i))]− c log(τ(ε(k)iε(j)i))∣∣∣∣∣+ν∑i=1|ni|∣∣∣∣∣[c log(τ(γ(k)iγ(j)i))]− c log(τ(γ(k)iγ(j)i))∣∣∣∣∣ .Since [ · ] denotes the nearest integer function, it is clear that |[ c ] − c| ≤ 1/2 forany integer c,|ατ − cΛτ | ≤ 12+12r∑i=1|ai|+ 12ν∑i=1|ni|≤ 12(1 +r∑i=1|ai|+ |u1 − r1|ν∑i=1|ai1|+ · · ·+ |uν − rν |ν∑i=1|aiν |).Applying Lemma 6.4.3, this becomes|ατ − cΛτ | ≤ 12+(wε1 + wε2)2maxw:L→Clog+ |w(z)|++ log(p1)|u1 − r1|((wγ11 + wγ21)2+12 log(p1)ν∑i=1|ai1|)+ · · ·+ log(pν)|uν − rν |((wγ1ν + wγ2ν)2+12 log(pν)ν∑i=1|aiν |).In the notation of (6.12), this inequality reduces to|ατ − cΛτ | ≤ 12+ wε maxw:L→Clog+ |w(z)|+ν∑l=1wγl log(pl)|ul − rl|.210Now the following upper bound for |Λτ | implies the statement. On using powerseries definition of exponential function, we obtainΛτ (1 +∑n≥2(Λτ )n−1/n!) = Λτ +∑n≥2(Λτ )n/n! = eΛτ − 1 = τ(λ).If Λτ ≥ 0 then 1 +∑n≥2(Λτ )n−1/n! > 1 which implies that |Λτ | ≤ |τ(λ)|.Suppose now that Λτ < 0. Our assumption hτ (z) ≥ log+(2|λ0|) means that|τ(λ)| ≤ 1/2 and thus |Λτ | = − log(τ(λ) + 1) ≤ − log(1/2) = log 2. Therefore,the absolute value of∑n≥2(Λτ )n−1/n! is at most∑n≥2|Λτ |n−1/n! =∑n≥1|Λτ |n/(n+ 1)! ≤ 12∑n≥1|Λτ |n/n! ≤ 12elog 2 − 1/2 = 1/2.More precisely, for any even N ≥ 2, we obtain|∑n≥2(Λτ )n−1/n!| = |∑n≥1(Λτ )n/(n+ 1)!|≤ |∑N≥n≥1(Λτ )n/(n+ 1)!|+ 1N+2 |∑n>N(Λτ )n/n!|≤ |∑N≥n≥1(Λτ )n/(n+ 1)!|+ 1N+2e|Λτ |≤ |∑N≥n≥1(Λτ )n/(n+ 1)!|+ 2N+2 := kN .We now give an upper bound for kN . Since Λτ < 0, we obtain∑n≥2(Λτ )n−1/n! =∑N≥n≥1(Λτ )n/(n+ 1)! =∑N≥n≥2, 2|n|Λτ |n(n+1)! − |Λτ |n−1n!=∑N≥n≥2, 2|n|Λτ |n−1n! (|Λτ |n+1 − 1) = |Λτ |2 ( |Λτ |3 − 1) +∑N≥n≥4, 2|n|Λτ |n−1n! (|Λτ |n+1 − 1)≥ log 22 ( log 24 − 1) +∑N≥n≥4, 2|n(log 2)n−1n! (3/4(log 2)n+1 − 1) := −kN .The last inequality follows by distinguishing two cases whether |Λτ | ≤ 3/4 · log 2or not; note that ln(2)/2 · (ln(2)/4 − 1)/(− ln(2) · 3/8) ≥ 1. Now, on using that211−kN is negative, it follows that|1 +∑n≥2(Λτ )n−1/n!| ≥ 1− |∑n≥2(Λτ )n−1/n!| ≥ 1− kNand thus|Λτ | ≤ κτ |τ(x)|, κτ = 11−kN |τ(λ0)|, cτ = log+(2|λ0|).The constant κτ depends on N which can be taken arbitrarily as long as N ≥ 2 iseven. Further, the value kN can be slightly improved when one finds the maximumof the functions xn−1( xn+1 − 1) on the interval [0, log 2] for each even n ≥ 2. Thisis our reason for taking κτ = 32 . Currently this is not the optimal choice of κτ , butit suffices for our present case.Finally, we have|ατ | ≤ 12+ wε maxw:L→Clog+ |w(z)|+ν∑l=1wγl log(pl)|ul − rl|+ cκτe−hτ (z).To summarize the results of this section, let m = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+νbe any solution of (6.2) with corresponding vector n = (n1, . . . , nν). Take l,h ∈Rν+m such that 0 ≤ l ≤ h and suppose hv(z) ≤ hv for all v ∈ S∗. ByLemma 6.4.1, we deduceqf (n) ≤ 1log(2)2ν∑k=1log(pk)2|uk − rk|2 ≤ 1log(2)2ν∑k=1h2k =: bγ . (6.13)212For l = 1, 2, Lemma 6.4.3 gives us|al|2 ≤(wεl maxw:L→Clog+ |w(z)|+ν∑k=1wγlk log(pk)|uk − rk|)2(6.14)≤([L : K]wεl maxw:L→Chw +ν∑k=1wγlkhk)2=: bεl . (6.15)Finally, suppose in addition thathτ (z) ≥ lτ > cτ .Then by Lemma 6.4.4, we obtain∣∣∣∣∣α0 +r∑i=1aiαεi +ν∑i=1niαγi∣∣∣∣∣2(6.16)≤(12+ wε maxw:L→Clog+ |w(z)|+ν∑k=1wγk log(pk)|uk − rk|+ cκτe−hτ (z))2(6.17)≤(12+ [L : K]wε maxw:L→Chw +ν∑k=1wγkhk + cκτe−lτ)2=: bε∗l . (6.18)It is of particular importance to note that the assumptions hτ (z) ≥ lτ and hv(z) ≤hv for all v ∈ S∗ are not arbitrary. Indeed, for the vectors l,h, these conditionsimply precisely that (x˜, y˜) ∈ Στ (l,h), where (x˜, y˜) are solutions to (6.2) corre-sponding to m.We are finally in position to define the ellipsoid corresponding to Στ (l,h). Fix anyε∗l ∈ {ε1, . . . , εr}. For each εl in {ε1, . . . , εr} such that εl 6= ε∗l , we associate thebound bεl . For εl, we associate the value bε∗l .Letx = (x1, . . . , xν , xε1 , . . . , xεr) ∈ Rν+r.213Then we define the ellipsoid Eτ ⊆ Rr+ν byEτ = {qτ (x) ≤ (1 + r)(bγbε1 · · · bεr); x ∈ Rr+ν} (6.19)whereqτ (x) = (bε1 · · · bεr)(qf (x1, . . . , xν) +r∑i=1bγbεix2εi)andqf (y) = (Ay)TD2Ay.We associate to this ellipsoid a matrix. More precisely, we let M = Mτ be thematrix defining the ellipsoid Eτ . Explicitly, this is the matrixM =√bε1 · · · bεrDA 0 . . . 0 00√bγbε1. . . 0 00 0√bγbε2. . . 0...... 0. . ....0 0 . . . . . .√bγbε∗.Note that we never need to compute M , but rather MTM so that we only everwork with integral matrices. In this case,MTM = bε1 · · · bεrATD2A 0 . . . 0 00bγbε1. . . 0 00 0bγbε2. . . 0...... 0. . ....0 0 . . . . . .bγbε∗.6.4.2 The non-Archimedean ellipsoidWe now restrict our attention to those pv ∈ {p1, . . . , pν} and define the corre-sponding ellipsoid. As before, let m = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+ν be any214solution of (6.2) with corresponding vector n = (n1, . . . , nν). Take l,h ∈ Rν+msuch that 0 ≤ l ≤ h and suppose hv(z) ≤ hv for all v ∈ S∗.Now, Lemma 6.4.1 and Lemma 6.4.3 still hold here. In particular, we let bγ , bεl bedefined as in (6.13) and (6.14), respectively, where l = 1, . . . , r. We do not distin-guish any ε∗l . Instead, we will see later that the condition hv(z) ≥ lv correspondingto the set Σv(l,h) will be used elsewhere.We define the ellipsoid Ev ⊆ Rν+r byEv = {qv(x) ≤ (1 + r)(bγbε1 · · · bεr); x ∈ Rr+ν}, (6.20)whereqv(x) = (bε1 · · · bεr)(qf (x1, . . . , xν) +r∑i=1bγbεix2εi)andqf (y) = (Ay)TD2Ay.Similar to the Archimedean case, we let M = Mv be the matrix defining theellipsoid Ev. Explicitly, this is the matrixM =√bε1 · · · bεrDA 0 . . . 