Thue equationsbyAmir GhadermarziB.Sc., Sharif University of Technology, 2005M.Sc., Sharif University of Technology, 2008A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYinThe Faculty of Graduate and Postdoctoral Studies(Mathematics)THE UNIVERSITY OF BRITISH COLUMBIA(Vancouver)January 2014c? Amir Ghadermarzi 2014AbstractIn this dissertation, we are mainly interested in effective methods to solveparametrized Thue equations. After briefly talking about the different effec-tive methods, two parametrized families of cubic Thue equations are com-pletely solved by using Pade? approximation and linear forms in logarithms.The Thue inequality ??x3 + pxy2 + qy3?? ? k,is studied by using Bombieri?s method. We find all solutions under someconditions on k, p and q. As an application of Thue equations, we find theintegral points on the Mordell curves Y 2 = X3 + k for all nonzero integersk with |k| ? 107. Our approach uses a classical connection between theseequations and cubic Thue equations.iiPrefaceChapter 4 is a joint work with Prof. Bennett, under the title ?Mordellequations: a classic approach?, submitted for publication .iiiTable of ContentsAbstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iiiTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . ivList of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viiAcknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . viii1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1 Effective methods . . . . . . . . . . . . . . . . . . . . . . . . 62.1.1 Linear forms in logarithms . . . . . . . . . . . . . . . 62.1.2 Pade? approximation method . . . . . . . . . . . . . . 152.1.3 Bombieri?s method . . . . . . . . . . . . . . . . . . . 182.2 Parametrized families of Thue equations . . . . . . . . . . . 213 Families of cubic Thue equations . . . . . . . . . . . . . . . . 273.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Solutions of type I,II? . . . . . . . . . . . . . . . . . . . . . . 293.2.1 Gap principle . . . . . . . . . . . . . . . . . . . . . . 293.2.2 Proof of lemmas 3.3, 3.4 . . . . . . . . . . . . . . . . 343.2.3 Applying the main lemma . . . . . . . . . . . . . . . 393.2.4 Solution of type I . . . . . . . . . . . . . . . . . . . . 393.2.5 Solutions of type II? . . . . . . . . . . . . . . . . . . . 433.3 Solutions of type II,I? . . . . . . . . . . . . . . . . . . . . . . 463.3.1 Solution of type II . . . . . . . . . . . . . . . . . . . . 463.3.2 Solutions of type I? . . . . . . . . . . . . . . . . . . . 553.4 Solutions of type III,III? . . . . . . . . . . . . . . . . . . . . . 68ivTable of Contents3.4.1 Solutions of type III . . . . . . . . . . . . . . . . . . . 683.4.2 Solutions of type III? . . . . . . . . . . . . . . . . . . 723.5 Small values of t . . . . . . . . . . . . . . . . . . . . . . . . . 763.5.1 Application of diophantine approximation . . . . . . 763.5.2 Solutions of type I . . . . . . . . . . . . . . . . . . . . 763.5.3 Solutions of type II? . . . . . . . . . . . . . . . . . . . 803.5.4 Solutions of type II . . . . . . . . . . . . . . . . . . . 853.5.5 Solution of type I? . . . . . . . . . . . . . . . . . . . . 853.5.6 Solutions of type III . . . . . . . . . . . . . . . . . . . 853.5.7 Solutions of type III? . . . . . . . . . . . . . . . . . . 864 Mordell curves . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 874.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 894.3 Finding representative forms . . . . . . . . . . . . . . . . . . 934.3.1 Forms of positive discriminant . . . . . . . . . . . . . 944.3.2 Forms of negative discriminant . . . . . . . . . . . . . 954.3.3 Reducible forms . . . . . . . . . . . . . . . . . . . . . 974.4 Running the algorithm . . . . . . . . . . . . . . . . . . . . . 984.5 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . 994.5.1 Number of solutions . . . . . . . . . . . . . . . . . . . 994.5.2 Number of solutions by rank . . . . . . . . . . . . . . 1014.5.3 Hall?s conjecture and large integral points . . . . . . . 1025 Bombieri method . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 1045.2 Preliminiaries . . . . . . . . . . . . . . . . . . . . . . . . . . 1055.3 The Thue-Siegel inequality . . . . . . . . . . . . . . . . . . . 1115.4 Dyson?s lemma, proof of Thue-Siegel?s principle . . . . . . . 1155.4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 1155.4.2 Dyson?s lemma . . . . . . . . . . . . . . . . . . . . . . 1195.4.3 The auxiliary polynomials . . . . . . . . . . . . . . . 1225.5 The Thue inequality??x3 + pxy2 + qy3?? < k . . . . . . . . . . 1285.5.1 Irrationality measure . . . . . . . . . . . . . . . . . . 1305.5.2 Proof of theorems 5.7, 5.8 . . . . . . . . . . . . . . . . 1345.6 Proof of theorem 5.9 . . . . . . . . . . . . . . . . . . . . . . . 1395.6.1 Continued fractions and Legendre?s theorem . . . . . 1395.6.2 Applying Legendre?s theorm for small values of y . . 144Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146vTable of ContentsAppendicesA Non-vanishing of ?r,g . . . . . . . . . . . . . . . . . . . . . . . . 153viList of Tables4.1 Number of Mordell curves with Nk integral points for positivevalues of k with 0 < k ? 107 . . . . . . . . . . . . . . . . . . . 1004.2 Number of Mordell curves with Nk integral points for negativevalues of k with |k| ? 107 . . . . . . . . . . . . . . . . . . . . 1004.3 Number of Mordell curves with Nk integral points for positivek ? 107 6-th power free . . . . . . . . . . . . . . . . . . . . . 1014.4 Number of Mordell curves with Nk integral points for negativek, |k| ? 107 6-th power free . . . . . . . . . . . . . . . . . . . 1014.5 Hall?s conjecture extrema for |k| ? 107 . . . . . . . . . . . . . 1024.6 Solutions to (4.1.1) with X > 1012 for |k| ? 107 . . . . . . . . 103viiAcknowledgementsFirst and foremost, I would like to express my heartfelt gratitude to my PhDsupervisor, Professor Michael Bennett, for his support, encouragement andpatience. I will never be able to thank him enough for the support and helpoffered during my program. Without his help this thesis would not exist.I would like to thank the members of my supervisory committee, Pro-fessor Vinayak Vatsal and Professor Greg Martin, for their valuable time,and the courses they offered. Also, I want to thank Professor Stephen Choifor providing me with the original copies of [84] and [15]. I am grateful tomy friends Erick Wong, Vasu Tewari and Carmen Bruni for helping me withcollecting and generating the data used in the chapters 4 and 5. Last, butnot the least, I would like to thank my family to whom I owe where I amright now.viiiChapter 1Introduction1.1 IntroductionIn 1909, Axel Thue [81] had proved the following theorem:Theorem 1.1. (Thue) Let F (x, y) ? Z[x, y] be an irreducible homogeneouspolynomial of degree d ? 3 and m a nonzero integer, then the equationF (x, y) = m, (1.1.1)has only finitely many solutions.An equation of the form (1.1.1) is named a Thue equation in honourof Axel Thue. Thue?s proof of the finiteness of the number of solutions isbased on an improvement of Liouville?s theorem of rational approximationto algebraic numbers. To understand Thue?s proof, consider this simpleobservation. LetF (x, y) = a0xd + a1xd?1y + ? ? ?+ adyd = mbe a Thue equation of degree d and ?1, ? ? ??n the roots of f(x, 1) = 0 in asplitting field . Then we havea0(x? ?1y) ? ? ? (x? ?dy) = m????xy? ?1???? ? ? ?????xy? ?d???? =????myda0???? .Suppose (x, y) is a solution for f(x, y) = m, and j is an index, such that????xy? ?j???? = min1?i?d????xy? ?i???? .We call (x, y) a solution of type j, and for this solution we have????xy? ?i???? ?12mini 6=j|?i ? ?j | .11.1. IntroductionTherefore ????xy? ?j???? ?c|y|d,for some constant c. Now if there exists a constant C(?), such that????xy? ?j???? ?C(?)|y|?for some ? < d,for all x ? Z and y ? N, then we can find a bound on the y value of solutionsof type j and therefore the Thue equation has finitely many solutions. Thefirst result of this type is due to Thue; it is a general sharpening of Liouville?stheorem.Theorem 1.2 (Thue [81]). If ? is an algebraic number of degree at least 3and ? ? 0, then there is a constant C(?, ?), such that????? ?pq???? >C(?, ?)qd2+1+?, (1.1.2)for all p ? Z and q ? N.Definition 1.1. Let ? be an irrational number. We call ? an irrationalitymeasure for ?, if for every > 0, there exists a constant c(?, ), such that??????pq???? > c(?, )q??for all integers p, q ? 1. If c(?, ) is effectively computable, then ? is aneffective measure.An obvious upper bound for the irrationality measure is obtained by Li-ouville, who showed that every algebraic number of degree n has an effectiveirrationality measure n. According to Theorem 1.1.1, every algebraic num-ber of degree n has irrationality measure n2 +1. Siegel [72] improved the expo-nent to 2?n. Gel?fond and Dyson [33],[25] independently obtained a betterbound,?2n, and finally Roth proved a sharp bound of 2 for the irrationalitymeasure of all algebraic numbers. Unfortunately, none of these measures areeffective in the sense that they don?t supply an upper bound C = C(F,m)such that max(|x| , |y|) < C. However, we can obtain some upper boundson the number of solutions. Siegel proved the first important result in thisdirection, for the case d = 3 and binomial case F (x, y) = axd? byd = m. Heasked whether it is possible to find a bound that depends on d (the degree21.1. Introductionof F ) and m but is independent of F . In 1983, J.H. Evertse answered thisquestion. Let Nd,m be the number of solutions for equation (1.1.1). Evertse[29] obtained the boundNd,m < 715((d3)+1)2+ 6 ? 72(d3)(?(m)+1)for the number of primitive solutions of 1.1.1 , where ?(m) is the numberof distinct prime factors of m. Bombieri and Schmidt proved the followingresultTheorem 1.3. If F is an irreducible binary form of degreed d , the numberof solutions toF (x, y) = 1is bound above by c0d, where c0 is an absolute constant, and it has at most215d solutions if d is large enough [13].Remark 1.1. Consider the Thue equationF (x, y) = xd + c(x? y)(x? 2y)(x? dy) = 1.It has at least d solutions (1, 1) ? ? ? (1, d). Thus the upper bound c1d is thebest possible, except for the determination of c1 [13].Bombieri and Schmidt[13] showed that if Nd is an upper bound for num-ber of solutions of F (x, y) = 1, then Nd,m < dvNd , where v is the numberof distinct prime factors of m. Therefore the number of primitive solutionsfor F (x, y) = m does not exceedC1d1+v.Later, some improvements were made on the upper bound for the numberof solutions for Thue equations and Thue inequalities. We will mention twoof them.Theorem 1.4. (Stewart [76]). Let F be an irreducible binary form of degreed, and m a given integer . The number of solutions of the Thue inequality|F (x, y)| ? mis at mostdm2/d(1 + logm1/d).31.1. IntroductionTheorem 1.5. (Akhtari [1]). The Thue equation|F (x, y)| = 1has at most 11d? 2 solutions provided that disc(F ) is large in terms of d.In case of cubic, which is the focus of our interest, better bounds aregiven by several authors . The first result is due Siegel. He proved thatthe number of solutions for F (x, y) = 1 is at most 18, provided that thediscriminant is sufficiently large . Later on, Gel?man [22] showed that 18can be replaced by 10. Nagell [66] and Delone [23] showed that if F (x, y) isan irreducible binary form with negative discriminant, then F (x, y) = 1 hasat most 5 solutions . This bound is sharp, since the Thue equationx3 ? xy2 + y3 = 1has discriminant -23 and 5 solutions (1, 0), (0, 1), (?1, 1), (1, 1), (4,?3). Cu-bic Thue equations with positive discriminant have been treated in severalpapers . Evertse [28] proved that the number of solutions of the cubic Thueequation F (x, y) = 1 with positive discriminant (not necessary large) is atmost 12. Bennett [9] improved the result to 10. They used Pade? approxi-mation to find a gap between solutions of certain type. This result is almostsharp, since the cubic formF (x, y) = x3 + x2y ? 2xy2 ? y3 = 1has 9 solutions.(1, 0), (0, 1), (?1, 1), (?1,?1), (2, 1), (?1,?2), (5,?4), (4, 9), (?9,?5).But presumably, the ?truth? of matter is the following conjecture by Petho?[69].Conjecture 1. If F is a binary cubic form with DF > 0, then NF = 9 ifDF = 49, NF = 6 if DF ? {81, 229, 257, 361} and Nf ? 5, otherwise .Finally, Okazaki [67] proved that when discriminant of F is greater than5 ? 1065, the Thue equation F (x, y) = 1 has at most 7 solutions. Neverthe-less, all of these results are ineffective in the sense that they don?t providean algorithm to compute all the solutions of (1.1.1) , or give an upper boundto the height of solutions. In this thesis we are mainly interested in effectivemethods of solving families of Thue equations . In Chapter 2, we introduce41.1. Introductionthe basic definitions and preliminaries of effective methods for solving fami-lies of Thue equations and a survey on the previous results and families thathave been solved will be given. In Chapter 3, two families of cubic Thueequations will be completely solved using the methods introduced in Chap-ter 2. As far as we know, the only algorithmic method to find integer pointson elliptic curves is to reduce them to finitely many Thue equations. InChapter 4, we will follow[22] ,[65] to give an algorithmic approach to findingall solutions for small values of k up to a bound for the Mordell?s equationsy2 = x3 + k . We solve the Mordell?s equation with 0 < |k| < 107. Thenumerical result in Chapter 4 includes an estimation for Hall?s conjecture,distribution of number of solutions and also curves with large integer points. In Chapter 5, we will closely follow [14] , to explain elements of Bombieri?smethod for finding the irrationality measure of algebraic numbers of degree3. We will explicitly find the constants in section 9 of [14] to solve inequality??x3 + pxy2 + qy3?? ? k. (1.1.3)5Chapter 2Background2.1 Effective methodsThere are three methods to improve Liouvill?s bound effectively or solve aThue equation . The first method is the most general and is based on Baker?stheory of linear forms in logarithms of algebraic numbers [4]. The secondmethod is Pade? approximation method. It is originally due Thue [82] . Bakerwas the first to use this method to obtain an effective irrationality measureof some algebraic numbers [2]. This method is only applicable to certaintypes of equations . The third method has been developed by Bombieri [11]and Bombieri and Mueller [12] . They combined elements of the noneffectivemethods of Thue and Siegel with an improvement of Dyson?s lemma [25].In this chapter we will discuss the basic elements of these effective methods.2.1.1 Linear forms in logarithmsIn 1968, Baker [3] has shown the first general improvement of Liouville?sbound for approximating algebraic numbers . He succeeded in showing aneffective upper bound for the solutions of any Thue equations.Theorem 2.1 (Baker[3]). Let k > d+ 1 and (x, y) be an integer solution of(1.1.1), thenmax {|x| , |y|} < Celogk|m|,where C is an effectively computable constant depending only on n, k andthe coefficients of F (x, y).The bounds that we find using Baker?s theorem are large, but later re-finements have led to the construction of algorithms for solving the Thueequation [42],[85]. Using the work of Baker, a theoretical algorithm could begiven to find the solutions. Later on, some equations had been solved by ap-plying Baker?s method in an efficient way. In 1989, Tzanakis and de Wegerproduced, for the first time, a general practical algorithm to solve any Thueequation. In this thesis we are mainly interested in solving parametrized62.1. Effective methodsThue equations. Since the techniques of solving a parametrized Thue equa-tion are similar to those of solving a single Thue equation, we first give anoverview of the latter (see [85],[75]). Before explaining the algorithm wewill briefly talk about the lower bound for linear forms in logarithms. By alinear form in logarithms, we mean a linear form? = b1 log ?1 + ? ? ?+ bn log ?nwhere ?1, ?2, ? ? ? , ?n are algebraic numbers, not zero or one, and bi?s belongto Z. Baker proved that if ? 6= 0 then |?| cannot get arbitrarily close tozero.Lower bound of Linear form in logarithmsIn this section we briefly give some results on lower bounds for linear formsin logarithms.Definition 2.1. For an algebraic number ? with minimal polynomiald?i=0aixiand conjugates ?1, ? ? ? , ?d we define the Weil height of ? byh(?) =1dlog(add?i=1max (1, |?i|)).The first classic result that can be used in the general case is:Theorem 2.2 (Baker-Wu?stholz [5]). Let ?1, ? ? ? , ?n be algebraic numbers,not 0 or 1. Let d = [Q(?1, ? ? ? , ?n) : Q] and set:? = b1 log ?1 + ? ? ?+ bn log ?n 6= 0, bi ? Z.If B ? max{|b1|, ? ? ? , |bn|} thenlog|?| ? ?c(n, d)h1 ? ? ?hn logB,wherec(n, d) = 18(n+ 1)!nn+1(32d)n+2 log(2nd).One of the best results on forms in three logarithms is the following:72.1. Effective methodsTheorem 2.3 (Matveev [55]). Let ?1, ?2, ?3 be logarithms of nonzero alge-braic numbers linearly independent over Q. Let b1, b2, b3 be rational integerswith b1 6= 0 set:? = b1?1 + b2?2 + b3?3.Define ?j = e?j for j = 1, 2, 3 and set D = [Q(?1, ?2, ?3) : Q] , ? =[R(?1, ?2, ?3) : R]. Let A1, A2, A3 to be positive real numbers which sat-isfyAj ? max {Dh(?j), |?j | , 0.16} (1 ? j ? 3) .Assume thatB ? max{1, |b1|A1A3, |b2|A2A3, |b3|A3A3}.Define alsoC1 =5 ? 1656?e3 (7 + 2?)(3e2)? (20.2 + log(35.5D2 log(eD))).Thenlog |?| > ?C1D2A1A2A3 log (1.5eDB log(eD)) .Sometimes if we are lucky enough, we can reduce the number of loga-rithms to two, and get much better bounds. As an example we can mentiona result by Laurent, Mignotte and Nesterenko:Theorem 2.4 (Laurent-Mignotte-Nesterenko; [50]). Let ?1, ?2 be multiplica-tively independent positive algebraic numbers, set:? = b2 log ?2 ? b1 log ?1, b1, b2 ? N.Let D = [Q(?1, ?2) : Q] and for i = 1, 2 define:hi ? max{h(?i),log ?iD,1D}.Also:b? ?b1Dh2+b2Dh1.If log|?| 6= 0 thenlog|?| ? (?24.34)h1h2D4(max{log b? + 0.14,21D,12})2.82.1. Effective methodsMignotte [58] in an unpublished paper provides a refined method tofind lower bounds for forms in three logarithms. In this method, he firstfinds a lower bound for linear form in all cases except the cases that hecalled ?degenerate?, which means there exists a linear relation between thecoefficients. This makes them related by coefficients that are not too large,whereby we can reduce our form to two logarithms.Lemma 2.1. Assume ?1, ?2, ?3 ? (1,+?) are multiplicatively independentreal numbers. SetD =[Q(?1, ?2, ?3) : Q][R(?1, ?2, ?3) : R]and consider the linear form:? = b2 log?2 ? b1 log?1 ? b3 log?3,where b1, b2, b3 ? N are co-prime. Putd1 = gcd(b1, b2), d3 = gcd(b3, b2), b2 = d1b?2 = d3b??2.Let a1, a2 and a3 be real numbers such that:ai ? max {4, 5.296`i ? log |?i|+ 2Dh(?i)} , `i = |log?i|a1a2a3 ? 100.Put? = a1a2a3b? =(b?1a2+b?2a? 1)(b??3a2+b??2a3)logB = max{0.882 + log b?,10D},then eitherlog |?| > (?790.95)?D2 logB> (?307187)D5 log2B3?i=1max{0.55, hi,`iD}or one of the following two holds:? there are nonzero integers r0, s0 such that r0b2 = s0b1 with:|r0| ? 5.61a23?D logB|s0| ? 5.61a13?D logB.92.1. Effective methods? there are r1, s1, t1, t2 ? Z such that:r1s1 6= 0(t1b1 + r1b3) s1 = r1b2t2,gcd(r1, t1) = gcd(s2, t2) = 1,which also satisfy|r1s1| ? (5.61)?a33?D logB,|s1t1| ? (5.61)?a13?D logB,|r1t2| ? (5.61)?a23?D logB,where ? = gcd(r1, s1). Moreover, when t1 = 0 we can take r1 = 1 andwhen t2 = 0 we can take s1 = 1.AlgorithmIn this subsection we will explain the algorithmic approach to solving a singleThue equation first introduced by Tzanakis and de Weger [85]. Following[75], AssumeF (x, y) = m (2.1.1)is a Thue equation of degree n ? 3. First, we consider a simple case in whichF (X, 1) has no real roots.Lemma 2.2. Suppose F (x, 1) has no real roots , then any solutions (X,Y )of equation (2.1.1) satisfy|Y | ?|m|min1?i?n|Img(?i)|.Proof. Suppose (X,Y ) is a solution and ?i is chosen so that |X ? ?iY | isless than m , then we have|Img(?i)Y | ? |X ? ?iY | ? |m| ,and the result follows.Therefore we assume F (X, 1) has s real roots (s ? 1 ) and t pairs ofcomplex conjugate roots, so n = s + 2t. Let ? be a root of F (X, 1) andK = Q(?) . We order the roots in the standard way.?i ? R if 1 ? i ? s,?i = ?i+t if s+ 1 ? i ? s+ t.102.1. Effective methodsWe set?i = X ? y?iFor any solution (X,Y ) of (2.1.1), let i0 be the index such that????xy? ?j0???? = min1?i?n????xy? ?i???? .Or equivalently, |?i| = min1?i?n |?i| . From now on we will refer to thesesolutions as solutions of type i0. Let r = s + t ? 1 be the rank of units ofK = Q(?1). Let 1,1, ? ? ? , 1,r be a system of fundamental units of K. Denoteby i,1 ? ? ? , i,r the conjugates of 1,1, ? ? ? 1,r In Galois closure L of K, thereexist elements ?i in Q(?i) of norm m , and integers u1, ? ? ?ur such that?i = ?iu1i,1 ? ? ? uri,r.For a solution of type i, we take distinct indices labelled by i, j, k. The indicesj and k are arbitrary. We obtain three linear equations,X ? ?iY = ?i,X ? ?jY = ?j ,X ? ?kY = ?kUsing this, we can obtain?i(?j ? ?k) + ?j(?k ? ?i) + ?k(?i ? ?j) = 0.Dividing both sides by ?j(?k ? ?i) , we get?1?1 + ?2?2 + 1 = 0,where?1 =?i(?j ? ?k)?j(?k ? ?i), ?2 =?k(?i ? ?j)?j(?k ? ?i),?1 =l=r?l=1(i,lj,l)ul, ?2 =l=r?l=1(k,lj,l)ul.Suppose that |?1?1| is very small; then |?2?2| have to be very close to 1.Thus we have a small value for? = log(?2?2),112.1. Effective methodsby taking the principal value of the logarithm, we can write? = log(??2) +r?l=1ul log(k,lj,l)+ u02pi??1,for some u0 ? Z, we setU = maxi 6=0(|ui|). (2.1.2)If we can find a bound (hopefully small) on U then we can loop through allpossible ui?s and solve our problem. First we can see|u02pi| =?????arg(??2?2)? arg(??2)?r?i=1uiarg(k,lj,l)?????? pi (2 + rU) .Therefore for U > 2 , |u0| ?(r+1)U2 . Next we find the equation:|y| |?j ? ?i| ? |x? ?jy|+ |x? ?iy| ? 2 |x? ?jy| ,= 2 |?j | ,which implies:1|x? ?iy|?2|y| |?i ? ?j0 |.Now one can write:|x? ?iy| =?i 6=i1|x? ?jy|??i 6=i2|y| |?j ? ?i|?2n?1|y|n?1???i 6=j01|?j ? ?i|???c1|y|n?1< c1.Remark 2.1. The constant c1 that appears above is effective and only de-pends on Q(?1).Remark 2.2. Using Lagrange?s theorem for y > (2c1)1n?2 (which is usuallya small number) we find that xy is a convergent of ?j.122.1. Effective methodsBy settingc2 = maxl1 6=l2 6=l3 6=l1?????l2 ? ?l3?l3 ? ?l1???? ,we have|?1?1| ? c1c2 |?i| = c3 |?i| .Therefore, to make sure ?1?1 is small , we need to find a bound on ?i.Regarding this matter we recall the following lemma, which relates the sizeof a particular conjugate of a unit to the maximum exponent of a unit .Lemma 2.3. Let K be a number field with r fundamental units i ? K andlet ui ? Z for 1 ? i ? r. Define U = max |ui| and? =r?i=1uiiLet I = {i1, ? ? ? , ir} ? {1, ? ? ? , s + t} denote any set of r distinct indices;then, the following matrix is invertibleUI =???log(i1,1) . . . log(i1,r)......log(ir,1) . . . log(ir,r)???and there exists an index t ? I such that|log |?t|| = max1?l?r+1|log |?l||then|log |?t|| ? U/??U?1I??? ,where ?.?? denotes the infinity norm of a matrix [75].Set ?i as ?i/?i , and c4 to be the maximum of????U?1I????? over all possiblesubsets I . Applying this lemma, there exists an index t such that|log |?t|| ? c4U.Then choose a constant c5 to be a positive real number with c5 < c4/(n?1).There are 3 different cases132.1. Effective methods? Case I: |?i| > e?c5U and |?t| ? ec4U . We have inequalities|?t| < |m|?l 6=t|?l|?1 < |m| |?i|?(n?1) < |m| e(n?1)c5U .Thus we obtain the inequalityec4U ? |?t| = |?i| / |?i| ? c6 |m| e(n?1)c5U ,where c6 = max1?l?n |?l|?1 . Taking the logarithm from both sides,we haveU ?log(c6 |m|)c4 ? (n? 1)c5= U1.? Case II: |?i| > e?c5U and |?t| ? e?c4U . We have inequalitiese?c4U ? |?t| = |?i| / |?i| ? c7 |m| e?c5U ,where c7 = min1?l?n |?l|?1 . Taking the logarithm from both sides, wehaveU ?log(c7)c4 ? c5= U2.? Case III |?i| ? e?c5U . Usually, since ?i is chosen to be the smallestconjugate of ?, it can be very small . In practice we normally dealwith this case. We have???e? ? 1??? = |?1?1| ? c3e?c5U .For any complex number z if z ? 1 ? 12 , then |log z| ? 2 |z ? 1| .Therefore we have eitherU ?log(2c3)c5= U3,or ???e? ? 1??? = |?1?1| ? 1/2.Which means that? ? 2c3e?c5U . (2.1.3)We can use theory of linear forms in logarithms as in section 2.1.1, tofind a lower bound for ? and determine a constant c8 such that forU ? 3142.1. Effective methodslog |?| > ?c8 log((r + 1)U/2).Comparing this lower bound with the upper bound (2.1.3), we haveU <c8 logU + log(2c3) + c8 log ((r + 1)/2)c5.By the lemma of Petho? and de Weger [85] we can obtain a betterboundU ? U4 =2c5(log(2c3) + c8 log ((r + 1)/2) + c8 log(c8c5)).The bound U4 could be rather large. We can use techniques such asthe LLL algorithm [75] to reduce it to a more manageable size, suchas U5. Therefore we haveU ? max (3, U1, U2, U3, U5) .As we had shown, all of these constants are effectively computable.Therefore one can obtain an effective bound on U and so on Y , tosolve the Thue equation.2.1.2 Pade? approximation methodUnlike Baker?s method, when using Pade? approximation we don?t need toknow features of extension, fundamental units or even the roots. And whenPade? approximation is applicable (which is not always the case) it typicallygives better results than linear forms in logarithms.In order to know what we mean by Pade? approximation, we provide thiselementary observation: Given a power seriesf(x) =??i=0rixi, ri ? Q,for any fixed n there exist nonzero polynomials Pn(x), Qn(x) ? Q[x], ofdegree at most n, such thatPn(x)?Qn(x)f(x) = cnx2n+1 + ? ? ? .We call the pairs (Pn(x), Qn(x)) Pade? approximations to f(x). By the aboveargument, the Pade? approximation to any such function always exists. Butin practice, it is necessary to know some features of the polynomials suchas the size of denominators of the coefficients and bounds on the heights ofpolynomials Pn and Qn.152.1. Effective methodsAn Example by Thue and SiegelThe idea of using Pade? approximation for approximating irrational numbersis originally due to Thue and Siegel. Thue was interested in solving:??(a+ 1)x3 ? ay3?? = 1,which can be rewritten as:????(1 +1a)x3 ? y3???? =1|a|,which in turn is equivalent to:?????3?1 +1a?yx?????=1|a||x|g(x, y),where g(x, y) is a degree two polynomial in x, y and 3?1 + 1a . Thus one seesthat solving the equation is intimately related to the question of approxi-mating irrational numbers. More general, Pade? approximations tof(x) = d?1 + xare given by:Pn(x) = F(?1d? n,?n,?2n;?x),Qn(x) = F(1d? n,?n,?2n;?x),where F is a hypergeometric polynomial. Plugging in 1a for x and clearingdenominators, and multiplying by an we obtainpn ? qnd?1 +1a= c?n(1a)n+1+ ? ? ? pn, qn ? Z.It means we can find a sequence of integers pn and qn such that:pnqn?? d?1 +1a.Using Pade? approximation we have the following lemma.162.1. Effective methodsLemma 2.4. Let ? ? R. Suppose that there are real numbers k0, `0 > 0 andE,Q > 1 such that for all n ? N there exist rational integers pn and qn with|qn| < k0Qn and |qn? ? pn| ? `0E?n satisfying pnqn+1 6= pn+1qn. Then forp, q ? Z with |q| ? 12`0 , we have:????? ?pq???? >1c |q|?+1,where:c = 2k0Q (2`0E)? and ? =logQlogE.Remark 2.3. Although the sequences pn, qn obtained by Pade? approximationconverge to ?, they are not as strong as principle convergents. However,using lemma 2.4 we still get an effective rational approximation.Remark 2.4. From Lemma 2.4 it is clear that the smaller the constant Q,the better the result. Finding better bounds on the common denominatorimproves the irrationality measure.As an example in our first equation, the Pade? approximation for 3?1 + xhas the following properties? |Pn(z)| < 4n, |Qn(z)| < 4n.? |En(z)| < 4n(1? |z|)? 12 (2n+1) .? If z 6= 0 then Pn(z)Qn(z) 6= Pn+1(z)Qn+1(z).Using the same method with better approximations, Baker proved anirrationality measure of 2.955 for irrationality measure of 3?2 with effectiveconstant 10?6 (see [2]). Subsequently he used this to solve Thue equationx3 ? 2y3 = n. Using better bounds on denominators Chudnovsky improvedthe measure to 2.42971 (see [16]). Bennett [8] proved an irrationality mea-sure of 2.5 and constant 14 . The following proposition, which is a versionof Thue?s ?Fundamental theorem?, is very important in order to give Pade?approximation for more general Thue equations rather than binomial onesProposition 2.1. Let P ? Q[x] be a polynomial of degree n ? 2, and U aquadratic polynomial in Q[x] with disc(U) 6= 0 such thatUP ?? ? (n? 1)U ?P ? +n(n? 1)2U ??P = 0172.1. Effective methodsholds. Let Y1 = 2UP ? ? nU ?P and ? =disc(U)4 , and define the followingpolynomials :a =n2 ? 16(??U ? + 2?), c =n2 ? 16(??(U ?X ? 2U) + 2?X),b =n2 ? 16(??U ? ? 2?), d =n2 ? 16(??(U ?X ? 2U)? 2?X),z =12(Y12n??+ P), u = z ? P, w =zu.Then for r ? N there are rational polynomials Ar, Br ? Q[X] given by(??)rAr = a??n,r(z, u)? b??n,r(u, z),(??)rBr = c??n,r(z, u)? d??n,r(u, z),such that for any root ? of P the polynomialCr = ?Ar ?Bris divisible by (X ? ?)2r+1 [52] .2.1.3 Bombieri?s methodThe proof of the finiteness theorem of Thue equation is based on the idea thatif there is a rational number which is an exceptionally good approximationto a given real number ? then all other approximations cannot be too closefrom some point onward. In essence, Thue has proven the following:Theorem 2.5. (Thue) Let ? be an algebraic number of degree r and leth, k be given positive numbers. There is an effectively computable constantG0 = G(?, h, k) such that if there exist p0, q0 with??????p0q0???? < q?r/2?1?k0 , q0 > G0then we can effectively determine a G = G(?, h, k.q0) such that??????pq???? > q? r2(k+1)?1?hfor all q > G.182.1. Effective methodsThe problem is the existence of such a good approximation. The constantG0 in the Thue theorem is far too large and thus no pair (?, p0/q0) has beenfound to satisfy the theorem?s hypothesis [11]. By using a significant resultobtained by Dyson, Bombieri [25] succeeded to remove the requirement thatq0 should be large. The main idea is to construct auxiliary polynomialP (x, y) which vanishes at the point (?1, ?2) to high order, P vanishes at(p0/q0, p/q) only to a low order and h(p) is not too large. One way toensure that P vanishes at (p0, q0) to a low order is to make the unpleasantrestriction that q0 is sufficiently large . An alternative way is to use Dyson?slemma to find vanishing order of (p0/q0, p/q) by using information on thedegree of P and the vanishing degree at (?1, ?2).To express Dyson?s lemma, we first define a notation [11]. Let K bean algebraically closed field and ?? = (??1, ? ? ? ??n) be m points in Kn. Weassume that for i = 1, 2 ? ? ?n, the m numbers ?1i, ?2i, ? ? ? ?mi are distinct. Inthis case the set of points ?? will be called admissible. Let vi > 0 i = 1, ? ? ?nbe real numbers, and d1, ? ? ? dn be positive numbers, and let t? ? = 1, ? ? ?mbe real numbers. We define?(d, v; t1, ? ? ? , tm|?1, ? ? ? , ?m),and abbreviate ?(d; t?) to be the vector space of all polynomials P =P (x1, ? ? ?xn) in n variables, satisfyingdegxiP ? di,for i = 1, .., n?IP (??) = 0,for?I =?i1+...in?xi11 ...?xinnand all indices I = (i1, ? ? ? ?n) withv1i1d1+ ? ? ? vnindn< t?,we also define?n(t) =? 10? ? ?? 10v1x1+???+vnxn?tdx1 ? ? ? dxn.According to the above notation, following is Bombieri?s version of Dyson?slemma192.1. Effective methodsLemma 2.5. Let n = 2 and ?1, ? ? ? ?m be admissible then if ?(d; t?) 6= 0, wehave???2(t?) ? 1 + max(m? 22, 0)d2d1.Bombieri [11] and Bombieri and Mueller [12] describe the auxiliary poly-nomials in a quantitative fashion. As an example, assume k(?1) = k(?2) =K and they have degree r over k. As well, t and v satisfy0 < t <?2rt ? v ? t?1.Then there exists a polynomial p ? k[x1, x2] withdegxip ? di d2 ? ?d1and?i1+i2?xi11 ?xi22p(?1, ?2) = 0for all (i1, i2)withv?1i1d1+ vi2d2< t.Now let us define? =?2? rt2 + (r ? 1)?.According to Dyson?s lemma, there are (i?1, i?2) with?i?1+i?2?xi?11 ?xi?22p(?1, ?2) 6= 0andv?1i?1d1+ vi?2d2< ?,for the polynomial with these properties, using Siegel?s box principal lemma,one can show thatlog h(p) ? A1d1 +A2d2 + o(d1 + d2),where Ai = rt22?rt2(log h(?i) + 12).Remark 2.5. The sharp bound of A1 appears to be a mystery. For algebraicnumbers of the form ?1 = r??, Bombieri and Mueller [12] found a betterbound for A1 and therefore a better irrationality measure .202.2. Parametrized families of Thue equationsThe last step is to find a Thue Siegel type principle using the informationwe have about p. Let? =1i?1!i?2!?i?1+i?2?xi?11 ?xi?22p(?1, ?2).So we have, ? 