0 00√bγbε1. . . 0 00 0√bγbε2. . . 0...... 0. . ....0 0 . . . . . .√bγbε.As before, we never need to compute M , but rather MTM so that we only ever215work with integral matrices. In this case,MTM = bε1 · · · bεrATD2A 0 . . . 0 00bγbε1. . . 0 00 0bγbε2. . . 0...... 0. . ....0 0 . . . . . .bγbε.6.5 The Archimedean sieve: the real caseLet τ : L → C be an embedding. We take l,h ∈ Rm+ν with 0 ≤ l ≤ h andlτ ≥ log 2. Let c be a constant the size of elτ and let α0, αε1, . . . , αεr, αγ1, . . . , αγνbe defined as in (6.10) and (6.11).Define the (ν + r)× (ν + r)-dimensional matrix Aτ asAτ =1 0 . . . . . . 0 00 1 . . . . . . 0 0....... . . . . .......0 0 . . . . . . 1 0αγ1 . . . αγν αε1 . . . αεrand consider the lattice defined by its columns. Let w = (0, . . . , 0, α0) be a vectorof length (ν + r). We now consider the translated lattice Γτ defined by Aτx+w,where x is an arbitrary coordinate vector.Let Eτ = Eτ (h, lτ ) be the ellipsoid constructed in (6.19). Letm = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+νbe any solution of (6.2). We say that m is determined by some y ∈ Γτ ify = (y1, . . . , yr+ν) =(n1, . . . , nν , a1, . . . , ar−1, α0 +r∑i=1aiαεi +ν∑i=1niαγi)216where the missing element al corresponds to ε∗l .Lemma 6.5.1. Let m = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+ν be any solution of (6.2)which lies in Στ (l, h). Then m is determined by some y ∈ Γτ ∩ Eτ .Proof. Lety =(n1, . . . , nν , a1, . . . , ar−1, α0 +r∑i=1aiαεi +ν∑i=1niαγi).Then y ∈ Γτ and (6.16) implies that y2ε∗l ≤ bε∗l . Further qf (y1, . . . , yν) ≤ bγ by(6.13) and (6.14) provides that y2εl ≤ bεl for l = 1, . . . , r with εl 6= ε∗l . It followsthatqτ (y) = (bε1 · · · bεr)(qf (y1, . . . , yν) +r∑i=1bγbεiy2εi)≤ (1 + r)(bγbε1 · · · bεr).This proves that y ∈ Eτ and hence the statement follows.We now explicitly determine Γτ ∩Eτ . Suppose that y ∈ Γτ ∩Eτ . Let M = Mτ bethe matrix defining the ellipsoid Eτ . Since y ∈ Γτ ∩Eτ , there exists x ∈ Rr+ν suchthat y = Aτx+w and ytM tMy ≤ (1 + r)(bγbε1 · · · bεr). We thus have(Aτx+w)tM tM(Aτx+w) ≤ (1 + r)(bγbε1 · · · bεr).As Aτ is clearly invertible, with matrix inverseA−1τ =1 0 . . . . . . 0 00 1 . . . . . . 0 0....... . . . . .......0 0 . . . . . . 1 0−αγ1αεr . . . −αγναεr−αε1αεr . . . 1αεr,217we can find a vector c such that Aτc = −w. Indeed, this vector isc = A−1τ w =00...0− α0αεr.Now,(1 + r)(bγbε1 · · · bεr) ≥ (Aτx+w)tM tM(Aτx+w)= (Aτ (x− c))TMTM(Aτ (x− c))= (x− c)T (MAτ )TMAτ (x− c)= (x− c)TBTB(x− c)where B = MAτ . That is, we are left to solve(x− c)TBTB(x− c) ≤ (1 + r)(bγbε1 · · · bεr).Now, finding all vectors satisfying this inequality amounts to computing all solu-tions to (6.2) contained in Στ (l,h). The set of vectors x can be found using theFincke-Pohst algorithm outlined in Section 3.6.2.6.6 The non-Archimedean SieveLet v ∈ {1, . . . , ν}. We take vectors l,h ∈ Rν+r with 0 ≤ l ≤ h andlvlog(p)≥ max(1p− 1 , ordpv(δ1))− ordpv(δ2)and then consider the translated lattice Γv ⊆ Zν+r defined below. We say thatm = (n1, . . . , nν , a1, . . . , ar) ∈ Rr+ν is determined by some y ∈ Γv if the en-tries of y are a (fixed) permutation of the entries of m. Let Ev be the ellipsoidconstructed in (6.20).218Lemma 6.6.1. Any (x˜, y˜) ∈ Σv(l, h) is determined by some y ∈ Γv ∩ Ev.In the remainder of this section, we prove this lemma.We begin by applying the results of Section 3.5. In particular, we consider theformΛv =1+ν+r∑i=1biαiwhereb1 = 1, b1+i = ni for i ∈ {1, . . . , ν},b1+ν+i = ai for i ∈ {1, . . . , r},andα1 = logpl δ1, α1+i = logpl(γ(k)iγ(l)i)for i ∈ {1, . . . , ν},α1+ν+i = logpl(ε(k)iε(l)i)for i ∈ {1, . . . , r}.We apply Lemma 3.5.2 by which∑νj=1 njavj can be computed directly providedordpv(δ1) 6= 0. In doing so, we assume for the remainder of this chapter thatordpv(δ1) = 0. Furthermore, we apply Lemma 4.5.4 to obtain a small boundon∑νj=1 njavj when ordpv(α1) < min2≤i≤1+ν+rordpv(αi). Again, in doing so, weassumeordpv(α1) ≥ min2≤i≤1+ν+rordpv(αi)for the remainder of this chapter.We now set some notation and give some preliminaries for the pl-adic reductionprocedures. Let I be the set of all indices i′ ∈ {2, . . . , 1 + ν + r} for whichordpv(αi′) = min2≤i≤1+ν+rordpv(αi).Following [50], we are always in the case where there exists an index i′ ∈ I suchthat αi/αi′ ∈ Qpl for i = 1, . . . , 1 + ν + r. Thus, let iˆ denote this index. We219defineβi = −αiαiˆi = 1, . . . , 1 + ν + r,andΛ′v =1αiˆΛv =1+ν+r∑i=1bi(−βi).Now, we have βi ∈ Zpv for i = 1, . . . , 1 + ν + r.Lemma 6.6.2. Suppose ordpv(δ1) = 0 andv∑i=1niali >1pv − 1 − ordpv(δ2).Thenordpv(Λ′v) =v∑i=1niali + ordpl(δ2)− ordpl(αiˆ).Proof. Immediate from Lemma 3.5.3 and Lemma 3.5.4.We now describe the pv-adic reduction procedure. Recall that lv is a constant suchthatlvlog(p)≥ max(1pv − 1 , ordpv(δ1))− ordpv(δ2).Now, let µ be the largest element of Z≥0 at mostµ ≤ lvlog(p)− ordpl(αiˆ) + ordpl(δ2).For each x ∈ Zpl , let x{µ} denote the unique rational integer in [0, pµl − 1] suchthat ordpl(x− xµ) ≥ µ (ie. x ≡ x{µ} (mod pµl )).Let Γv be the (ν+r)-dimensional translated lattice determined byAvx+w, whereAv is the diagonal matrix having iˆth row(β{µ}2 , · · · , β{µ}iˆ−1 , pµl , β{µ}iˆ+1, · · · , β{µ}1+ν+r)∈ Zν+r.220Here, pµl is the (ˆi, iˆ) entry of Av. That is,Av =1. . . 