6= 0 . We proceed to estimate log |?|w for all different valua-tions w of K over k in terms of h(p) and height of ?1 and ?2. Using the fact?w log |?| = 0, one can prove effective versions of Thue-Siegel?s principle.As an example we can mention this result [12] :Theorem 2.6. Let K = Q(?1) = Q(?2) be an algebraic number field ofdegree n ? 3, and let ?, t and ? be positive real numbers satisfying the con-ditions: 0 <?2? nt2 < ? < t <?2n ? min (?, ??1). Then there existeffective constants c, c0 such that one of the following inequalities hold:?????1 ?p1q1???? ? (cH(?1))? 4n?(2?nt2)(t??) (|p1|+ q1)? 2?(t??) ,?????2 ?p2q2???? ? (cH(?2))? 4n?(2?nt2)(t??) (|p2|+ q2)? 2?(t??) ,(cH(?1)c0H(?2)) 2n(2?nt2) > q2 (|p1|+ q1)? nnt2+?2?2 .2.2 Parametrized families of Thue equationsDefinition 2.2. A family of Thue equations :pn(t)xn + pn?1(t)xn?1y + ...+ p0(t)yn = p(t),where p and each pi are integral polynomials in the variable t is called aparametrized family of Thue equations.The goal is to find a set of solutions expressed in terms of the parametert, for which all the solutions of all the members of the family, or at least forvalues of t bigger than some bound, are expressed in terms of t. Thomas[80] formulates this as follows:Definition 2.3. The set ? : {(g1(t), h1(t)) , ..., (gs(t), hs(t))} is called a setof solutions if there exists an effectively computable positive integer T? suchthat for all t > T? , all the solutions of the family are given by the elementsof ?. A family is stably solvable if the family has a complete solution set.212.2. Parametrized families of Thue equationsNote that all families are not stable solvable. The study of parametrizedfamilies of Thue equations dates back to Thue [82]. He showed that theequation(t+ 1)xn ? tyn = 1has only the solutions x = y = 1 if t is large enough in relation to the primen ? 3 . Although the Baker method provides a general algorithm for solvinga single Thue equation, it was not until 1993 that parametrized families ofThue equations were considered by this method. Thomas [79], studied thefamilyx3 ? (t? 1)x2y ? (t+ 2)xy2 ? y3 = ?1 t ? Z.Thomas [79] and Mignotte [56] solved this equation completely. It has onlysolutions ?(1, 0),?(0, 1) ? (1,?1) for all t ? Z, as well as several extrasolutions for t = 0, 1, 3. Since then, several families have been considered bymultiple authors. We will mention some of the results. For a survey, referto [42] and [44].? Mignotte, Petho? and Lemmermeyer provided a complete solution forthe following family [59]??x3 ? (t? 1)x2y ? (t+ 2)xy2 ? y3?? ? 2t+ 1.? In [61] Mignotte and Tzanakis proved that the only integer solutionof the familyx3 ? tx2y ? (t+ 1)xy2 ? y3 = x(x+ y)(x? (t+ 1)y)? y3 = 1are: (1, 0), (0,?1), (1,?1)(?t ? 1,?1), (1,?t) when t ? 3.33 ? 1023.Later, Mignotte [57] proved the same result for all t ? 3.? Considering the family:x(x? tay)(x? tby)? y3 = 1Thomas [80] proved that if0 < a < b and t ?(2 ? 106 ? (a+ 2b)) 4.85b?athen there are no nontrivial solutions. In fact, this family is in a specialcase of families that Thomas termed split families; we will discuss themin more detail later.222.2. Parametrized families of Thue equations? Wakabayashi [87] proved that the only integer solutions of the family:x3 ? t2xy2 + y3 = 1are the trivial ones: (0, 1), (1, 0), (1, t2), (t, 1), (?t, 1), if t ? 1.35 ? 1014.? Togbe [83] proved that if t > 1 the only integral solutions ofx3+(t8 + 2t6 ? 3t5 + 3t4 ? 4t3 + 5t2 ? 3t+ 3)x2y++ (2? t3)t2xy2 ? y3 = ?1are (?1, 0), (0,?1).? The quartic familyx4 ? tx3y ? x2y2 + txy3 + y4 = ?1was solved by Petho? [70] for large values of t; Mignotte, Petho?, andRoth [60] solved it completely. The only solutions are:?{(0, 1), (1, 0), ), (1, 1), (1,?1), (a, 1), (1,?a)} for |t| /? {2, 4}? the equation x4 ? tx3y ? 3x2y2 + txy3 + y4 = ?1 has been solved fort ? 9.9 ? 1027 by Petho? [70].? x4?tx3y?6x2y2 +txy3 +y4 ? ?{1, 4} was completely solved by Lettleand Petho? [51].? Wakabayashi [86] proved that??x4 ? t2x2y2 + y4?? =??x2(x? t)(x+ t) + y4?? ? t2 ? 2has only trivial solutions with |y| ? 1 for t ? 8.?x(x2 ? y2)(x2 ? t2y2)? y5 = ?1For t > 3.6 ? 1019, all solutions have been founded by Heuberger [41].Wakabayashi [88] used a combination of the Pade? approximation methodand linear forms in logarithms.Given a parametrized family of Thue equations, one should try to performthe same steps as in the case of single Thue equation. To do so we needto express every quantity in the algorithm in terms of the parameter t. Aswe see in the case of a single Thue equation, we require information about232.2. Parametrized families of Thue equationsthe unit group of the corresponding number field. Hanrot [37] mentionsthat we don?t need to know the fundamental units; a suitable system ofindependent units is enough. In the case of cubic forms, fortunately, thereis a rich theory of finding fundamental units (see [78] and [10]). Moreover,to solve parametrized families of Thue equations, we need an extra toolthat is crucial for finding integral solutions: stable growth, developed byThomas [80] . The idea is to exclude the trivial solutions. The concept oftrivial solutions is vague. We call a solution trivial if the quantity U of theequation(2.1.2) is not small. In other words, the solutions that can be foundwith checking small values of U are called trivial. If we exclude them, wehave to prove a stable growth condition for the remaining possible solutionsof the family of parametrized equations .Definition 2.4. We say that a parametrized family of Thue equations hasstable growth if for each nontrivial solution we haveU > Ctg log t,for some positive constant C and g ? N.The lower bound for U hopefully contradicts the upper bound that wefind from previous estimations. As of now, we don?t know of any condi-tion that guarantees stable growth in the general case. For the cubic case,Thomas [80] proved the existence of stable growth for some families of splitcubic forms. In the next chapter, while studying two families of parametrizedcubic Thue equations with five solutions, we will discuss a couple of differentmethods to prove the stable growth condition.Using Pade? approximation method, some results on families of Thueequations were obtained. G.V. Chudnovsky [16] estimated more preciselythe common denominators of hypergeometric polynomials of 3?1 + x, trans-forming cubic Thue equations to diagonal forms. He proved the followingresultTheorem 2.7. Let a be an integer (positive or negative) with a ? ?3 (mod 9)and let ? be the real zero of f(x) = x3 + ax+ 1, whose absolute value is thesmallest among the three zeros of f . Then for any > 0, ? has effectiveirrationality measurelog((G1 +?G21 +D1)2/ |D1|)log((G1 +?G21 +D1)2/ |D1| ?) + ,242.2. Parametrized families of Thue equationswhereD1 = ?427a3 ? 1, G1 =227a6 +23a3 + 1 ? = e?3pi/6/?3[16].Wakabayashi [87] Using Pade? approximation, find all the solutions off(x) = x3 +axy2 + by3 < a+ b with the condition a ? 360b4 Using the Pade?approximation method, Lettle, Petho? and Voutier [52] completely solvedthis family of Thue equations:?F 3t (x, y) = x3 ? tx2y ? (t+ 3)xy2 ? y3,?F 4t (x, y) = x4 ? tx3y ? 6x2y2 + txy3 + y4,?F 6t (x, y) = x6 ? 2tx5y ? (5t+ 15)x4y2 ? 20x3y3+5tx2y4 + (2t+ 6)xy5 + y6.They used the proposition 2.1 with the quadratic polynomialU = x2 + x+ 1.These families possess special features, and are called the simple families ofThue equations. For these families, we have? There exists an A =(a bc d)? PGL2(Q) such that ? : z ? az+bcz+dpermutes the zeros of F (x, 1) transitively.? Let ?A permute the roots. Then there is some r ? Q? such thatFA = rF .? The roots of F (x, 1) generate a cyclic number field of degree deg(F ).The Galois action on the roots is given by ?A.Up to equivalence, the only simple forms in Q[x, y] are?F 3t with t ? Z.?F 4t with t ? Z \ {?3, 0, 3}.252.2. Parametrized families of Thue equations?F 6t with t ? Z \ {?8,?3, 0, 5}.Let F (x, y) be an irreducible binary cubic form with integral coefficients andpositive discriminant. Evertse (1983) proved that the number of solutions ofthe cubic Thue equation F (x, y) = 1 with positive discriminant is at most 12.Bennett (2000) improved the result to 10. They used Pade? approximationto find a gap between solutions of a certain type. In Chapter 3 we will makesome small refinements to the gap principal proved by Evertse under certainconditions. We will then use these refinements to find solutions of a certaintype to some families of parametric cubic Thue equations.According to Bombieri?s method explained in section 2.1.3, effective mea-sure of irrationality has been proved for some algebraic numbers.Theorem 2.8. Let d ? 40 and let ? be the positive root of xd ? axd?1 + 1, where a ? A(d) and A(d) is effectively computable . Then, ? has effectiveirrationality measure ? = 39.2574[11].Refining the method Bombieri and Mueller [12] provedTheorem 2.9. Let d ? 3 , and let a and b be coprime positive integers ,and let ? = log|a?b|log b . Let ? ? Q(d?a/b)be of degree d. If ? < 1? 2/d , thenfor any > 0 ? has effective irrationality measure? =21? ?+ 6(d5 log dlog b)1/3+ .26Chapter 3Families of cubic Thueequations3.1 IntroductionE. Thomas [80] introduced families of Thue equations of the formFt(x, y) =n?i=1(x? Piy)? yn = ?1, Pi ? Z[t],which he called split families. The trivial solutions for these families are{(?Pi, 1) , (?1, 0)}. Thomas conjectured that ifP1 = 0,0 < degP2 < ? ? ? < degPnAll Pi are monicthen these families have exactly four solutions for large values of pa-rameter t. He also proved his conjecture for n = 3 under some technicalhypothesis [80].Other authors (see [36], [41],[45] and [43]), proved the conjecture underdifferent conditions. In 2007 Ziegler [90] found two split families of cubicThue equations with at least 5 solutions for all values of t that disproveThomas?s conjecture. These two families areF (x, y) = x3 ? (t4 ? t)x2y + (t5 ? 2t2)xy2 + y3 = 1 (3.1.1)with solutions(1, 0), (0, 1), (t, 1), (t4 ? 2t, 1) and (1? t3, t8 ? 3t5 + 3t2) (3.1.2)andF ?(x, y) = x3 ? (t4 + 4t)x2 + (t5 + 3t2)xy2 + y3 = 1273.1. Introductionwith solutions(1, 0), (0, 1), (t, 1), (t4 + 3t, 1) and (t9 + 3t6 + 4t3 + 1, t8 + 3t5 + 3t2).In this chapter we will use the techniques from Chapter 2 to completelysolve these families. Basically we will prove that these families have exactly5 solutions for all values of t > 0.To the forms F (x, y) and F ?(x, y) we relatively associate polynomialsP (x) and P ?(x) which are defined byP (x) = F (x, 1) = x3 ? (t4 ? t)x2 + (t5 ? 2t2)x+ 1andP ?(x) = F ?(x, 1) = x3 ? (t4 + 4t)x2 + (t5 + 3t2)x+ 1.P (x) as a polynomial in x has three real roots; we denote them by ?, ??, ???and similarly P ?(x) as a polynomial in x has three real roots; we denote themby ?, ??, ???.By studying the sign of P (x) and p?(x) we can deduce the followingbounds for the roots:?1t5?2t8?4t11<? < ?1t5?2t8?3t11,t+1t5+3t8<?? < t+1t5+4t8,t4 ? 2t?2t8<??? < t4 ? 2t?1t8,(3.1.3)and?1t5+3t8?8t11<? < ?1t5+3t8?7t11,t+1t5?2t8<?? < t+1t5?1t8,t4 + 3t?1t8<??? < t4 + 3t?1t9.(3.1.4)It is easy to check that if |y| ? 1 then both F (x, y) = 1 and F ?(x, y) = 1have no nontrivial solutions. So we may assume |y| > 1. Let (x, y) be asolution to equation F (x, y) = 1 or F ?(x, y) = 1 with |y| > 1; then wecategorized the solutions based on the root that they are closer to. Therefore,for the solutions of F (x, y) = 1, by studying the sign we define these typeof solutions:283.2. Solutions of type I,II?solution of type I if: ?2t5+1t6<xy< ?1t6,solution of type II if : t+1t6<xy< t+2t5,solution of type III if: t4 ? 2t?2t8<xy< t4 ? 2t?1t9,(3.1.5)and similarly for the solutions of F ?(x, y) = 1 letsolution of type I? if ?2t5+2t8<xy< ?1t6,solution of type II? if t+1t6<xy< t+2t5?1t6,solution of type III? if t4 + 3t?1t8<xy< t4 + 3t?1t9.(3.1.6)In this chapter we assume t > 10. We will use Pade? approximationmethod to find all solutions of type I and II?. The remaining types of so-lutions are treated by Baker?s method. Using these methods we will provethat the forms have exactly 5 solutions for large and intermediate valuesof t. Finally for the small values of t we use diophantine techniques andcomputer search.3.2 Solutions of type I,II?In this section we use Pade? approximation method to prove a gap principlefor solutions of a cubic Thue equation. The main result is that there existsat most one ?suitably large? solution of a certain type for a cubic Thueequation. We will show that the unique ?large solutions? for F (x, y) = 1and F ?(x, y) = 1 are relatively (1 ? t3, t8 ? 3t5 + 3t2) and (t9 + 3t6 + 4t3 +1, t8 + 3t5 + 3t2) and there is no other solution for these types. During thecourse of the proof it will become clear what we mean by large solution.3.2.1 Gap principleThe key step to proving our gap principal is to reduce the problem to a diag-onal form over an imaginary quadratic field, since the Pade? approximationmethod is well suited to diagonal forms. It is a classic method originating293.2. Solutions of type I,II?in works of Eisenstein, Hermite, Arndt and Berwick; see Dickson [24] forreferences.preliminaries of the proof For a cubic form F , Define the associatedquadratic form, the Hessian H = HF and cubic form GF byH(x, y) = ?14(?2F?x2?2F?y2?(?2F?x?y)2)2,H(x, y) = Ax2 +Bxy + Cy2,H(x, y) = y2(A(xy)2 +B(xy) + C),GF (x, y) =?F?x?H?y??F?y?H?x.The forms HF and GF are related to F by the identity4H(x, y)3 = G(x, y)2 + 27DFF (x, y)2.Working in number field Q(???) and a fixed choice of square root, fromabove identityG(x, y)? 3???F (x, y)2are cubic forms in M [x, y] with coeeficents conjugate to one another andno common factors; they are cubes of linear forms ?(x, y) and ?(x, y) inM [x, y] with complex conjugate coefficients that (see Evertse [28])satisfythese equations:?(x, y)3 ? ?(x, y)3 = 3???F (x, y),?(x, y)3 + ?(x, y)3 = G(x, y),?(x, y)?(x, y) = H(x, y),?(x, y)?(1, 0)and?(x, y)?(0, 1)?M [x, y],For any rational pairs x0, y0, the numbers ?(x0, y0) and ?(x0, y0) arecomplex algebraic integers. The pair of forms ? and ? satisfying aboveproperties is called a pair of resolvent forms. We say that a pair of rationalintegers (x, y) is related to a pair of resolvent forms (?, ?) if????1??(x, y)?(x, y)???? = min0?k?2????e2kpii3 ??(x, y)?(x, y)???? .303.2. Solutions of type I,II?By this definition if (x, y) is related to the pair (?, ?) and ? ? 72000 thenfrom inequality (5.4) of [9] we have????1??(x, y)?(x, y)???? <1.012??|?(x, y)|3. (3.2.1)The next lemma by Bennett [9] shows that two different solutions relatedto the same set of resolvent can?t be very close to each other.Lemma 3.1. If (x1, y1) and (x2, y2) are distinct solutions related to (?, ?),with|?(x2, y2)| ? |?(x1, y1)| ?1?2?14and ? ? 72000, then|?(x2, y2)| > 0.987 |?(x1, y1)|2 .For our purpose we need a stronger gap between solutions. Under somestronger conditions we have the following lemma:Lemma 3.2. with the same hypothesis as above lemma if |x1y2 ? x2y1| ? 2then|?(x2, y2)| > 1.975 |?(x1, y1)|2 .Proof. By inequality (3.2.1) for ? > 72000,???1? ?(x,y)?(x,y)??? < 1.012??|?(x,y)|3Also frominequality ?2?1 ? ?1?2 = ????(x1y2 ? x2y1), using lemma (3.1), we have?? ?|?2?1 ? ?1?2|2?12|?1| |?2|(????1??1?1????+????1??2?2????).So2?? < |?1| |?2|(1.012??)(|?1|?3 + |?2|?3).It follows that|?1|?3 + |?2|?3 > 1.976 |?1|?1 |?2|?1=? |?1|3 + |?2|3 > 1.976 |?1|2 |?2|2 .Let |?2| = k |?1|2 then|?1|3 + k3 |?1|6 > 1.976k2 |?1|6 .313.2. Solutions of type I,II?Therefore(1.976k2 ? k3)|?1|3 < 1.Since|?| ?1?2?14 > 11.58,the above inequality holds for k > 1.975 So|?2| > 1.975 |?1|2as desired.Using the above notation we have these two lemmas:Lemma 3.3. Assume that the Thue equation F (x, y) = 1 with ?F > 72000has a solution (x1, y1) , such that 1 = (x1, y1) satisfy ?231< 122 . Let(x2, y2) be another solution related to the same resolvent form with 2 > 1then 2 > (2.4?)