01β{µ}2 · · · β{µ}iˆ−1 pµl β{µ}iˆ+1· · · β{µ}1+ν+r10. . .1.Additionally,w is the vector whose only non-zero entry is the iˆth element, β{µ}1 ,w = (0, . . . 0, β{µ}1 , 0, . . . , 0)T ∈ Zν+r.Of course, we must compute the βi to pl-adic precision at least µ in order toavoid errors here. Let y = (n1, . . . , nν , a1, . . . , ar) ∈ Rν+r denote a solutionto (6.2).Lemma 6.6.3. Suppose ordpv(δ1) = 0 andν∑i=1niavi >1pv − 1 − ordpv(δ2).Then the following equivalence holds:ν∑i=1niavi ≥ µ− ordpv(δ2) + ordpv(αiˆ) if and only if ordpv(Λ′v) ≥ µif and only if y ∈ Γv.Proof. By Lemma 6.6.2, the assumption means thatordpv(Λ′v) =ν∑i=1niavi + ordpv(δ2)− ordpv(αiˆ).221Now, supposeν∑i=1niavi ≥ µ− ordpv(δ2) + ordpv(αiˆ).We thus haveordpv(Λ′v) =ν∑i=1niavi + ordpv(δ2)− ordpv(αiˆ)≥ µ− ordpv(δ2) + ordpv(αiˆ) + ordpv(δ2)− ordpv(αiˆ)= µ.Conversely, suppose ordpv(Λ′v) ≥ µ. Thenµ ≤ ordpv(Λ′v) =ν∑i=1niavi + ordpv(δ2)− ordpv(αiˆ).That is,ν∑i=1niavi ≥ µ− ordpv(δ2) + ordpv(αiˆ).Hence, it follows thatν∑i=1niavi ≥ µ − ordpv(δ2) + ordpv(αiˆ) if and only ifordpv(Λ′v) ≥ µ.Now, suppose y = (n1, . . . , nν , a1, . . . , ar) ∈ Rν+r is a solution to (6.2). Supposefurther thatν∑i=1niavi ≥ µ− ordpv(δ2) + ordpv(αiˆ) so that ordpv(Λ′v) ≥ µ. Letλ =1pµvν+r+1∑i=1bi(−β{µ}i )and consider the (ν + r)-dimensional vectorx = (n1, . . . , niˆ−1, λ, niˆ+1, . . . , nν , a1, . . . , ar).We claim x ∈ Zν+r. That is, λ ∈ Z, meaning that ∑ν+r+1i=1 bi(−β{µ}i ) is divisible222by pµv , or equivalently,ordpv(ν+r+1∑i=1bi(−β{µ}i ))≥ µ.Indeed, sinceordpv(β{µ}i − βi)≥ µ for i = 1, . . . , 1 + ν + r,by definition, it follows that β{µ}i and βi share the first µ − 1 terms and thusordpv(βi) = ordpv(β{µ}i ). Now, to compute this order, we only need to concernourselves with the first non-zero term in the series expansion of∑ν+r+1i=1 bi(−β{µ}i ).Since β{µ}i and βi share the first µ− 1 terms, it follows that showingordpv(ν+r+1∑i=1bi(−β{µ}i ))≥ µis equivalent to showing thatordpl(Λ′l) ≥ µ.Of course, this latter inequality is true by assumption. Thus λ ∈ Z.Then, computing Avx+w yieldsAvx+w =b2...biˆ−1b∗biˆ+1...bν+r+1,223whereb∗ = b2β{µ}2 + · · ·+ biˆ−1β{µ}iˆ−1 + λpµl + biˆ+1β{µ}iˆ+1+ · · ·+ bν+r+1β{µ}1+ν+r + β{µ}1 .Now,λpµv = pµv1pµvν+r+1∑i=1bi(−β{µ}i ) =ν+r+1∑i=1bi(−β{µ}i ),henceb2β{µ}2 + · · ·+ biˆ−1β{µ}iˆ−1 + biˆ+1β{µ}iˆ+1+ · · ·+ bν+r+1β{µ}1+ν+r + λpµl + β{µ}1= biˆ(−β{µ}iˆ )= biˆwhere the last equality follows from the fact that−βi =αiˆαiˆ= 1.Thus,Avx+w =b2...biˆ−1biˆbiˆ+1...bν+r+1=n1...nνa1...ar= y.and y ∈ Γv.Definecpv = log pv(max(1pv−1 , ordpv(δ1))− ordpv(δ2)).Corollary 6.6.4. Assume that