?15471 .Lemma 3.4. Main Lemma(Gap Principal) Assume that the Thue equa-tion F (x, y) = 1 with ?F > 72000 has a solution (x1, y1) , such that1 = (x1, y1) satisfy ?231< 122?2/45. Let (x2, y2) be another solution re-lated to the same resolvent form with 2 > 1 then |x2y1 ? x1y2| = 1.In the rest of this subsection we will prove these two lemmas.Auxiliary polynomials To prove the lemmas 3.3 and the main lemma3.4 we apply arguments due to Siegel [73] with refinements by Gel?man [22]and Evertse [28]. A hypergeometric function is a power series of the formF (?, ?, ?, z) = 1 +??n=1?(?+ 1) ? ? ? (?+ n? 1)?(? + 1) ? ? ? (? + n? 1)?(? + 1) ? ? ? (? + n? 1)n!zn.Following (Bennett [9] ,Evertse [28]) we defineAr,g(z) =r?m=0(r ? g + 13m)(2r ? g ?mr ? g)(?z)mandBr,g(z) =r?g?m=0(r ? 13m)(2r ? g ?mr ? g)(?z)m.323.2. Solutions of type I,II?For r ? N and g ? {0, 1} we haveAr,g =(2r ? gr)F(?13? r + g,?r,?2r + g, z),Br,g =(2r ? gr)F(13? r,?r + g,?2r + g, z),where F is the standard hypergeometric function. Define Cr,g to be thegreatest common divisor of the numerators of the coefficients of Ar,gWe recall a lemma by Evertse [28].Lemma 3.5. Let r, g be integers with r ? 1, g ? {0, 1} :(i) There exist a power series Fr,g(z) such that for all complex numbers zwith |z| < 1Ar,g(z)? (1? z)13Br,g(z) = z2r+1?gFr,g(z)(Where (1? z)13 =??m=0( 13m)(?z)m) and if |z| < 1|Fr,g(z)| ?(r?g+ 13r+1?g)(r? 13r)(2r+1?gr) (1? |z|)?12 (2r+1?g) .(ii) If |1? z| ? 1, then|Ar,g(z)| ?(2r ? gr).Let z1 = 1??3?3 then|z1| =3??|?1|3 < (10)?9. (3.2.2)Consider the complex sequences of ?r,g by?r,g =?2?2Ar,g(z1)??1?1Br,g(z1).Using inequality (3.2.2) and lemma (3.5) we conclude the following lemmaSee Bennett, [9]. It shows that the nonvanishing of ?r,g enables us to find arelation between the size of 2 and 1:333.2. Solutions of type I,II?Lemma 3.6. If ?r,g 6= 0 thenc1(r, g)?g3 |?1|3r+1?g |?2|?2 + c2(r, g)?r? g6 |?2| |?1|?3r?2(1?g) > 1 (3.2.3)where we may takec1(r, g) =1?r4r and c2(r, g) =1?r(2.252)rfor r ? 9 , and as follows , for 1 ? r ? 8.(r, g) c1(r, g) c2(r, g) (r, g) c1(r, g) c2(r, g)(1, 1) 2.6 1.3 (6, 1) 42.1 2.4(1, 0) 1.1 0.7 (6, 0) 66.8 2.7(2, 0) 6.1 2.4 (7, 1) 547.0 16.9(3, 0) 1.1 0.3 (7, 0) 39.5 0.9(4, 0) 14.2 1.8 (8, 1) 745.9 13.0(5, 0) 9.2 0.7 (8, 0) 236.9 3.1To use this key inequality we need to prove nonvanishing of ?r,g for somevalues of r, g:Lemma 3.7. ?r,g 6= 0 for (r, g) = (1, 1), and (r, 0) for 2 ? r ? 15.Proof. This is lemma 6.3 of [9] for (1, 1), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0); forremaining cases we mimic Bennett?s proof (see appendix A).Also by lemma (6.4)of [9] we haveLemma 3.8. If r ? N and h ? {0, 1}, then at least one of{?r,0,?r+h,1}is nonzero.3.2.2 Proof of lemmas 3.3, 3.4Proof of Lemma 3.3Proof. We prove a more general formula; if the hypothesis of lemma 3.3satisfies, then|?2| > (2.4?)?r |?|3r+2 ,343.2. Solutions of type I,II?for all 1 ? r ? 15.By lemma 3.7 , ?r,0 = 0 for all 1 ? r ? 15 we can apply lemma 3.2.3 withg=0 for these values of r. Since |x2y1 ? x1y2| > 1 from lemma 3.2, we have|?2| > 1.975 |?1|2and so since c1(1, 0) = 1.1 we obtainc1(1, 0)???41?? |?2|?2 < 1.1(11.975)2 < 0.29.Also since ?1,0 6= 0, we can apply lemma (3.6) to conclude thatc2(1, 0)??2 |?1|?5 > 0.71=? |?2| > 1.014??1 |?1|5 .Now we use nonvanishing of ?2,0; since c2(1, 0) = 6.1, we havec1(2, 0)???71?? |?2|?2 < 6.1(?2?3)(11.014)2< 6.1 ?122(11.014)2< 0.27.By lemma 3.2.3 since ?2,0 = 0 we havec2(2, 0)??2?2 |?1|?8 > 0.73=? |?2| > 0.3??2 |?1|8 .Arguing similarly for (r, 0), r ? 9 we have?2 > c3(r, g)??r |?|3r+2 ,Where c3(r, g) is(r, g) c3(r, g) (r, g) c3(r, g)(1, 0) 1.014 (6, 0) 0.37(2, 0) 0.3 (7, 0) 1.11(3, 0) 3.23 (8, 0) 0.322(4, 0) 0.555 (9, 0) 0.002(5, 0) 1.422353.2. Solutions of type I,II?Note that for all these values we have c3(r, g) > (2.4)?r. Now we useinduction to prove the lemma for 9 ? r ? 15. Suppose it is true for some9 ? r < 15, thenc1(r + 1, 0) |?1|3r+4 |?2|?2 <4r+1?r + 1(2.4?)2r |?1|?3r<4?r + 1(21.16)r(122)r< 0.9.Since ?r+1,0 6= 0, By lemma (3.6)|?2| >0.1c2(r + 1, 0)??r?1 |?1|3r+5=? |?2| > (2.4?)?r?1 |?1|3r+5 .So the result holds for r + 1 and it completes the proof.The next lemma shows that if the hypothesis of lemma 3.4 satisfies then|?2| is arbitrarily large in relation to |?1| .Lemma 3.9. Assume the Thue equation F (x, y) = 1 with ?F > 72000 hasa solution (x1, y1), such that ?1 = ?(x1, y1) satisfy ?2?31< 122?2/15and for anyother solutions (x2, y2) related to the same resolvent forms with ?2 > ?1 wehave |x2y1 ? x1y2| > 1; then|?2| > (2.3?)?r |?1|3r+2 ??16 , (3.2.4)for all r ? 1.Proof. We use the same steps as the proof of lemma 3.3 From lemma (3.2),since c1(1, 0) = 1.1 we obtainc1(1, 0)???41?? |?2|?2 < 1.1(11.975)2 < 0.29.Also since ?1,0 6= 0 , we can apply lemma (3.6) to conclude thatc2(1, 0)??2 |?1|?5 > 0.71=? |?2| > 1.014??1 |?1|5 .363.2. Solutions of type I,II?Now we use nonvanishing of ?2,0, and since c2(1, 0) = 6.1 we havec1(2, 0)???71?? |?2|?2 < 6.1(?2?3)(11.014)2< 6.1 ?122?2/15(11.014)2< 0.07.By lemma 3.2.3, since ?2,0 = 0, we havec2(2, 0)??2?2 |?1|?8 > 0.93=? |?2| > 0.38??2 |?1|8 .Arguing similarly for (r, 0), r ? 9, we have, for r ? 9|?2| > c3(r, g)??r |?1|3r+2 .Where c3(r, g) is:(r, g) c3(r, g) (r, g) c3(r, g)(1, 0) 1.014 (6, 0) 0.37(2, 0) 0.38 (7, 0) 1.11(3, 0) 3.33 (8, 0) 0.322(4, 0) 0.555 (9, 0) 0.002.(5, 0) 1.427Note that for all these values, we have c3(r, g) > (2.3)?r. To prove thelemma for r > 9 we use induction in two steps: For r ? 15, we will showthat|?2| > (2.3?)?r |?1|3r+2 .The inequality holds for r ? 9. Supposing it is true for some 9 ? r < 15,thenc1(r + 1, 0) |?1|3r+4 |?2|?2 <4r+1?r + 1(2.3?)2r |?1|?3r<4?r + 1(21.16)r(122?215)r< 0.001.Since ?r+1,0 6= 0, by lemma (3.6)373.2. Solutions of type I,II?|?2| >0.999c2(r + 1, 0)??r?1 |?1|3r+5=? |?2| > (2.3?)?r?1 |?1|3r+5 .Now assume (3.2.4) holds for some r ? 15, then:c1(r + 1, 0) |?1|3r+4 |?2|?2 <4r+1?r + 1(2.3?)2r |?1|?3r ?13<4?r + 1(21.16)r(122?215)r?13(Since r ? 15) <(21.1622)15??5/3 < 0.001.If ?r+1,0 6= 0, then by lemma (3.6)|?2| >0.999c2(r + 1, 0)??r?1 |?1|3r+5=? |?2| > (2.3?)?r?1 |?1|3r+5 .Otherwise if ?r+1,0 = 0 then both ?r+1,1,?r+2,1 are nonzero.Usingsame process for ?r+1,1c1(r + 1, 1)?13 |?1|3r+4 |?2|?2 <4r+1?r + 1(2.3?)2r |?1|?3r ?23|?|<4?r + 1(21.16)r(122?215)r ?23|?|< 0.001.Using this bound for ?2 and lemma 3.2.3 for ?r+2,1 we can conclude|?2| >0.999?r + 1(2.252)r+1??r?56 |?1|3r+3 .It follows383.2. Solutions of type I,II?c1(r + 2, 1)?1/3 |?1|3r+6 |?2|?2 <4r+2?r + 2(2.252)2r+2(r + 1)(0.999)2(?2?31)r?2<81.31(r + 1)?r + 2(20.29)r(122?2/15)r?2(Since r ? 15) < 0.37.And finally applying lemma 3.6 we conclude|?| >0.73c2(r + 2, 1)??r?11/6 |?|3r+6 .Since |?|3 > ?2 we obtain|?2| > (2.3?)?r?1 |?1|3r+5 ??16as desired.Corollary 3.1. The main lemma (lemma 3.4) is the immediate consequenceof this lemma. Assume the conditions of lemma 3.9 satisfy. Since???31?? >2.3?, there is no upper bound for ?2. Therefore F (x, y) = 1 can not have asolution (x2, y2) such that ?2 > ?1 and |x2y1 ? x1y2| > 1.3.2.3 Applying the main lemmaIn this section we will prove that X1 = (x1, y1) = (1 ? t3, t8 ? 3t5 + 3t2)is the only solution of type I for F (x, y) = 1 for all values of t > 8586, and X ?1 = (x?1, y?1) = (t9 + 3t6 + 4t3 + 1, t8 + 3t5 + 3t2) is the uniquesolution of type II? for F ?(x, y) = 1. for all values of t > 8586. The outlineof the proof for solutions of type I is as follows: (the proof for type II?is completely the same). First we will show that any solution of type Ibelongs to the same resolvent as X1.Then we will show that X1 satisfies theconditions of the lemma 3.4. Next step is to show that if there exist anyother solutions X2 = (x2, y2) of type I, we have ?X2 > ?X1 . Moreover, wehave |x2y1 ? x1y2| ? 2, which contradicts lemma 3.4.3.2.4 Solution of type INote thatH(x, y) = Ax2 +Bxy + Cy2is a quadratic. For the solution (1? t3, t8 ? 3t5 + 3t2) we have393.2. Solutions of type I,II?A = t8 ? 5t5 + 7t2,B = ?t9 + 3t6 ? 2t3 + 6,C = t10 ? 4t7 + 7t4 ? 3t,DF = t18 ? 10t15 + 41t12 ? 90t9 + 102t6 ? 40t3 ? 27?F = 3t18 ? 30t15 + 123t12 ? 270t9 + 306t6 ? 120t3 ? 81.(3.2.5)Also using estimates (3.1.5) for solutions of type Iy2(t10?4t7+9t4) > H(x, y) > y2(t10?4t7+7t4) ? 4(t10?4t7+7t4). (3.2.6)Moreover,H(0, 1) = t10 ? 4t7 + 7t4 ? 3t.The following lemma relates the arguments made in the previous section tosolutions of type I.Lemma 3.10. Assume the solution (1? t3, t8 ? 3t5 + 3t2) is related to pairof resolvents (?, ?). Then any solution (x, y) of type I and the solution (0,1)are also related to same pair of resolvents.Proof. First we can observeH(1? t3, t8 ? 3t5 + 3t2) =t26 ? 10t23 + 47t20 ? 130t17++ 225t14 ? 249t11 + 180t8 ? 88t5 + 25t2.(3.2.7)Let (?, ?) be the pair of resolvent forms that (x1, y1) = (1 ? t3, t8 ?3t5 + 3t2) is related to; then from inequality (3.2.1), and considering ?F >72000 for all t ? 2, we conclude????1??1(x, y)?1(x, y)???? <1.012??F|?(x, y)|3< t?27.If (x2, y2) is any other solution of type I then by estimations (3.1.3)x2y2?x1y1=x2y1 ? x1y2y1y2< 2t?5=? x2y1 ? x1y2 < 2t?5y1y2.403.2. Solutions of type I,II?Also?2?2??1?1=?2?1 ? ?2?1?1?2=???F (x1y2 ? x2y1)?1?2=??????2?2??1?1???? <2??F t?5y1y2|?1| |?2|.|?1| |?2| =????H1???????H2???=y1?A(xy)2+B(xy)+ C y2?A(xy)2+B(xy)+ C.=??????2?2??1?1???? < 2??Ft5?A(xy)2+B(xy)+ C?A(xy)2+B(xy)+ C<4t9t5t5t4<4t4.So????1??(x2, y2)?(x2, y2)???? <????1??(x1, y1)?(x1, y1)????+?????2?2??1?1???? <1t27+4t4<pi3for all t ? 2 =? (x2, y2) is also related to pair (?, ?) for all t ? 2.For solution (x2, y2) = (0, 1)?????2?2??1?1???? =?2?1 ? ?2?1|?1| |?2|<2t9t3t12t4<2t4,which means (0, 1) is also related to same pair of resolvent forms.Lemma 3.11. If (x2, y2) is a solution of type I different from (x1, y1) =(1? t3, t8 ? 3t5 + 3t2) then |?(x2, y2)| > |?(x1, y1)| and |x1y2 ? x2y1| ? 2.Proof. Let (x2, y2) be a solution of type I such that (x2, y2) 6= (x1, y1) =(1 ? t3, t8 ? 3t5 + 3t2). First assume |?(x2, y2)| < |?(x1, y1)|. From theestimates (3.2.6) for solutions of type I we have:H(x, y) > 4(t10 ? 4t7 + 7t4) =? H(x2, y2) > H(0, 1)=? |?(x2, y2)| > |?(1, 0)| .By using lemma (3.1),413.2. Solutions of type I,II?|?(x2, y2)| > 0.987 |?(0, 1)|2|?(x1, y1)| > 0.987 |?(x2, y2)|2=? |?(x1, y1)| > (0.987)3 |?(0, 1)|4 .But from inequality (3.2.7) (0.987)3 |?(0, 1)|4 > |?(x1, y1)| . This contra-diction leads us to conclude that |?(x2, y2)| > |?(x1, y1)| which by usinglemma (3.1)|?(x2, y2)| >0.987 |?(x1, y1)|2|H(x2, y2)| >(0.987)2 |H(x1, y1)|2 (3.2.8)=? y22(A(x2)2 +B(xy) + C) > 1048.Using estimations (3.1.5 )for xy , we get|y| > t19. (3.2.9)On the other hand, if (x2, y2) is an integral solution for F (x, y) = 1, thenx32 ? (t4 ? t)x22y2 + (t5 ? 2t2)x2y22 + y32 = 1,Multiplying both sides by x31:(x1x2)3 ? (t4 ? t)(x2x1)2(x1y2) + (t5 ? 2t2)(x2x1)(x1y2)2 = x31 ? (x1y2)3Since xi, yi and t are integers x2|x31 ? (x1y2)3.If x1y2 ? x2y1 = 1 then x1y2 = 1 + x2y1 using value of x1x2??(t9 ? 3t6 + 3t3) =? |x2| < t9.Therefore from estimations (3.1.5) for solutions of type I, |y2| < t14. Similarlyif x1y2 ? x2y1 = ?1 then x1y2 = ?1 + x2y1x2??(t9 ? 3t6 + 3t3 ? 2) =? |x2| < t9.Again using estimations (3.1.5), |y2| < t14, but by (3.2.9) |y| > t19. Thiscontradiction completes the proof.423.2. Solutions of type I,II?Lemma 3.12. For t > 8586 there is no integral solution of type I, otherthan (1? t3, t8 ? 3t5 + 3t2).Proof. By (3.2.5) and (3.2.7) we can see for t > 8586 ,?2F?31< 122?2/15F. andalso ?F > 72000. By lemma 3.11, if there exist any other solution (x2, y2)of type l, we have |?(x2, y2)| > |?(x1, y1)| . Therefore by the lemma 3.4(mainlemma), we have |x2y1 ? x1y2| = 1 which contradicts lemma 3.11.3.2.5 Solutions of type II?For solutions of type II? we prove this lemma:Lemma 3.13. For t > 8586 there is no integral solution of type II?, otherthan (t9 + 3t6 + 4t3 + 1, t8 + 3t5 + 3t2).For the proof we will follow exactly the same steps as in the previoussection. Using the same notation as we used in 3.2.1 for the solution (t9 +3t6 + 4t3 + 1, t8 + 3t5 + 3t2) we haveH(x, y) = Ax2 +Bxy + Cy2,whereA = t8 + 5t5 + 7t2,B = ?t9 ? 7t6 ? 12t3 ? 9,C = t10 + 6t7 + 12t4 + 12t,DF ? = t18 + 10t15 + 41t12 + 90t9 + 102t6 + 40t3 ? 27, (3.2.10)?F ? = 3t18 + 30t15 + 123t12 + 270t9 + 306t6 + 120t3 ? 81.Also using estimates (3.1.6) for solutions of type II?y2(t10+4t7+9t4) > H(x, y) > y2(t10+4t7+7t4) ? 4(t10?4t7+7t4) (3.2.11)H(t, 1) = t10 + 4t7 + 7t4 + 3t,H(t9 + 3t6 + 4t3 + 1, t8 + 3t5 + 3t2) =t26 + 10t23 + 47t20 + 130t17 (3.2.12)+ 225t14 + 234t11 + 120t8.Similar to lemma (3.10) we have433.2. Solutions of type I,II?Lemma 3.14. Assume the solution (x1, y1) = (t9+3t6+4t3+1, t8+3t5+3t2)is related to a pair of resolvents (?, ?), then any solution (x2, y2) of type II?is also related to the same pair of resolvents.Proof. Sincex2y2?x1y1=x2y1 ? x1y2y1y2< 2t?5.The same argument as proof of lemma (3.10)is valid.Lemma 3.15. If (x2, y2) is a solution of type II?, different from (x1, y1) =(t9+3t6+4t3+1, t8+3t5+3t2) then |?(x2, y2)| > |?(x1, y1)| and |x1y2 ? x2y1| ?2.Proof. Let (x2, y2) be a solution of type II? such that (x2, y2) 6= (x1, y1) =(t9 +3t6 +4t3 +1, t8 +3t5 +3t2). First assume |?(x2, y2)| < |?(x1, y1)|. Fromthe estimates (3.2.11) for solutions of type II? we have:H(x, y) > 4(t10 +4t7 +7t4) =? H(x2, y2) > H(t, 1) =? |?(x2, y2)| > |?(t, 1)|and by using lemma (3.1)|?(x2, y2)| > 0.987 |?(t, 1)|2 ,|?(x1, y1)| > 0.987 |?(x2, y2)|2 ,=? |?(x1, y1)| > (0.987)3 |?(0, 1)|4 .But by inequality (3.2.12) (0.987)3 |?(t, 1)|4 > |?(x1, y1)| . This contradic-tion leads us to conclude that |?(x2, y2)| > |?(x1, y1)| which by using lemma(3.1)|?(x2, y2)| >0.987 |?(x1, y1)|2|H(x2, y2)| >(0.987)2 |H(x1, y1)|2 (3.2.13)=? y22(A(x2)2 +B(xy) + C) > 1048.Using estimations (3.1.5 )for xy , we get|y2| > t19 =? x2 > t20. (3.2.14)If (x2, y2) is an integral solution for F ?(x, y) = 1 thenx32 ? (t4 + 4t)x22y2 + (t5 + 3t2)x2y22 + y32 = 1.443.2. Solutions of type I,II?Multiplying both sides by x31:(x1x2)3 ? (t4 + 4t)(x2x1)2(x1y2) + (t5 + 3t2)(x2x1)(x1y2)2 = x31 ? (x1y2)3.Since xi, yi and t are integersx2??(x31 ? (x1y2)3 ? (t5 + 3t2)(x2x1)(x1y2)).If x1y2 ? x2y1 = 1 then x1y2 = 1 + x2y1 we havex22??x31 ?(3y1 + (t5 + 3t2)y1)x2 ? 1.By plugging the values of x1 and y1 it is easy to check thatx31 ?(3y1 + (t5 + 3t2)y1)x2 ? 1 6= 0.Therefore we have??x22?? ???x31 ?(3y1 + (t5 + 3t2)y1)x2 ? 1?? . (3.2.15)On the other hand since t > 10 ,x1 < t10, y1 < t9. Hence from inequality(3.2.14), we have:x31 <x223,??(3y1 + (t5 + 3t2)y1)?? <x23=???x31 ?(3y1 + (t5 + 3t2)y1)x2 ? 1?? <2x223,which contradicts (3.2.15). Similarly if x1y2 ? x2y1 = ?1 then x1y2 =?1 + x2y1x22??x31 +(3y1 + (t5 + 3t2)y1)x2 ? 1.By same argument, this cannot happen.Proof of theorem 3.13Proof. By (3.2.10) and (3.2.12) for t > 8586 ,?2F ??31< 122?2/15F ?. and also?F ? > 72000. By lemma (3.15) if there exist any other solutions (x2, y2) oftype II?, we have |?(x2, y2)| > |?(x1, y1)| . Therefore by the lemma 3.4(mainlemma), we have |x2y1 ? x1y2| = 1 which contradicts lemma 3.15.453.3. Solutions of type II,I?3.3 Solutions of type II,I?In this section we use linear forms in logarithms to find all solutions of typeII and I? . As we mentioned in Chapter 2, for this purpose we need to have asystem of fundamental units and the crucial step is to prove a ?stable growth? condition. Fortunately Thomas [80] introduced a system of fundamentalunits for the split familes of Thue equations. By Lemma 4.11 of [80], t ? ?and ? form a pair of fundamental units in Q(?) and similarly t ? ? and ?form a pair of fundamental units in Q(?). If (x, y) is a solution of a Thueequation F (x, y) = 1 then we have(x? y?)(x? y??)(x? y???) = 1.Therefore x? y? is a unit in Q(?). And it means that there exist m,n ? Zsuch that x?y? = (t??)m?n. Similarly for the solution (x, y) of F ?(x, y) = 1there exists a probably different pair m,n such that (x? y?) = (t??)m?n.Moreover we havey3(xy? ?)(xy? ??)(xy? ???) = 1 (3.3.1)andy3(xy? ?)(xy? ??)(xy? ???) = 1. (3.3.2)The main idea in this section to prove the ?stable growth? condition isto use some congruence conditions. Let (x, y) be a solution for F (x, y) = 1thenx? y? = ?(t? ?)n??m for m,n ? Z. (3.3.3)= (?1)n(?2 ?(n1)t? +(n2)t2 ?(n3)t3??1 + ? ? ?)?(n?m?2).The idea is to find a congruence condition for both sides of the aboveequation in different rings say Z/tZ?,Z/t3Z? and Z/t5Z?.3.3.1 Solution of type IIFor solutions of type II we will prove thatLemma 3.16. If t > 1119749, then there are no non-trivial solutions oftype II.463.3. Solutions of type II,I?Let (x, y) be a solution of type II of F (x, y) = 1 so t + 1t6 <xy <t + 2t5 . First we will show that y > 0. We rewrite the equation (3.3.1) asy2(xy ? ?)(xy ? ???)(x ? y??) = 1. So by (3.1.3) and (3.1.5) We can concludethat x? y?? < 0.Consider the equationx? y??x? y???=(t? ??t? ???)n( ?????)?m,from estimations (3.1.3) it is clear that RHS is positive. Therefore LHS isalso positive =? x? y??? = y(xy ? ???)< 0 ,but xy ? ??? < 0 =? y > 0.Using equation (3.3.3) and estimations (3.1.3) we can obtain0 >xy ? ???xy ? ?=x? ???yx? ?y=(t? ???t? ?)n(????)?m.So n , m have different parities mod (2). Taking absolute values of bothsides we have:??? ? xyxy ? ?=(??? ? tt? ?)n( ???|?|)?m.Taking logarithms from both sides and using estimations (3.1.3) we obtainthat :log(t3 ? 4) < n log(??? ? tt? ?)?m log(???|?|)< log(t3 ? 3),t3 ? 4 <??? ? tt? ?< t3 ? 3, t9 ? 3t6 <???|?|< t9 ? 2t6.(3.3.4)Considering (3.3.4) , if m = 0 then n = 1 which with positive sign givesus the solution (t, 1). If m < 0 then n < 0 so |x?y??| = |(t???)n(??)?m| > 1.On the other hand from (3.3.1) we have |x? y??| = 1???y2(xy??)(xy???????, hence bythe estimations (3.1.5) we have |x? y??| < 1. This contradiction leads us toconclude that m > 0 ? n > 0 and from first inequality of (3.3.4) n > m;moreover:n?m >??log(???|?|)log(????tt??) ? 1?? > m(log t6(t3 ? 3)log(t3 ? 3)? 1)> mlog t6log(t3 ? 3)> 2m.SO n > 3m , n?m?2 > 0? n?m?2 = 3k+i with i ? {0, 1, 2}and k > 0.473.3. Solutions of type II,I?Linear form in logarithmsFrom Siegel?s identity:(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (? ? ??)(x? ???y) = 0,using this, corresponding to the solutions of type II we define a linear formin logarithms as follows:? = log??? ? ???? ? ?+ nlog????t? ?t? ???????+m log???????????? (3.3.5)= log(1 +(? ? ???)(x? ??y)(?? ? ?)(x? ???y)), (3.3.6)where (?????)(x???y)(????)(x????y) < 0.We will use the (3.1.3) to find an upper bound for |?| .Upper bound for ?(??? ? ?)(x? ??y)(?? ? ?)(x? ???y)=??? ? ??? ? ?(t? ??t? ???)n(?????)m(since m <n3), <??? ? ??? ? ?(t? ??t? ???)n(?????)n3< (t3 ? 2)((1t3 ? 3)3)n(t3 ? 2)n3<(1.01(t3 ? 2)3)n(t3 ? 2)n3 +1< (t3 ? 2)(?8n3 +1)(1.01)n.If 0 < x < 12 then |log(1? x)| < 2x, So |?| < 2(t3 ? 2)(?8n3 +1)(1.01)n=? log |?| < log(2) + n log(1.01) +(?8n3+ 1)log(t3 ? 2). (3.3.7)Lower bound for ANext step is to prove the ?stable growth? condition for solutions of typeII. Let A = max{??n??,??m??}, in this case since n > m > 0, A is equal to n.Using some congruence arguments we find a lower bound for A. Consideringequation (3.3.3),we will write RHS of (3.3.3) as linear combination of 1, ?, ?2483.3. Solutions of type II,I?and calculate the coefficient of ?2 in the rings Z/tjZ[?] for j ? 1, 3, 5,in whichthey all should be zero. Since n?m? 2 > 0? n?m? 2 = 3k+ i with i ?{0, 1, 2}and k ? 0. For each value of i we do the factorization in a differentring and find the lower bound for A.i = 0, Factoring in Z/tZ[?]Lemma 3.17. For i = 0 there is no integral solution.Proof. Let i = 0?3 = (t4 ? t)?2 ? (t5 ? 2t2)? ? 1 ? ?1 (mod t).Considering (3.3.3) LHS is congruent to x? y?.RHS ? (?1)n?2(?3)k?i? (?1)n?2(?1)k? (?1)n+k?2=? (?1)n?+k ? 0 (mod t). which is not possible so there is no solutionfor i = 0.i = 1 Factoring in Z/t3Z[?] In this ring we can write this list of congru-ences:?3 ? ?t?2 + 2t2? ? 1,t2?4 ? ?t2?,t2?5 ? ?t2?2.(3.3.8)Using congruences (3.3.8), we have(?3)k? (?1)k(1 + kt?2 ? 2kt2? +(k2)t2?4),(?3)k? ? (?1)k(? + kt?3 ? 2kt2?2 +(k2)t2?5),? (?1)k(?(3k +(k2))t2?2 + ? ? kt).Let i = 1; then for RHS of (3.3.3), we obtain:493.3. Solutions of type II,I?RHS? (?1)n(?2 ? nt? +(n2)t2)(?3)k?,? (?1)n+k(?2 ? nt? +(n2)t2)(?(3k +(k2))t2?2 + ? ? kt),? (?1)n+k(?(3k +(k2))t2?4 + ?3 ? (n+ k) t?2 +(kn+(n2))t2?),? (?1)n+k((n+ k + 1) t2?2 +(kn+(n2)+ 3k +(k2)+ 2)t2? ? 1).Considering coefficient of ?2, leads us to conclude that(k + n+ 1)t ? 0 (mod t3) =? k + n+ 1 ? 0 (mod t2)But k + n + 1 > 0 so k + n + 1 ? t2. Using value of k one can obtain43A > t2 =? A > 34 t2. Hence we have the following lemma :Lemma 3.18. For i = 1 there is no solution unless A > 34 t2.Factoring in Z/t5Z[?]Lemma 3.19. For i = 2 there is either no nontrivial integral solution orA >(45?2t3 + 161100 +1475).Proof. In the ring Z/t5Z[?] ,we have this list of congruences:?3 ? (t4 ? t)?2 + 2t2? ? 1,t3?3 ? ?t4?2 ? t3,t4?5 ? ?t4?2.(3.3.9)Finally using these congruences we have:(?3)k? (?1)k(Mt4 +((k3)+ 2k(k ? 1) +(k2))t3?((k2)+ 2k)t2? + kt?2 + 1),for some M ? Z.Let i = 2 then n ?m ? 2 = 3k + 2 with k ? 0. By equation (3.3.3)wecan write:503.3. Solutions of type II,I?RHS ?(?4 ? nt?3 +(n2)t2?2 ?(n3)t3? +(n4)t4)(?3)k.Using above calculation and congruences (3.3.9) one can see the coeffi-cient of ?2 in (t? ?)n (?)?m is(n+ nk + 4k + 3 +(n2)+(k2))t2=?(4k +(k2)+ kn+(n2)+ n+ 3)? 0 (mod t3).Since LHS is positive=?(4k +(k2)+ kn+(n2)+ n+ 3)> t3=? (k + n)2 + (k + n) + 6k + 6 ? 2t3.From inequalities (3.3.4) it is easy to see n4 < m, so we can obtain2516A2 ?712A > 2t3 +149=? A >(45?2t3 +161100+1475).Lemma 3.20. There is no solution of type II other than the trivial onesunlessA >(45?2t3 +161100+1475).Proof. Follows from lemmas (3.17), (3.18), (3.19).Large values of tLower bound for ? in type II By lemma 3.20, if there exists a non-trivial solution of type II, then we have:A >(45?2t3 +161100+1475).Recall that:? = log??? ? ???? ? ?+ n log????t? ?t? ???????+m log????????????513.3. Solutions of type II,I?We will apply theorem (2.3) to find a lower bond for ?. Put ?1 = ?????????? , ?2 =???t??t??????? , ?3 =?????????? , b1 = 1, b2 = n, b3 = m,D = 6.First, using (3.1.3) we find an upper bound for the weil heights of ?i?s:h(??? ? ????? ? ?)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?))< 6 log th(????)=16log(????)2< 3 log th(t? ?t? ???)=16log(t? ???t? ??)2< 3 log t.(3.3.10)Therefore we can take A1 = 36 log(t), A2 = 18 log(t), A3 = 18 log(t)and B = n; by the above calculation we have:log|?| > ?8.344 ? (10)15 log(t)3 log(68.3n).Comparing this with (3.3.7) :?8.344(10)15 log(t)3 log(68.3n)< log(2) + n log(1.01) +(?8n3+ 1)log(t3 ? 2).(3.3.11)SinceA >(45?2t3 +161100+1475),the above inequality is only valid for t < (1.197) ? (10)13. We can summarizethe argument for solutions of type II in this lemmaLemma 3.21. For t > 1.197 ? (10)13 there is no solution of type II otherthan the trivial ones.Intermediate values of tAs we have seen there is no nontrivial solution of type II, if t < 1.197 ?(10)13.This lower bound for t is still huge and treating the smaller value of t is verytime consuming, thus we should try to find a better lower bound for t. Theidea is to reduce the linear form in three logarithms used in the previous523.3. Solutions of type II,I?section to a linear form in two logarithms with the aim of getting a betterbound for intermediate values of t. To do so, in this section we assumet < 1.197 ? (10)13 and A >(45?2t3 + 161100 +1475). Moreover, from inequality(3.3.11) one can obtainn < 4.8 ? 1019, m < 1.6 ? 1019.Lemma 3.22. For t > 10, if there exists a non-trivial solution of type IIthen the following inequality holds:1m(n? 3m? 1) <log(1 + 11t3)log(t3 ? 4).Proof. We haven log(??? ? tt? ?)? log(??? ? xyxy ? ?)= m log(???|?|).Since??? ? xyxy ? ?<??? ? tt? ?,we can conclude(n? 1) log(??? ? tt? ?)< m log(???|?|).And from estimates (3.1.3)(n? 1) log(t3 ? 4) < m log(t9 ? 2t6). (3.3.12)On the other hand:(t9 ? 2t6) <(1 +11t3)(t3 ? 4)3. (3.3.13)Now combining (3.3.12) and (3.3.13) we get:(n? 1) log(t3 ? 4) < m log((1 +11t3)(t3 ? 4)3),which is equivalent ton? 3m? 1 < m(log(1 + 11t3)log(t3 ? 4)).533.3. Solutions of type II,I?Lemma 3.23. Let t > 1119749. If there exists a non-trivial solution of typeII, then n = 3m+ k with k ? 3.Proof. Substituting 3m+ k for n in lemma 3.22, we have the following:k ? 1m<log(1 + 11t3)log(t3 ? 4).We know m < 1.6? 1019; now, if k > 4, these two give us:31.6? 1019<log(1 + 11t3)log(t3 ? 4).which implies3.125? 10?20 <log(1 + 11t3)log(t3 ? 4).The right hand side is a strictly decreasing function in the variable t and sowe easily deduce that t < 1119749. So for t > 1119749 we have k ? 3.Lemma 3.24. If there exists a non-trivial solution such that n = 3m + kwith k ? 3 then t < 1119749.Proof. We have:? = log??? ? ???? ? ?+ n log????t? ?t? ???????+m log????????????= log??? ? ???? ? ?+ (3m+ k) log????t? ?t? ???????+m log????????????=[log??? ? ???? ? ?+ k log????t? ?t? ???????]+[3m log????t? ?t? ???????+m log????????????]= log(??? ? ???? ? ?????t? ?t? ???????k)+m log(????t? ?t? ???????3 ????????????)= ? log?????(?? ? ???? ? ??) ????t? ???t? ?????k? ?? ??1?????+m log?????????t? ?t? ???????3 ????????????? ?? ??2?????= m log?2 ? log?1,where ?1, ?2 > 1. To find a lower bound for this new form of ? we usethe lemma 2.4543.3. Solutions of type II,I?To apply the lemma for log |?| we take b2 = m, b1 = 1, D = 6, using(3.1.3), (3.5.8)|log?1| /D < h(?1) < (3k + 6) log t < 15 log t|log?2| /D < h(?2) < 12 log t.Thus we can take the values of A1, A2 of the lemma 2.4 to be real numberssuch that logA1 = 18 log t, logA2 = 12 log t. Then log b? + 0.21 > 30/D = 5for t > 3384; from lemma 2.4 we obtain:log |?| ? ?5.02? 106 (log(t))2(max{log b? + 0.38, 5})2.Comparing this with (3.3.7) :?4.176? 106 (log(t))2(max{log b? + 0.38, 5})2< log(2) + n log(1.01) +(?8n3+ 1)log(t3 ? 2).Since A >(45?2t3 + 161100 +1475), the above inequality is not valid fort > 1063847; this completes the proof.From lemmas 3.23, 3.24 we conclude:Corollary 3.2. If t > 1119749 then there are no non-trivial solutions oftype II.3.3.2 Solutions of type I?In this section for solutions of type I? we will prove the following lemma:Lemma 3.25. If t > 712080 then there are no non-trivial solutions of typeII?.The proof is quite similar to the proof of solutions of type II, thereforewe can leave out some very similar details. Recall that if (x, y) is a solutionof type I? of F ?(x, y) = 1 then ? 1t5 +2t8 <xy < ?1t6 . Therefore by (3.1.4) and(3.3.3) we conclude0 <xy ? ???xy ? ?? =x? ???yx? ??y=(t? ???t? ??)n(?????)?m,553.3. Solutions of type II,I?logxy ? ???xy ? ?? = n log(t? ???t? ??)?m log(?????). (3.3.14)Also from estimates (3.1.4) we have :t3 + 2 <xy ? ???xy ? ?? < t3 + 3, (3.3.15)t3 + 2 <?????< t3 + 3,t9 + 3t6 <(t? ???t? ??)< t9 + 5t6.In equation (3.3.14 ) if n = 0 then m = ?1, which implies x = 0, y = 1.If n > 0 then m > 0 so??x ? y???? > 1. On the other hand with the sameargument as the solutions of type II,??x ? y???? < 1. By (3.3.2 )and (3.3.3),we conclude that n < 0,m < 0. Let n? = n,m? = m be positive numbers,then we can rewrite (3.3.14 )as:logxy ? ???xy ? ?? = m? log(?????)? n? log(t? ???t? ??). (3.3.16)Using above equation we haven???log(t????t???)log(?????)?? < m? =?83n? < m?.Also sincexy????xy??? <(?????)(m? ? 1)log(?????)< n???log(t????t???)log(?????)?? .By estimates (3.3.15 )(m? ? 1)< n???log(t????t???)log(?????)?? < 3n?.To summarize, we have(83)n? < m? ? 3n?.563.3. Solutions of type II,I?Linear form in logarithmsFrom Siegel?s identity:(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (?? ??)(x? ???y) = 0,we obtain that :? = log?? ????? ??+ n logt? ??t? ???+m log?????(3.3.17)= log(1 +(?? ? ???)(x? ?y)(?? ??)(x? ???y)). (3.3.18)Moreover , we have :????(?? ? ???)(x? ?y)(?? ??)(x? ???y)???? =??? ? ???? ? ?(??? ? tt? ?)n? ( ????)m?(Since n? <38m?)<??? ? ???? ? ?(??? ? tt? ?) 38m? ( ????)m?<(t3 + 2) (t3 + 2) 38m?((1t3 + 1)3)m?<(t3 + 2) 38m?+1(1.01(t3 + 2)3)m?< (1.01)m? (t3 + 2)?218 m?+1.If 0 < |x| < 12 then |log(1 + x)| < 2x, so |?| < 2 (1.01)m? (t3 + 2)?213 m?+1=? log |?| < log(2) +m? log(1.01) +(?21m?8+ 1)log(t3 + 2). (3.3.19)Lower bound for A in type I?Let A = max{??n??,??m??}; in this case since m < n < 0, A is equal to ?m.Using some congruence arguments we find a lower bound for A. We writeboth sides of equation (3.3.3) as a linear combination of 1,? and ?2 andcalculate the corresponding coefficients in the rings(Z/tiZ)[?] for i = 1, 2, 3.x? y? = ?(t? ?)n??m for m,n ? Z.573.3. Solutions of type II,I?We consider the positive sign with m even (considering the negative signgives the same results)1x? y?= (t? ?)n??m where n? = ?n=? (x? y??)(x? y???) = (t? ?)n??m.=?xy(?)2 + (x2 ? (t4 + 4t)xy)(?)? y2 = (t? ?)n??m?(3.3.20)=(?1)n?(?2 ?(n?1)t?+(n?2)t2 ?(n?3)t3??1 + ? ? ?)?(n?+m??2).(3.3.21)In above equation n? = ?n,m? = m + 1, n? > 0 and m? < ?1 since ifm? = ?1 then m = ?2, n = ?1, and one can check there is no solution withthis condition.So n? +m? ? 2 < 0.Factoring in (Z/tZ)[?]Consider the equation (3.3.20), LHS is congruent to xy(?)2 +x2(?)?y2 andRHS is congruent to (?1)n?(?2)(?(n?+m??2)). We putn? +m? ? 2 = ?3k + i for i ? {0, 1, 2}, and k ? N,since n? +m? ? 2 < 0?3 = (t4 + 4t)?2 ? (t5 + 3t2)?? 1 ? ?1 (mod t).??1 = ??2 + (t4 + 4t)?? (t5 + 3t)=???1 ? ??2 (mod t)=?(??1)3 ? ??6 ? ?(?3)2 (mod t)=?(??1)3 ? ?1 (mod t).So we can concludeRHS ? (?1)n??2?(n?+m??2) ? (?1)n?((??1)3)k?i ? (?1)n??2(?1)k?i.We have three cases:583.3. Solutions of type II,I?i = 0Lemma 3.26. For i = 0 there is no integral solution.Proof. Comparing coefficients of 1,?, ?2 in RHS and LHS give?????xy ? 1 (mod t)x2 ? 0 (mod t)y2 ? 0 (mod t)not possible. (3.3.22)So there is no solution in this case.i = 1 In this case k is odd, so RHS ? (?1)n??2(?1)? ? (?1)n?+1?3 ?(?1)n?+2. Again by comparing coefficients of powers of ? we obtain?????xy ? 0 (mod t)x2 ? 0 (mod t)?y2 ? (?1)n?(mod t)??????x ? 0y2 ? (?1)n?+1y3 ? 1?{x ? 0 (mod t)y ? 1 (mod t).(3.3.23)i = 2 In this case k is even. Therefore RHS ? (?1)n??2(?1)k?2 ?(?1)n??4 ? ??????xy ? 0 (mod t)x2 ? 1 (mod t)?y2 ? 0 (mod t)??????x2 ? 1 (mod t)x3 ? 1 (mod t)y ? 0 (mod t)?{x ? 1 (mod t)y ? 0 (mod t).(3.3.24)Factoring in (Z/t2Z)[?]In this ring we have this list of congruences:?3 ? (4t?2 ? 1),?6 ? ?8t?2 + 1,?4 ? 4t?3 ? ? ? 4t(4t?2 ? 1)? ? ? ?4t? ?,t?5 ? (t?3)?2 ? ?t?2,??1 ? ??2 + 4t?,(??1)3 ? ??6 + 12t?5 ? ?4t?2 ? 1.(3.3.25)593.3. Solutions of type II,I?In equation (3.3.20 ) LHS is congruent to xy(?2) + (x2 ? 4txy)?? y2.Also by (3.3.25 ) for RHS we can obtainRHS ? (?1)n?(?2 ? n?t?)(?(1 + 4t?2))k?i? (?1)(n?+k)(?2 ? n?t?)(1 + 4kt?2)?i? (?1)(n?+k)(4kt?4 + ?2 ? n?t?)?i? (?1)(n?+k)(?2 ? (n? + 4k)t?)?i.let i = 1 thenRHS ? (?3 ? (n? + 4k)t?2) (mod t2)? ?(n? + 4k ? 4)t?2 ? 1 (mod t2).Therefore we have:?????xy ? ?(4k + n? ? 4)t (mod t2)x2 ? 4txy ? 0 (mod t2)y2 ? 1 (mod t2).(3.3.26)In regards to the results we found, (mod t), we can conclude{x ? ?(4k + n? ? 4)t (mod t2)y ? 1 (mod t2).(3.3.27)i = 2 . In this case k is even. Therefore using (3.3.25 ) we haveRHS ? (?1)(n?+k)(?4 ? (n? + 4k)t?3) (mod t2)? (?1)(n?+k)(4t?3 ? ?+ (n? + 4k)t?3) (mod t2)? (?1)(n?+k)((n? + 4k + 4)t?3 ? ?) (mod t2)? (n? + 4k + 4)t+ ? (mod t2).By comparing the coefficients of both sides of (3.3.20) we have:?????xy ? 0 (mod t2)x2 ? 4txy ? 1 (mod t2)y2 ? ?(n? + 4k + 4)t (mod t2)=?{x ? 1 (mod t2)y ? 0 (mod t2).(3.3.28)603.3. Solutions of type II,I?Factoring in (Z/t3Z)[?]In this ring we can write this list of congruences:?3 ? 4t?2 ? 3t2?? 1,?4 ? 4t?3 ? 3t2?2 ? ?,t2?4 ? ?t2?,t?4 ? 4t2?3 ? t? ? ?4t2 ? t?,?6 ? 16t2?4 ? 8t?2 + 6t2?+ 1 ? ?8t?2 ? 10t2?+ 1,??1 ? ??2 + 4t?? 3t2,(??1)3? ??6 + 12t?5 ? 9t2?4 ? 48t2?4,? ?4t?2 + 19t2?? 1,((??1)3)k? (?1)k(4kt?2 ? 19kt2?+ 16(k2)t2?4 + 1),? (?1)k(4kt?2 ? k(8k + 11)t2?+ 1).(3.3.29)i = 2:Lemma 3.27. For i = 2 there is no integral solution unless A > 34(t2 ? 4).Proof.LHS ? xy?2 +(x2 ? 4txy)?? y2,using congruencies (3.3.30) we have:RHS? (?1)n?(?2 ? n?t?+(n?2)t2)((??1)3)k?2? (?1)(n?+k)(?2 ? n?t?+(n?2)t2)(4kt?4 ? kt2?3(8k + 11) + ?2)? 4kt?6 ? (8k + 11 + 4n?)kt2?5 + ?4 ? n?t?3 +(n?2)t2?2? ?16kt2?2 ? 4kt?3 + (8k + 11 + 4n?)kt2?2 + ?4 ? n?t?3 +(n?2)t2?2? ?(n? + 4k ? 4)t?3 +(8k2 ? 5k + 4n?k +(n?2)? 3)t2?2 ? ?.613.3. Solutions of type II,I?By comparing coefficients of constant term on both sides of (3.3.20 ) weobtain:y2 ? ?(n? + 4k + 4)t (mod t3).Using congruences (3.3.28) y2 ? 0 (mod t3)=? n? + 4k + 4 ? 0 modt2 , But n? + 4k + 4 > 0 so n? + 4k + 4 ? t2 by thevalue of k we haveA >34(t2 + 4).Factoring in (Z/t5Z)[?]In this ring we can write this list of congruences:?3 ? (t4 + 4t)?2 ? 