hpv(z) > max(0, cpv). Then the following equiva-224lence holds:hpv(z) ≥ log pv(µ− ordpv(δ2) + ordpv(αiˆ))if and only if y ∈ Γv.Proof. Recall from Proposition 6.1.3 thathpv(z) =log(pv)|uv − rv|0 .Since hpv(z) > 0, it follows that hpv(z) = log(pv)|uv−rv|. Hence the assumptionbecomeslog(pv)|uv − rv| = hpv(z) > log pv(max(1pv−1 , ordpv(δ1))− ordpv(δ2)),or equivalently,ν∑j=1njavj >(max(1pv−1 , ordpv(δ1))− ordpv(δ2)).Moreover, the conclusion is equivalent tolog(pv)|uv−rv| ≥ log pv(µ− ordpv(δ2) + ordpv(αiˆ))if and only if y ∈ Γv,or,ν∑j=1njavj ≥(µ− ordpv(δ2) + ordpv(αiˆ))if and only if y ∈ Γv,which is the previous lemma.We now prove Lemma 6.6.1.Proof of Lemma 6.6.1. If (n1, . . . , nν , a1, . . . , ar) ∈ Rr+ν is a solution of (6.2),then, by definition, it corresponds to a solution (x˜, y˜) ∈ Σv(l,h). Hence hv(z) >225lv, where lv is a constant such thatlvlog(pv)≥ max(1pv − 1 , ordpv(δ1))− ordpv(δ2).That is,hv(z) > lv ≥ log(pv)(max(1pv − 1 , ordpv(δ1))− ordpv(δ2))= cp.Now, recall that l ≥ 0 so that lv ≥ 0. It thus follows thathv(z) > lv ≥0cp =⇒ hv(z) > max(0, cp).In other words, the conditions of Corollary 6.6.4 are satisfied.Now, recall that µ is the largest element of Z≥0 at mostµ ≤ lvlog(pv)− ordpv(αiˆ) + ordpv(δ2).That islvlog(pv)≥ µ+ ordpv(αiˆ)− ordpv(δ2)so thathv(z) > lv ≥ log(pv)(µ+ ordpl(αiˆ)− ordpv(δ2)).Now, by Corollary 6.6.4, we must have y ∈ Γv. This shows that (x˜, y˜) is deter-mined by y = m′ ∈ Γv, which proves Lemma 6.6.1.Finally, suppose that y ∈ Γv∩Ev. LetM = Mv be the matrix defining the ellipsoid226Ev. That isM =√bε1 · · · bεrDA 0 . . . 0 00√bγbε1. . . 0 00 0√bγbε2. . . 0...... 0. . ....0 0 . . . . . .√bγbε.Recall that Avx+w defines the lattice Γv. In particular, since y ∈ Γv ∩ Ev, thereexists x ∈ Rr+ν such that y = Avx+w and yTMTMy ≤ (1 + r)(bγbε1 · · · bεr).We thus have(Avx+w)TMTM(Avx+w) ≤ (1 + r)(bγbε1 · · · bεr).As Av is clearly invertible, with matrix inverseA−1v =1. . . 01−β{µ}2pµl· · · −β{µ}iˆ−1pµl1pµl−β{µ}iˆ+1pµl· · · −β{µ}1+ν+rpµl10. . .1,we can find a vector c such that Avc = −w. Indeed, this vector is c = A−1v (−w),227wherec =0...0−β{µ}1p{µ}0...0.Now,(1 + r)(bγbε1 · · · bεr) ≥ (Avx+w)TMTM(Avx+w)= (Avx−Avc)TMTM(Avx−Avc)= (x− c)T (MAv)TMAv(x− c)= (x− c)TBTB(x− c)where B = MAv. That is, we are left to solve(x− c)TBTB(x− c) ≤ (1 + r)(bγbε1 · · · bεr).As in Section 6.5 finding all vectors satisfying this inequality amounts to comput-ing all solutions to (6.2) contained in Σv(l,h). The set of vectors x can be foundusing the Fincke-Pohst algorithm outlined in Section 3.6.2.228Bibliography[1] M. K. Agrawal, J. H. Coates, D. C. Hunt and A. 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