3t2?? 1,?4 ? (t4 + 4t)?3 ? 3t2?2 ? ?,t2?4 ? 13t4?2 ? t2?? 4t3,t?4 ? 13t3?2 ? (12t4 + t)?? 4t2,?6 ? 119t4?2 ? 8t?2 ? 10t2?? 40t3 + 1,??1 ? ??2 + (t4 + 4t)?? (t5 + 3t2),(??1)3? ?4t?2 + 19t2?? 24t3 ? 1? 10t4?2,((??1)3)k? (?1)k(Lt4?2 + 4kt?2 ? k(8k + 11)t2?+Mt3 + 1).(3.3.30)whereL = 464 + 130k ? 512(k3)+ 287(k2)+ 304k(k ? 12)M = ?26(k2)? 64(k3)+ 24k.i = 1Lemma 3.28. For i = 1 there is no integral solution unlessA >2411(?2t3 +100254356+1922).623.3. Solutions of type II,I?Proof.LHS ? xy?2 +(x2 ? 4txy)?? y2.Using congruences (3.3.30 ) One can check that the coefficient of ?2 in aboveis congruent to (n?+4k?4)t (mod t4) where the coefficient of ? is congruentto(8k2 ? 5k + 4n?k +(n?2)? 3)t2 (mod t5), and the constant coefficient iscongruent to 1 (mod t3). By comparing coefficients of both sides of (3.3.20)?????????xy ? ?(n? + 4k ? 4)t+ Ct4 (mod t5),x2 ? 4txy ?(8k2 ? 5k + 4n?k +(n?2)? 3)t2 (mod t5),y2 ? 1 + C ?t3 + C ??t4 (mod t5).(3.3.31)=? x2 ? ?(n? + 4k ? 4)2t2 (mod t5).Using the congruency we found for x from lost equivalency in (3.3.31 )we have(n? + 4k ? 4)2 + 4(n? + 4k ? 4) ?(8k2 ? 5k + 4n?k +(n?2)? 3)(mod t3).Simplifying both sides, we getn?2 + 16k2 + 8kn? ? 22k ? 7n? + 62? 0 (mod t3)=?(n? + 4k ? 72)2+ 6k ? 2542? 0 (mod t3).But(n? + 4k ? 72)2+ 6k ? 254 > 0 so(n? + 4k ?72)2+ 6k ?254? 2t3.Plug in the maximum value of k based on n we have:(119n? ?1922)2?100254356> 2t3=? n >911(?2t3 +100254356+1922)633.3. Solutions of type II,I?=? A = m? >2411(?2t3 +100254356+1922).Corollary 3.3. If A < 2411(?2t3 + 100254356 +1922)there is no solution of typeI? other than the five trivial solutions.Proof. Obvious by lemmas (3.26 ),(3.27), (3.28 ).Lower Bound for ? in type I?During this section we assume A > 2411(?2t3 + 100254356 +1922).Remember ? = log?? ????? ??+ n logt? ??t? ???+m log?????.Put ?1 = ????????? , ?2 =???t???t??????? , ?3 =??????????? , b1 = 1, b2 = n, b3 = m,D = 6.Using estimations (3.1.4 )h(??? ? ??? ? ?)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?))(3.3.32)<23log(t9 + 6t6),h(?????)=16log(????)2<13log(t9 + 7t6),h(t? ??t? ???)=16log(t? ???t? ??)2<13log(t9 + 5t6).Therefore we can take real numbers Ai as:A1 =4 log(t9 + 6t6),A2 =2 log(t9 + 5t6),A3 =2 log(t9 + 7t6),B =m?log(t9 + 7t6)log (t9 + 5t6)> (1.0007)m?We apply theorem (2.3) to find a lower bound for log |?|.log |?| > ?1.1446 ? 1013 logA1A2A3 log(68.3m?).643.3. Solutions of type II,I?Comparing with (3.3.19 ):?1.1446 ? 1013 log(t9 + 6t6)log(t9 + 5t6)log(t9 + 7t6)log(68.3m?)(3.3.33)< log(2) +m? log(1.01) +(?21m?8+ 1)log(t3 + 2).Since A > 2411(?2t3 + 100254356 +1922),the above inequality is not valid fort > 2.981 ? 1012. So we can conclude thatLemma 3.29. for t > 6.005 ? 1012 there is no solution of type I? other thanthe trivial ones.Intermediate values of tIn this section we assume t < 2.981 ? 1012 and A > 2411(?2t3 + 100254356 +1922).From (3.3.33 ), we can see m? < 4.54 ? 1019 and n? < 1.703 ? 1019.Lemma 3.30. For t > 10, if there exists a non-trivial solution of type II?then the following inequality holds:=? 3n? ?m+ 1 <(n? +13) log(t3t3?6)log (t3 + 3).Proof. Using (3.3.16 )and (3.3.15 ) we haven? log(t? ???t? ??)+ logxy ? ???xy ? ?? = m? log(?????)=?(n? +13)log(t? ???t? ??)< m? log(?????)=?(n? +13) log(t????t???)log(?????) < m?=?(n? +13)log t9 + 3t6log t3 + 2< m?.On the other hand653.3. Solutions of type II,I?t9 + 3t6 >(t3 + 2)3(1?6t3)=?(n? +13)(3 +log(1? 6t3)log (t3 + 3))< m?=? 3n? ?m? + 1 <(n? +13) log(t3t3?6)log (t3 + 3).Lemma 3.31. For t > 712080, if there exists a non-trivial solution of typeII? then m? = 3n? ? k with k ? 5.Proof. Substituting 3n? ? k for m? in Lemma (3.30) we have the following:k + 1(n? + 13) <log(t3t3?6)log (t3 + 3).We know n? < 1.703 ? 1019; now, if k ? 7 these two would give us:71.703 ? 1019<log(t3t3?6)log (t3 + 3).The right hand side is a strictly decreasing function in the variable t andso we easily deduce that t < 712080.Lemma 3.32. If there exists a non-trivial solution such that m? = 3n? ? kwith k ? 5 then t < 685450.663.3. Solutions of type II,I?Proof.? = log??? ? ??? ? ?+ n? logt? ???t? ??+m? log?????= log??? ? ??? ? ?+ n? logt? ???t? ??+(3n? ? k)log?????=[log??? ? ??? ? ?+ k log?????]+[n? logt? ???t? ??+ 3n? log?????]= log(??? ? ??? ? ?(?????)k)+ n? logt? ???t? ??(?????)3= log?????(??? ? ??? ? ?)(?????)k? ?? ??2?????? n? log?????(t? ??t? ???)(?????)3? ?? ??1?????= log?2 ? n? log?1.Where ?1, ?2 > 1. We apply lemma (2.4) for log |?| , with b2 = 1, b1 =n?, D = 6, using (3.1.4), (3.5.16) we have :|log?1|D< h(?1) <13log(t9 + 5t6)+ 3(13)log(t9 + 7t6)<43log(t9 + 7t6),|log?2|D< h(?2) <23log(t9 + 6t6)+ k(13)log(t9 + 7t6)<73log(t9 + 7t6).We can take A1, A2 as real numbers bigger than 1, such that, logA1 =43 log(t9 + 7t6)and logA2 = 73 log(t9 + 7t6). then b? = 4n?+756 log(t9+7t6) . Solog b? + 0.38 > 30/D = 5 for t > 1910; from lemma (2.4) we obtain:log |?| ? ?7.21728? 104(log(t9 + 7t6))2 (max{log b? + 0.38, 5})2.Comparing this with (3.3.19 ) :?7.21728? 104(log(t9 + 7t6))2 (max{log b? + 0.38, 5})2< log(2) +m? log(1.01) +(?21m?8+ 1)log(t3 + 2).673.4. Solutions of type III,III?Since A > 2411(?2t3 + 100254356 +1922), the above inequality is not valid fort > 685450. This completes the proof.The above lemmas (3.31)and (3.32) lead us to conclude:Corollary 3.4. If t > 712080 then there are no non-trivial solutions of typeII?.3.4 Solutions of type III,III?For this type of solution we again use Baker?s method. Proving the ?stablegrowth? condition is much easier in this case. Let (x, y) be an integralsolution of type III . By the same argument as in the beginning of thesection 3.3, there exist some integers m,n such thatx? y??? = (t? ???)n(???)?m.First we will show that if there exists a nontrivial solution of type III thenn 6= m and n and m have the same parity . It helps us to prove a strong?stable growth? condition just by using a simple calculus argument. Theargument for solutions of type III? is completely similar.3.4.1 Solutions of type IIIDuring this section we will prove that for t > 361165 there is no integralsolution of type III other than the trivial ones.Recall that if (x, y) is a nontrivial solution of type III then t4 ? 2t ? 2t3 <xy < t4 ? 2t? 1t9 . First we will show that for such a solution we have y > 0;we will rewrite the equation (3.3.1) asy2(xy? ?)(xy? ??)(x? y???) = 1.From the estimation (3.1.3) and (3.1.5) we have y2(xy ? ?)(xy ? ??) > 0,therefore We can conclude that x? y??? > 0.Consider the equationx? y??x? y???=(t? ??t? ???)n( ?????)?m.By estimations (3.1.3), it is easy to check that RHS is positive. ThereforeLHS is also positive =? x ? y?? = y(xy ? ??)> 0 . On the other hand,xy ? ?? > 0 =? y > 0.683.4. Solutions of type III,III?Using equation (3.3.3) and estimations (3.1.3), we can obtain that0 <xy ? ?xy ? ?? =x? ?yx? ??y=(t? ?t? ??)n( ???)?mSince both(t??t???)and(???)are negative So n , m have same parities(mod 2). Taking absolute values of both sides we havexy ? ?xy ? ?? =(t? ??? ? t)n( |?|??)?m.Taking logarithms from both sides and using estimations 3.1.3, we getlog(1 +1t3) < n log(t? ??? ? t)+m log(??|?|)< log(1 +2t3). (3.4.1)Also we have:t6 ? 4t3 + 16 <t? ??? ? t<t6 ? 3t3 + 10,t6 ? 2t3 + 1 <??|?|<t6 ? 2t3 + 3,1 +1t3<??|?|t?????t<1 +2t3.(3.4.2)Considering (3.4.1), we can deduce n 6= 0,m 6= 0 Also if m < 0 thenn > 0 and if m > 0 then n < 0.First assume m < 0, n > 0 =? |x? y???| > 1 but??x? y????? =1y2???xy ? ???????xy ? ?????< 1.So m > 0, n < 0. let n? = ?n from the second inequality of (3.3.8)m log(??|?|)< n? log(t? ??? ? t)+ log(1 +2t3)=? m log(??|?|)< (n? + 1) log(t? ??? ? t)=? m < (n? + 1).Since they have same parity (mod 2), it means m = n? or we have m+2 ? n?. First assume m = n?, then we have693.4. Solutions of type III,III?m log(??|?|)?m log(t? ??? ? t)< log(1 +2t3)=? m log( ??|?|t?????t)< log(1 +2t3).Using estimates of (3.4.2)m log(1 +1t3) < log(1 +2t3) =? m < 2.Therefore m = 1, n = ?1 which with a minus sign corresponds to solu-tion (t4 ? 2t, 1).Next we assume n? ? m+ 2. From first inequality of (3.3.8), we haven? log(t? ??? ? t)+ log(1 +1t3)< m log(??|?|)=? (m+ 2) log(t? ??? ? t)+ log(1 +1t3)< m log(??|?|)=? log((t? ??? ? t)2?(1 +1t3))< m log( ??|?|t?????t).Using estimations (3.4.2) we can concludelog(t12 ? 7t9)< m log(1 +2t3)(3.4.3)=?log(t12 ? 7t9)log(1 + 2t3) < m ? A? 2=?log(t12 ? 7t9)log(1 + 2t3) + 2 < A.Linear form in logarithmsFrom Siegel?s identity:(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (? ? ??)(x? ???y) = 0,703.4. Solutions of type III,III?we can find the linear form in logarithms corresponding to solutions oftype III:? = log??? ? ????? ? ?+ n log????t? ?t? ??????+m log??????????? (3.4.4)= log(1 +(? ? ??)(x? ???y)(??? ? ?)(x? ??y)), (3.4.5)where (????)(x????y)(?????)(x???y) < 0, so ? < 0.(?? ? ?)(x? ???y)(??? ? ?)(x? ??y)=?? ? ???? ? ?(t? ???t? ??)n( ?????)m(Since n < 0,m > 0) <(1t3 ? 3)((1t3 ? 3)3)n?<(1t3 ? 3)(3n?+1).If 0 < x < 12 then |log(1? x)| < 2x, So |?| < 2 ?(1t3?3)(3n?+1).=? log |?| < log(2)? (3n? + 1)(log(t3 ? 3)). (3.4.6)Lower bound for ? in type IIISo up to this point we have proved that if there exists a nontrivial solutionof type III, for F (x, y) = 1 then we haven? >log(t12 ? 7t9)log(1 + 2t3) + 2,n? > m+ 2.We will apply lemma (2.3) to find a lower bound for? = log??? ? ????? ? ?+ n log????t? ?t? ??????+m log??????????? .Letb1 = 1, b2 = n, b3 = m,n = 3, ?1 =??? ? ????? ? ?, ?2 = |t? ?t? ??|, ?3 = |???|.713.4. Solutions of type III,III?Now:h(??? ? ????? ? ?)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?))< 6 log th(????)=16log(????)2< 3 log th(t? ?t? ???)=16log(t? ???t? ??)2< 3 log t=? log|?| > ?8.344(10)15 log(t)3 log(68.3n?).Comparing with (3.4.6), we conclude? 8.344(10)15 log(t)3 log(68.3n?) < log(2)? (3n?+ 1)(log(t3 ? 3)). (3.4.7)Which leads to contradiction for t > 361165; therefore we haveLemma 3.33. For t > 361165, there is no integral solution of type III otherthan the trivial ones.Remark 3.1. From above inequality for t < 361166 we obtain n? < 7.5 ?1018.3.4.2 Solutions of type III?In this section we will prove thatLemma 3.34. For t > 361167, there is no integral solution of type III?other than the trivial ones.The proof is completely the same as solutions of type III. First we willshow that if (x, y) is an integral solution of type III? then y > 0. Usingequation (x ? y?)(x ? y??)(x ? y???) = 1, we conclude that x ? y??? > 0.Consider the equationx? y??x? y???=(t? ??t? ???)n( ?????)?m.Since RHS is positive, LHS is also positive =? x ? y?? = y(xy ? ??)> 0,but xy ??? > 0 =? y > 0. Let t4 +3t? 1t8 <xy < t4 +3t? 1t9 . Using equation(3.3.3) and estimations (3.1.4), we can obtain :0 <xy ? ?xy ? ?? =x? ?yx? ??y=(t? ?t? ??)n ( ???)?m.723.4. Solutions of type III,III?So n , m have same parities (mod 2). Taking absolute values of both sideswe get:xy ? ?xy ? ?? =(t? ??? ? t)n( |?|??)?m.Taking logarithms from both sides and using estimations 3.1.4 we obtain:log(1+1t3?2t6) < n log(t? ??? ? t)+m log(??|?|)< log(1+1t3?1t6). (3.4.8)Also we have:t6 + t3 + 1 <t? ??? ? t< t6 + 2t3 + 5, (3.4.9)t6 + 3t3 + 1 <??|?|< t6 + 3t3 + 3,1 +1t3?1t5<??|?|t?????t< 1 +1t3.Considering (3.4.8), we can deduce that n 6= 0,m 6= 0.Also ifm < 0 then n >0 and if m > 0 then n < 0.First assume m < 0, n > 0 =? |x? y???| > 1but by (3.3.2) |x? y???| < 1.so m > 0, n < 0. let n? = ?n from the second inequality of (3.3.30) we have:m log(??|?|)< n? log(t? ??? ? t)+ log(1 +1t3?1t6)=? m log(??|?|)< (n? + 1) log(t? ??? ? t)=? m < (n? + 1).since they have same parity (mod 2), it means m = n?or m+ 2 ? n?. Firstassume m = n?, then we havem log(??|?|)?m log(t? ??? ? t)< log(1 + 1/t3 ?1t6)=? m log( ??|?|t?????t)< log(1 +1t3?1t6).Using estimates of (3.4.9) we obtain:m log(1 +1t3?1t5) < log(1 +1t3?1t6) =? m < 2.733.4. Solutions of type III,III?Therefore m = 1, n = ?1 which with a minus sign corresponds to solution(t4 + 3t, 1).Next we assume n? ? m+ 2. From first inequality of (3.3.30), we have:n? log(t? ??? ? t)+ log(1 +1t3?2t6) < m log(??|?|),=? (m+ 2) log(t? ??? ? t)+ log(1 +1t3?2t6) < m log(??|?|),=? log((t? ??? ? t)2? (1 +1t3?2t6))< m log( ??|?|t?????t).Using estimations (3.4.9) we can conclude thatlog(t12 + 3t9 + 3t6) < m log(1 +1t3)=?log(t12 + 3t9 + 3t6)log(1 + 1t3 )< m ? A? 2=?log(t12 + 3t9 + 3t6)log(1 + 1t3 )+ 2 < A.(3.4.10)Linear form in logarithmsFrom Siegel?s identity :(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (?? ??)(x? ???y) = 0,we obtain? = log??? ? ????? ? ?+ nlog????t? ?t? ??????+m log??????????? (3.4.11)= log(1 +(?? ??)(x? ???y)(??? ? ?)(x? ??y)), (3.4.12)where (????)(x????y)(?????)(x???y) < 0. Moreover , we have:(?? ? ?)(x? ???y)(??? ? ?)(x? ??y)=?? ? ???? ? ?(t? ???t? ??)n( ?????)m(Since n < 0,m > 0) , <(1t3 + 1)((1t3 + 1)3)n?<(1t3 + 1)(3n?+1).743.4. Solutions of type III,III?If 0 < x < 12 then |log(1? x)| < 2x, So |?| < 2(1t3+1)(3n?+1)=? log |?| < log(2)? (3n? + 1)(log(t3 + 1)). (3.4.13)Lower bound for ? in type III?In this section we assume A >log(t12+3t9+3t6)log(1+ 1t3)+ 2 and n? ? m+ 2.To find alower bound for log|?| we apply lemma (2.3) . Recall that :? = log??? ? ????? ? ?+ n log????t? ?t? ??????+m log??????????? .Putb1 = 1, b2 = n, b3 = m,n = 3, ?1 =??? ? ????? ? ?, ?2 =????t? ?t? ?????? , ?3 =??????????? .We have:h(??? ? ??? ? ?)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?))<23log(t9 + 6t6),h(?????)=16log(????)2<13log(t9 + 7t6),h(t? ??t? ???)=16log(t? ???t? ??)2<13log(t9 + 5t6).Therefore we can take real numbers Ai as:A1 =4 log(t9 + 6t6),A2 =2 log(t9 + 5t6),A3 =2 log(t9 + 7t6),B =n?.=? log |?| > ?1.1446 ? 1013A1 ?A2 ?A3 log(68.3n?).Comparing with (3.4.13) we conclude that:?1.1446 ? 1013 log(t9 + 6t6)log(t9 + 5t6)log(t9 + 7t6)log(68.3n?)(3.4.14)< log(2)? (3n? + 1)(log(t3 + 1)).Which leads to contradiction for t > 361167.753.5. Small values of t3.5 Small values of tFor the remaining small values of t in each type of solution, we will use somediophantine techniques and run a computer search.3.5.1 Application of diophantine approximationThe following lemma of Mignotte [57], which is a variant of a result byBaker-Davenport, is essential for our computer search.Lemma 3.35. Let ? = ?? + ?? + ?, where ?, ? and ? are nonzero realnumbers and where ?, ? are rational integers, with |?| < A. Let Q > 0 be areal number; suppose that ?1 and ?2 satisfy?????1 ??????? <1100Q2and?????2 ??????? <1Q2.Let pq be a rational number with 1 ? q ? Q and????1 ?pq??? < 1q2 , and sup-pose q ||q?2|| ? 1.01A+ 2,(where ||?|| denote the distance to nearest integer)then|?| >|?|Q2. (3.5.1)The idea is to apply the above lemma to the corresponding linear form inlogarithms to each type of solution. We have all the data to apply the lemmato all types of solutions other than solutions of type I and II?. However sincewe used the Pade? approximation for solutions of type I and II?, applying thelemma requires some extra work.3.5.2 Solutions of type IDuring this section we assume 10 ? t < 8586 and let (x, y) be a solutionof type I. We have x? y? = (t? ?)n??m. The corresponding linear form inlogarithms for this solution is? = log? ? ???? ? ??+ n logt? ??t? ???+m log?????= log(1 +(?? ? ???)(x? ?y)(? ? ??)(x? ???y)).(3.5.2)Let A = max{1,m, n}. First, for all remaining values of t we find an absoluteupper bound for A by linear forms in logarithms; then, applying the lemma3.3, we find a lower bound for y value of nontrivial solution which leads usto find a lower bound for A. The bounds that we obtained in this way helpus treat the small values of t with the lemma 3.35.763.5. Small values of tFinding upper bound for A using linear forms in logarithmsFrom (3.3.3) we conclude0 <xy ? ???xy ? ?? =x? ???yx? ??y=(t? ???t? ??)n(?????)?m.Then by using estimations (3.1.3), we obtain:log(t3 ? 3) < n log(t? ???t? ??)?m log(?????)< log(t3). (3.5.3)By estimates (3.1.3) we have :t3 + 2 <xy ? ???xy ? ?? < t3 + 3, (3.5.4)t3 + 2 <?????< t3 + 3,t9 + 3t6 <(t? ???t? ??)< t9 + 5t6.in equation (3.5.3) if n = 0 then m = ?1, which implies x = 0, y = 1. Ifn > 0 then m > 0, so??x? y??? > 1. But it is easy to check that,??x? y??? < 1.Therefore we conclude that n < 0 and m < 0. Let n? = n,m? = m bepositive numbers; then we can rewrite (3.5.3)as:logxy ? ???xy ? ?? = m? log(?????)? n? log(t? ???t? ??). (3.5.5)Using above equation, we haven???log(t????t???)log(?????)?? < m? =?83n? < m?.From Siegel?s identity :(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (? ? ??)(x? ???y) = 0,we obtain? = log? ? ???? ? ??+ n logt? ??t? ???+m log?????= log(1 +(?? ? ???)(x? ?y)(? ? ??)(x? ???y)).(3.5.6)773.5. Small values of tMoreover, we have:????(?? ? ???)(x? ?y)(? ? ??)(x? ???y)???? =??? ? ???? ? ?(??? ? tt? ?)n? ( ????)m?(Since n? <38m?)<??? ? ???? ? ?(??? ? tt? ?) 38m? (????)m?<(t3 ? 3) (t3 ? 3) 38m?((1t3 ? 2)3)m?<(t3 ? 3) 38m?+1(1(t3 ? 3)3)m?<(t3 ? 3)?218 m?+1.If 0 < |x| < 12 then |log(1 + x)| < 2x, So |?| < 2(t3 ? 3)?213 m?+1=? log |?| < log(2) +(?21m?8+ 1)log(t3 ? 3). (3.5.7)Recall? = log? ? ????? ? ?+ nlog????t? ??t? ???????+mlog????????????? .To find a lower bound for the linear form in logarithms, we use theorem 2.3.Let ?1 = ????????? , ?2 =???t???t??????? , ?3 =??????????? , b1 = 1, b2 = n, b3 = m,D = 6.Using estimations (3.1.3), we have:h(? ? ???? ? ??)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?))< 6 log t,(3.5.8)h(?????)=16log(????)2< 3 log t,h(t? ??t? ???)=16log(t? ???t? ??)2< 3 log t.Therefore we can take A1 = 36 log(t), A2 = 18 log(t), A3 = 18 log(t) andB = m?. By above calculation we have:log|?| > ?8.344 ? (10)15 log(t)3 log(68.3m?).783.5. Small values of tComparing this with (3.5.7) :8.344(10)15 log(t)3 log(68.3m?)< log(2) +(?8m?3+ 1)log(t3 ? 3).(3.5.9)Since t < 8586, from above inequality we can conclude A < (4.03) ? 1018.Finding lower bound for A using gap principalSince t ? 10, by (3.2.5) and (3.2.7) we can see?2F?31< 122 ,?F > 72000. Theseinequalities, together with lemma (3.11), satisfy the conditions of lemma3.3. So if F (x, y) = 1 have another solution (x2, y2) of type I other than(x1, y1) = (1? t3, t8 ? 3t5 + 3t2) we have|?2| > 0.001??15F |?1|47( By (3.2.5) and (3.2.7), since t > 10) > t?3(t372)?15 (t252)47> t310=? H(x2, y2) > t620.On the other hand by (3.2.6):y2(t10) > H(x, y)=? y2 > t610.Now using equation (x? y?)(x? y??)(x? y???) = 1 and(3.3.3) we obtain:??(t? ?)n ??m?? =??????1y2(xy ? ??)(xy ? ???)??????.By (3.1.3), (3.1.5) and the lower bound we found for y2??????1y2(xy ? ??)(xy ? ???)??????<1t610 ? t ? t3< t?614.793.5. Small values of tAs for the left hand side,??(t? ?)n ??m?? =????1(t? ?)????n? ????m????(since1(t? ?)< 1 and n? <38m?)>????1(t? ?)????38m? ????m????(using estimates(3.1.3)) >(1t43) 38m? (1t5)m?= t?112 m?.=? 614 <112m? =? A = m? > 111. (3.5.10)Applying lemma 3.35We apply lemma 3.35 to the linear form :? = log? ? ???? ? ??+ n logt? ??t? ???+m log?????correspond to a nontrivial solution of type I.Put ? = log????????????? , ? = log????t? ??t? ??????? , ? = log? ? ???? ? ??, ? = m, ? = n,From(3.5.7)and (3.5.10), since t > 10, if the inequality (3.5.1)holds thenQ > 10870. We apply lemma 3.35 with the smaller value of Q being Q = 1060.Find ?1, ?2 with sufficient precision. Compute the denominators of principalconvergents of q < Q and check if the hypothesis of the lemma is satisfiedwith the smaller value of Q. Indeed this is the case for all 10 < t < 8586,and this contradiction completes the proof for small values of t. Using a codewritten in Pari/GP version, the verification takes less than half an hour.3.5.3 Solutions of type II?During this section we assume 10 ? t < 8586. Let (x, y) be a solution oftype I. We have x ? y? = (t ? ?)n??m. The corresponding linear form inlogarithms for this solution is? = log?? ????? ??+ n logt? ??t? ???+m log?????= log(1 +(?? ? ???)(x? ?y)(?? ??)(x? ???y)).(3.5.11)803.5. Small values of tLet A = max{1,m, n}. First for all remaining values of t we find an absoluteupper bound for A with linear forms in logarithms; then applying the lemma3.3 we find a lower bound for y value of a nontrivial solution which leads usto find a lower bound for A. The bounds that we obtained in this way helpus treat the small values of t by the lemma 3.5.1.Finding upper bound for A using linear forms in logarithmsFirst we will show that if (x, y) is an integral solution of this type then y > 0,Using equation (x?y?)(x?y??)(x?y???) = 1, we conclude that x?y? < 0.Consider the equationx? y??x? y???=(t? ??t? ???)n( ?????)?m.Since RHS is positive, LHS is also positive =? x ? y??? = y(xy ? ???)< 0,but xy ? ??? < 0 =? y > 0.Using equation (3.3.3) and estimations (3.1.4) we can obtain0 >xy ? ???xy ? ?=x? ???yx? ?y=(t? ???t? ?)n(????)?m.So n , m have different parities (mod 2). Taking absolute values of bothsides yields :??? ? xyxy ? ?=(??? ? tt? ?)n(???|?|)?m.Taking logarithms from both sides and using estimations 3.1.4 we have:log(t3 + 1) < n log(??? ? tt? ?)?m log(???|?|)< log(t3 + 2),t3 + 1 <??? ? tt? ?< t3 + 2, t9 + 6t6 <???|?|< t9 + 7t6 + 6.(3.5.12)Considering 3.3.4 , if m = 0 then n = 1 which, with positive sign, give usthe solution (t, 1), if m < 0 then n < 0 so |x ? y??| = |(t ? ??)n(??)?m| >1,but from(3.3.2) we have|x? y??| < 1. This contradiction leads us to conclude that m > 0? n > 0and from first inequality of (3.5.12) n > m; Moreover,mlog(???|?|)log(????tt??) < n =?83m < n.813.5. Small values of tFrom Siegel?s identity :(?? ? ???)(x? ?y) + (??? ? ?)(x? ??y) + (?? ??)(x? ???y) = 0,we obtain? = log??? ? ???? ? ?+ nlog????t? ?t? ???????+mlog???????????? (3.5.13)= log(1 +(?? ???)(x? ??y)(?? ? ?)(x? ???y)), (3.5.14)where (?????)(x???y)(????)(x????y) < 0 Using (3.1.3), we have:(??? ? ?)(x? ??y)(?? ? ?)(x? ???y)=??? ? ??? ? ?(t? ??t? ???)n(?????)m(since m <3n8), <??? ? ??? ? ?(t? ??t? ???)n(?????) 3n8< (t3 + 3)((1t3 + 1)3)n(t3 + 3)3n8<(1.01(t3 + 3)3)n(t3 + 3)3n8 +1< (t3 + 3)(?21n8 +1)(1.01)n.If 0 < x < 12 then |log(1? x)| < 2x, So |?| < 2(t3 ? 2)(?8n3 +1)(1.01)n=? log |?| < log(2) + n log(1.01) +(?21n8+ 1)log(t3 + 3) (3.5.15)Recall? = log??? ? ???? ? ?+ nlog????t? ?t? ???????+mlog???????????? .We can apply theorem (2.3) to find a lower bound for ?. Put ?1 = ?????????? ,?2 =???t??t??????? , ?3 =?????????? , b1 = 1, b2 = n, b3 = m,D = 6.823.5. Small values of tUsing estimations (3.1.4 )h(??? ? ???? ? ?)? 2h(??? ? ??)=23log((??? ? ??)(??? ? ?)(?? ? ?)),<23log(t9 + 6t6),h????t? ?t? ??????? =16log(????)2<13log(t9 + 7t6) (3.5.16)h???????????? =16log(t? ???t? ??)2<13log(t9 + 5t6).Therefore we can take real numbers Ai as:A1 =4 log(t9 + 6t6),A2 =2 log(t9 + 5t6),A3 =2 log(t9 + 7t6),B =n.We apply theorem (2.3) to find a lower bound for log |?|.log |?| > ?1.1446 ? 1013 ?A1 ?A2 ?A3 log (68.3n) .Comparing with (3.5.15) we obtain?1.1446 ? 1013 log(t9 + 6t6)log(t9 + 5t6)log(t9 + 7t6)log (68.3n)< log(2) + n log(1.01) +(?21n8+ 1)log(t3 + 3).Since n < 8586 we obtainA = n < 4.09 ? 1018.Lower bound for A using gap principalSince t ? 10, by (3.2.10 ) and (3.2.12), we can see for t > 10 ,?2F ??31<122 ,?F ? > 72000. These inequalities, together with lemma 3.15, satisfy theconditions of the lemma 3.3. So if F ?(x, y) = 1 has another solution (x2, y2)of type II? other than(x1, y1) = (t9 + 3t6 + 4t3 + 1, t8 ? 3t5 + 3t2) we have|?2| > 0.001??15 |?1|47( By (3.2.10) and (3.2.12), since t > 10) > t?3(t372)?15 (t252)47> t307=? H(x2, y2) > t614.833.5. Small values of tOn the other hand we havey2(t10 + 4t7 + 9t4) > H(x, y)=? y2 > t604.Now using equation (x? y?)(x? y??)(x? y???) = 1 and(3.3.3) we obtain:??(t? ??)n???m?? =??????1y2(xy ? ?)(xy ? ???)??????.By (3.1.3), (3.1.5) and the lower bound we found for y2??????1y2(xy ? ?)(xy ? ???)??????<1t604 ? t ? t4< t?609.As for the left hand side we have:??(t? ??)n???m?? =??(t? ??)??n????1??????m(since1??< 1 and m <38n)>??(t? ??)??n????1??????3n8(using estimates(3.1.4)) >(1t43)n( 1t163) 3n8= t??103 n.=? 609 <103n =? A = n > 182. (3.5.17)Applying lemma 3.35We apply lemma 3.35 to the linear form :? = log?? ????? ??+ n logt? ??t? ???+m log?????corresponding to a nontrivial solution of type II?.Put ? = log????????????? , ? = log????t? ??t? ??????? , ? = log?? ????? ??, ? = m, ? = n,From(3.5.7)and (3.5.17), since t > 10,if the inequality (3.5.1) holds we haveQ > 101421. We apply lemma 3.35 with a smaller value of Q, Q = 1060. Find843.5. Small values of t?1, ?2 with sufficient precision, then computing the denominators of principalconvergents of q < Q and check if the hypothesis of lemma is satisfied forthis smaller value of Q. This is the case for all 10 < t < 8586, and thiscontradiction completes the proof for small values of t. Using a code writtenin Pari/GP version, the verification takes less than half an hour.3.5.4 Solutions of type IIWe prove that there is no nontrivial solution of type II, for 10 ? t ? 1119749.We apply the lemma 3.35 for the corresponding linear forms in logarithms :? = log??? ? ???? ? ?+ nlog????t? ?t? ???????+mlog???????????? .Let A = max{m,n}. By inequality (3.20). Since t ? 10 we have A > 35.From inequality (3.3.11) since t ? 1119744 we have A < 1.35 ? 109. Andfinally from (3.3.7) since A > 35 if the the inequality (3.5.1) happens wehave Q > 10287. we apply lemma 3.35 with the smaller value of Q, Q = 1060.We checked the hypothesis of lemma for this smaller value of Q to get acontradiction. Using a code written in Pari/GP version,it takes less than 6hours to get contradiction for all values of 10 ? t ? 1119744.3.5.5 Solution of type I?To prove there is no nontrivial solution of type I? for 10 ? t ? 712018. Weperform exactly the same steps as the section 3.5.4 . The correspondinglinear form is? = log?? ????? ??+ n logt? ??t? ???+m log?????.Let A = max{m,n}. By inequality (3.3). Since t ? 10 we have A > 99. Frominequality (3.3.33) since t ? 712018 we have A < 1.85 ?109. And finally from(3.3.19) since A > 99, if the the inequality (3.5.1) happens then Q > 10772.we apply lemma 3.35 with the smaller value Q = 1060. Using a code writtenin Pari/GP version,it takes less than 5 hours to got contradiction for allvalues of 10 ? t ? 722018.3.5.6 Solutions of type IIITo prove there is no nontrivial solution of type III for 10 ? t ? 361165.We perform exactly the same steps as the section 3.5.4. The corresponding853.5. Small values of tlinear form is:? = log??? ? ????? ? ?+ n log????t? ?t? ??????+m log??????????? .Let A = max{m,n}. By inequality (3.4.3). Since t ? 10 we have A > 13827.From inequality (3.4.7) since t ? 361165 we have A < 3.62 ?1018. And finallyfrom (3.4.6) since A > 13827 if the the inequality (3.5.1) happens we haveQ > 1012439. we apply lemma 3.35 with Q = 1060. Using a code written inPari/GP version,it takes less than 4 hours to got contradiction for all valuesof 10 ? t ? 361165.3.5.7 Solutions of type III?To prove there is no nontrivial solution of type III? for 10 ? t ? 361167.We perform exactly the same steps as the section 3.5.4. The correspondinglinear form is:? = log??? ? ????? ? ?+ n log????t? ?t? ??????+m log??????????? .Let A = max{m,n}. By lemma 3.4.10. Since t ? 10 we have A > 27649.From inequality (3.4.14) since t ? 361167 we have A < 7.23 ? 1018. Andfinally from (3.4.13) since A > 27649 , if the inequality (3.5.1) happenswe have Q > 1024873. we apply lemma 3.35 with a smaller value of Q beingQ = 1060. Using a code written in Pari/GP version,it takes less than 4 hoursto got contradiction for all values of 10 ? t ? 361167.86Chapter 4Mordell curves4.1 IntroductionIf k is a nonzero integer, then the equationY 2 = X3 + k (4.1.1)defines an elliptic curve over Q. Such Diophantine equations have a longhistory, dating back (at least) to work of Bachet in the 17th century, and arenowadays termed Mordell equations, honouring the substantial contributionsof L. J. Mordell to their study. Indeed the statement that, for a given k 6= 0,equation (4.1.1) has at most finitely many integral solutions is implicit inwork of Mordell [63] (via application of a result of Thue [81]), and explicitlystated in [64].Historically, the earliest approaches to equation (4.1.1) for certain specialvalues of k appealed to simple local arguments; references to such work maybe found in Dickson [24]. More generally, working in either Q(??k) orQ( 3?k), one is led to consider a finite number of Thue equations of the shapeF (x, y) = m, where the m are nonzero integers and the F are, respectively,binary cubic or quartic forms with rational integer coefficients. Via classicalarguments of Lagrange (see e.g. page 673 of Dickson [24]), these in turncorrespond to a finite (though typically larger) collection of Thue equationsof the shape G(x, y) = 1. Here, again, the G are binary cubic or quarticforms with integer coefficients. In case k is positive, one encounters cubicforms of negative discriminant which may typically be treated rather easilyvia Skolem?s p-adic method (as the corresponding cubic fields have a singlefundamental unit). For negative values of k, one is led to cubic or quarticfields with a pair of fundamental units, which may sometimes be treated bysimilar if rather more complicated methods; see e.g. [30], [38] and [53].There are alternative approaches for finding the integral points on a givenmodel of an elliptic curve. The most commonly used currently proceeds viaappeal to lower bounds for linear forms in elliptic logarithms, the idea forwhich dates back to work of Lang [49] and Zagier [89] (though the boundsrequired to make such arguments explicit are found in work of David and874.1. Introductionof Hirata-Kohno, see e.g. [21]). Using these bounds, Gebel, Petho? and Zim-mer [31], Smart [74] and Stroeker, Tzanakis [77] obtained, independently, a?practical? method to find integral points on elliptic curves. Applying thismethod, in 1998, Gebel, Petho?, Zimmer [32] solved equation (4.1.1) for allintegers |k| < 104 and partially extended the computation to |k| < 105. Asa byproduct of their calculation, they obtained a variety of interesting infor-mation about the corresponding elliptic curves, such as their ranks, genera-tors of their Mordell-Weil groups, and information on their Tate-Shafarevicgroups.The only obvious disadvantage of this approach is its dependence uponknowledge of the Mordell-Weil basis over Q of the given Mordell curve (in-deed, it is this dependence that ensures, with current technology at least,that this method is not strictly speaking algorithmic). For curves of largerank, in practical terms, this means that the method cannot be guaranteedto solve equation (4.1.1).Our goal in this paper is to present (and demonstrate the results of) apractical algorithm for solving Mordell equations with values of k in a some-what larger range. At its heart are lower bounds for linear forms in complexlogarithms, stemming from the work of Baker [3]. These were first appliedin the context of explicitly solving equation (4.1.1), for fixed k, by Ellisonet al [27]. To handle values of k in a relatively large range, we will appeal toclassical invariant theory and, in particular, to the reduction theory of bi-nary cubic forms (where we have available very accessible, algorithmic workof Belabas [6], Belabas and Cohen [7] and Cremona [17]). This approachhas previously been outlined by Delone and Fadeev (see Section 78 of [22])and Mordell (see e.g. [65]; its origins lie in [62]). We use it to solve Mordell?sequation (4.1.1) for all k with 0 < |k| ? 107. We should emphasize that ouralgorithm does not provide a priori information upon, say, the ranks of thecorresponding elliptic curves, though values of k for which (4.1.1) has manysolutions necessarily (as long as k is 6-th power free) provide curves with atleast moderately large rank (see [34]).We proceed as follows. In Section 4.2, we will discuss the precise corre-spondence that exists between integer solutions to (4.1.1) and integer solu-tions to cubic Thue equations of the shape F (x, y) = 1, for certain binarycubic forms of discriminant ?108k. In Section 4.3, we indicate a methodto choose representatives from equivalence classes of forms of a given dis-criminant. Section 4.4 contains a brief discussion of our computation, whileSection 4.5 is devoted to presenting a summary of our data, including infor-mation on both the number of solutions, and on their heights.884.2. Preliminaries4.2 PreliminariesIn this section, we begin by outlining a correspondence between integer so-lutions to the equation Y 2 = X3 + k and solutions to certain cubic Thueequations of the form F (x, y) = 1, where F is a binary cubic form of dis-criminant ?108k. As noted earlier, this approach is very classical. To makeit computationally efficient, however, there are a number of details that wemust treat rather carefully.Let us suppose that a, b, c and d are integers, and consider the binarycubic formF (x, y) = ax3 + 3bx2y + 3cxy2 + dy3, (4.2.1)with discriminantD = DF = ?27(a2d2 ? 6abcd? 3b2c2 + 4ac3 + 4b3d).It is important to observe (as a short calculation reveals) that the set offorms of the shape (4.2.1) is closed within the larger set of binary cubicforms in Z[x, y], under the action of both SL2(Z) and GL2(Z). To such aform we associate covariants, namely the Hessian H = HF (x, y) given byH = HF (x, y) = ?14(?2F?x2?2F?y2?(?2F?x?y)2)and the Jacobian determinant of F and H, a cubic form G = GF defined asG = GF (x, y) =?F?x?H?y??F?y?H?x.Note that, explicitly, we haveH/9 = (b2 ? ac)x2 + (bc? ad)xy + (c2 ? bd)y2andG/27 = a1x3 + 3b1x2y + 3c1xy2 + d1y3,wherea1 = ?a2d+ 3abc? 2b3, b1 = ?b2c? abd+ 2ac2, c1 = bc2 ? 2b2d+ acdand d1 = ?3bcd+ 2c3 + ad2.Crucially for our arguments, these covariants satisfy the syzygy4H(x, y)3 = G(x, y)2 + 27DF (x, y)2.894.2. PreliminariesDefining D1 = D/27, H1 = H/9 and G1 = G/27, we thus have4H1(x, y)3 = G1(x, y)2 +D1F (x, y)2.If (x0, y0) satisfies the equation F (x0, y0) = 1 and D1 ? 0 (mod 4) (i.e.if ad ? bc (mod 2)), then necessarily G1(x0, y0) ? 0 (mod 2). We maytherefore conclude that Y 2 = X3 + k, whereX = H1(x0, y0), Y =G1(x0, y0)2and k = ?D14= ?D108.It follows that, to a given triple (F, x0, y0), where F is a cubic formas in (4.2.1) with discriminant ?108k, and x0, y0 are integers for whichF (x0, y0) = 1, we can associate an integral point on the Mordell curveY 2 = X3 + k.Conversely suppose, for a given integer k, that (X,Y ) satisfies equation(4.1.1). To the pair (X,Y ), we associate the cubic formF (x, y) = x3 ? 3Xxy2 + 2Y y3.Such a form F is of the shape (4.2.1), with discriminantDF = ?108Y2 + 108X3 = ?108kand covariants satisfyingX =G1(1, 0)2=G(1, 0)54and Y = H1(1, 0) =H(1, 0)9.In summary, there exists a correspondence between the set of integralsolutionsSk = {(X1, Y1), . . . , (XNk , YNk)}to the Mordell equation Y 2 = X3 + k and the set Tk of triples (F, x, y),where each F is a binary cubic form of the shape (4.2.1), with discriminant?108k, and the integers x and y satisfy F (x, y) = 1. Note that the forms Funder consideration here need not be irreducible.In the remainder of this section, we will show that there is, in fact, abijection between Tk under SL2(Z)-equivalence, and the set Sk. We beginby demonstrating the following pair of lemmata.Lemma 4.2.1. Let k be a nonzero integer and suppose that F1 and F2 areSL2(Z)-inequivalent binary cubic forms of the shape (4.2.1), each with dis-criminant ?108k, and that x1, y1, x2 and y2 are integers such that (F1, x1, y1)and (F2, x2, y2) are in Tk. Then the tuples (F1, x1, y1) and (F2, x2, y2) cor-respond to distinct elements of Sk.904.2. PreliminariesProof. Suppose that the tuples (F1, x1, y1) and (F2, x2, y2) are in Tk, so thatF1(x1, y1) = F2(x2, y2) = 1.Since, for each i, xi and yi are necessarily coprime, we can find integers miand ni such that mixi?niyi = 1, for i = 1, 2. Writing ?i = (xi niyi mi ), we thushaveFi ? ?i(x, y) = x3 + 3bix2y + 3cixy2 + diy3,for integers bi, ci and di, and hence, under the further action of ?i =(1 ?bi1 0),we observe that Fi is SL2(Z)-equivalent tox3 ? 3pixy2 + 2qiy3,wherepi =GFi(xi, yi)54and qi =HFi(xi, yi)9.If the two tuples correspond to the same element of Sk, necessarilyGF1(x1, y1) = GF2(x2, y2) and HF1(x1, y1) = HF2(x2, y2),contradicting our assumption that F1 and F2 are SL2(Z)-inequivalent.Lemma 4.2.2. Suppose that k is a nonzero integer, that F is a binary cubicform of the shape (4.2.1) and discriminant ?108k, and that F (x0, y0) =F (x1, y1) = 1 where (x0, y0) and (x1, y1) are distinct pairs of integers. Thenthe tuples (F, x0, y0) and (F, x1, y1) correspond to distinct elements of Sk.Proof. Via SL2(Z)-action, we may suppose, without loss of generality, thatF (x, y) = x3 + 3bx2y + 3cxy2 + dy3 and that (x0, y0) = (1, 0). If the triples(F, 1, 0) and (F, x1, y1) correspond to the same element of Sk, necessarilyGF (1, 0) = GF (x1, y1) and HF (1, 0) = HF (x1, y1),wherebyx31 + 3bx21y1 + 3cx1y21 + dy31 = 1, (4.2.2)(b2 ? c)x21 + (bc? d)x1y1 + (c2 ? bd)y21 = (b2 ? c) (4.2.3)anda1x31 + 3b1x21y1 + 3c1x1y21 + d1y31 = a1. (4.2.4)It follows that3(ba1 ? b1)x21y1 + 3(ca1 ? c1)x1y21 + (da1 ? d1)y31 = 0.914.2. PreliminariesSince (x1, y1) 6= (1, 0), we have that y1 6= 0 and so3(ba1 ? b1)x21 + 3(ca1 ? c1)x1y1 + (da1 ? d1)y21 = 0,i.e.?3(b2? c)2x21 +3(b2? c)(d? bc)x1y1 +(3bcd? b3d? c3?d2)y21 = 0. (4.2.5)If b2 = c, it follows that3bcd? b3d? c3 ? d2 = ?(d? b3)2 = 0,so that d = b3 and F (x, y) = (x + by)3, contradicting our assumption thatDF 6= 0. We thus have(b2 ? c)x21 + (bc? d)x1y1 =(3bcd? b3d? c3 ? d2)3(b2 ? c)y21and so(3bcd? b3d? c3 ? d2)3(b2 ? c)y21 = (bd? c2)y21 + b2 ? c,i.e.(?d2 + 6bcd? 4b3d? 4c3 + 3b2c2)y21 = 3(b2 ? c)2and soDF y21 = 81(b2 ? c)2.Since DF = ?108k, it follows that k = ?3m2 for some integer m, where2my1 = b2 ? c. From (4.2.5), we thus have?12m2x21 + 6(d? bc)mx1 + 3bcd? b3d? c3 ? d2 = 0,wherebyx1 =14m(d? bc? 2) .Substituting the expressions for x1 and y1 into equation (4.2.5), we conclude,from b2 6= c, that?34(d? bc? 2)2 +32(d? bc)(d? bc? 2) + 3bcd? b3d? c3 ? d2 = 0,whence, since the left-hand-side of this expression is just DF /27 ? 12, wefind that DF = 324. It follows that the triple (F, 1, 0), say, corresponds toan integral solution to the Mordell equation Y 2 = X3?3. Adding 4 to bothsides of this equation, however, we observe that necessarily X2 ?X + 1 ?3 (mod 4), contradicting the fact that it divides the sum of two squaresY 2 + 4. The lemma thus follows as stated.924.3. Finding representative formsTo conclude as desired, we have only to note that, for any ? ? SL2(Z),covariance implies that HF?? = HF ?? and GF?? = GF ??, and hence triples(F, x0, y0) and (F ? ?, ?(x0), ?(y0))in Tk necessarily correspond to the same solution to (4.1.1) in Sk.Remark 4.1. Instead of working with SL2(Z)-equivalence, we can insteadconsider GL2(Z)-equivalence classes (and, as we shall see in the next sec-tion, this equivalence is arguably a more natural one with which to work).Since H(x, y) and G2(x, y) are GL2(Z)-covariant, if two forms are equiva-lent under the action of GL2(Z), but not under SL2(Z), then we haveHF?? = HF ? ? and GF?? = ?GF ? ?.It follows that, in order to determine all pairs of integers (X,Y ) satisfyingequation (4.1.1), it is sufficient to find a representative for each GL2(Z)-equivalence class of forms of shape (4.2.1) and discriminant ?108k and, foreach such form, solve the corresponding Thue equation. A pair of integers(x0, y0) for which F (x0, y0) = 1 now leads to a pair of solutions (X,?Y ) toY 2 = X3 + k, whereX = H1(x0, y0) and Y = G1(x0, y0)/2,at least provided G1(x0, y0) 6= 0.4.3 Finding representative formsAs we have demonstrated in the previous section, to solve Mordell?s equationfor a given integer k, it suffices to determine a set of representatives forSL2(Z)-equivalence classes (or, if we prefer, GL2(Z)-equivalence classes) ofbinary cubic forms of the shape (4.2.1), with discriminant ?108k, and thensolve the corresponding Thue equations F (x, y) = 1. In this section, we willdescribe how to find distinguished representatives for equivalence classes ofcubic forms with a given discriminant. In all cases, the various notions ofreduction arise from associating to a given cubic form a particular definitequadratic form ? in case of positive discriminant, the Hessian defined earlierworks well. In what follows, we will state our definitions of reduction solelyin terms of the coefficients of the given cubic form, keeping the role of theassociated quadratic form hidden from view.934.3. Finding representative forms4.3.1 Forms of positive discriminantIn case of positive discriminant forms (i.e. those corresponding to negativevalues of k), there is a well-developed classical reduction theory, dating backto work of Hermite [39], [40] and later applied to great effect by Davenport(see e.g. [18], [19] and [20]). This procedure allows us to determine distin-guished reduced elements within each equivalence class of forms. We can, infact, apply this reduction procedure to both irreducible and reducible forms;initially we will assume the forms we are treating are irreducible, for reasonswhich will become apparent. We will follow work of Belabas [6] (see alsoBelabas and Cohen [7] and Cremona [17]), in essence a modern treatmentand refinement of Hermite?s method.Definition 4.1. An irreducible binary integral cubic formF (x, y) = ax3 + 3bx2y + 3cxy2 + dy3of positive discriminant is called reduced if we have? |bc? ad| ? b2 ? ac ? c2 ? bd,? a > 0, b ? 0, where d < 0 whenever b = 0,? if bc = ad, d < 0,? if b2 ? ac = bc? ad, b < |a? b|, and? if b2 ? ac = c2 ? bd, a ? |d| and b < |c|.The main value of this notion of reduction is apparent in the followingresult (Corollary 3.3 of [6]).Proposition 4.1. Any irreducible cubic form of the shape (4.2.1) with pos-itive discriminant is GL2(Z)-equivalent to a unique reduced one.To determine equivalence classes of reduced cubic forms with boundeddiscriminant, we will appeal to the following result (immediate from Lemma3.5 of Belabas [6]).Lemma 4.3.1. Let K be a positive real number andF (x, y) = ax3 + 3bx2y + 3cxy2 + dy3be a reduced form whose discriminant lies in (0,K]. Then we have1 ? a ?2K1/43?3944.3. Finding representative formsand0 ? b ?a2+13(?K ?27a24)1/2.If we denote by P2 the unique positive real solution of the equation?4P 32 + (3a+ 6b)2P 22 + 27a2K = 0,then9b2 ? P29a? c ? b? a.In practice, to avoid particularly large loops on the coefficients a, b, c andd, we will instead employ a slight refinement of this lemma to treat formswith discriminants in the interval (K0,K] for given positive reals K0 < K.It is easy to check that we have9b2 ? P29a? c ? min{3b2 ? (a2K0/4)1/33a, b? a}.To bound d, we note that the definition of reduction implies that(a+ b)c? b2a? d ?(a? b)c? b2a.The further assumption that K0 < DF ? K leads us to a quadratic equationin d which we can solve to determine a second interval for d. Intersectingthese intervals provides us with (for values of K that are not too large) areasonable search space for d.4.3.2 Forms of negative discriminantIn case of negative discriminant, we require a different notion of reduction, asthe Hessian is no longer a definite form. We will instead, following Belabas[6], appeal to an idea of Berwick and Mathews [54]. We take as our definitionof a reduced form an alternative characterization due to Belabas (Lemma4.2 of [6]).Definition 4.2. An irreducible binary integral cubic formF (x, y) = ax3 + 3bx2y + 3cxy2 + dy3of negative discriminant is called reduced if we have? d2 ? a2 > 3(bd? ac),954.3. Finding representative forms? ?(a? 3b)2 ? 3ac < 3(ad? bc) < (a+ 3b)2 + 3ac,? a > 0, b ? 0 and d > 0 whenever b = 0.Analogous to Proposition 4.1, we have, as a consequence of Lemma 4.3of [6] :Proposition 4.2. Any irreducible cubic form of the shape (4.2.1) with neg-ative discriminant is GL2(Z)-equivalent to a unique reduced one.To count the number of reduced cubic forms in this case we appeal toLemma 4.4 of Belabas [6] :Lemma 4.3.2. Let K be a positive real number andF (x, y) = ax3 + 3bx2y + 3cxy2 + dy3be a reduced form whose discriminant lies in [?K, 0). Then we have1 ? a ?(16K27)1/40 ? b ?a2+13(?K/3?3a24)1/21? 3b ? 3c ?(K4a)1/3+{3b2/a if a ? 2b,3b? 3a/4 otherwise.As in the case of forms of positive discriminant, from a computationalviewpoint it is often useful to restrict our attention to forms with discrimi-nant ? with ?? ? (K0,K] for given 0 < K0 < K. Also as previously, theloop over d is specified by the inequalities defining reduced forms and by thedefinition of discriminant.It is worth noting here that a somewhat different notion of reductionfor cubic forms of negative discriminant is described in Cremona [17], basedon classical work of Julia [48]. Under this definition, one encounters rathershorter loops for the coefficient a ? it appears that this leads to a slight im-provement in the expected complexity of this approach (though the numberof tuples (a, b, c, d) considered is still linear in K).964.3. Finding representative forms4.3.3 Reducible formsSuppose finally that F is a reducible cubic form of discriminant ?108k, asin (4.2.1), for which F (x0, y0) = 1 for some pair of integers x0 and y0. Then,under SL2(Z)-action, F is necessarily equivalent tof(x, y) = x(x2 + 3Bxy + 3Cy2), (4.3.1)for certain integers B and C. We thus haveDf = DF = 27C2(3B2 ? 4C) (4.3.2)(so that necessarily BC ? 0 (mod 2)). Almost immediate from (4.3.2), wehaveLemma 4.3.3. Let K > 0 be a real number and suppose that f is a cubicform as in (4.3.1). If we have 0 < Df ? K then?(K108)1/3? C ?(K27)1/2, C 6= 0andmax{0,(4C + 13)1/2}? B ?(K + 108C381C2)1/2.If, on the other hand, ?K ? Df < 0 then1 ? C ?(K27)1/2andmax{0,(?K + 108C381C2)1/2}? B ?(4C ? 13)1/2.One technical detail that remains for us, in the case of reducible forms,is that of identifying SL2(Z)-equivalent forms. If we havef1(x, y) = x(x2 + 3B1xy + 3C1y2) and f2(x, y) = x(x2 + 3B2xy + 3C2y2),withf1 ? ?(x, y) = f2(x, y) and ?(1, 0) = (1, 0), (4.3.3)where ? ? SL2(Z), then necessarily ? = ( 1 n0 1 ) , for some integer n, so thatB2 = B1 + n, C2 = C1 + 2B1n+ n2 and 3C1n+ 3B1n2 + n3 = 0.974.4. Running the algorithmAssuming n 6= 0, it follows that f1(x, 1) factors completely over Z[x], whereby,in particular, C1 = 3C0 for C0 ? Z and B21 ? 4C0 is a perfect square, sayB21 ? 4C0 = D20 (where D0 6= 0 if we assume that Df1 6= 0). We may thuswriten =?3B1 ? 3D02,whence there are precisely three pairs (B2, C2) satisfying (4.3.3), namely(B2, C2) = (B1, C1),(?B1 + 3D02,32D0(D0 ?B1))and (?B1 ? 3D02,32D0(D0 +B1)).Let us define a notion of reduction for forms of the shape (4.3.1) asfollows :Definition 4.3. A irreducible binary integral cubic formF (x, y) = x(x2 + 3bxy + 3cy2)of nonzero discriminant DF is called reduced if we have either? DF is not the square of an integer, or? DF is the square of an integer, and b and c are positive.From the preceding discussion, it follows that such reduced forms areunique in their SL2(Z)-class. Note that the solutions to the equationF (x, y) = x(x2 + 3bxy + 3cy2) = 1are precisely those given by (x, y) = (1, 0) and, if c | b, (x, y) = (1,?b/c).4.4 Running the algorithmWe implement the algorithm implicit in the preceding sections for findingthe integral solutions to equations of the shape (4.1.1) with |k| ? K, forgiven K > 0. The number of cubic Thue equations F (x, y) = 1 which weare required to solve is of order K. To handle these equations, we appealto by now well-known arguments of Tzanakis and de Weger [85] (which, asnoted previously, are based upon lower bounds for linear forms in complexlogarithms, together with lattice basis reduction); these are implemented984.5. Numerical resultsin several computer algebra packages, including Magma and Pari (Sage).We used the former despite concerns over its reliance on closed-source code,primarily due to its stability for longer runs. The main computational bot-tleneck in this approach is typically that of computing the fundamentalunits in the corresponding cubic fields; for computations with K = 107, weencountered no difficulties with any of the Thue equations arising (in par-ticular, the fundamental units occurring can be certified without relianceupon the Generalized Riemann Hypothesis). We are unaware of a compu-tational complexity analysis, heuristic or otherwise, of known algorithmsfor solving Thue equations and hence it is not by any means obvious howtimings for this approach should compare to that based upon lower boundsfor linear forms in elliptic logarithms, as in [31]. Using a ?stock? version ofMagma, our approach does have the apparent advantage of not crashing forparticular values of k in the range under consideration (i.e with K = 107).4.5 Numerical resultsThe full output for our computation is available at the weblinkhttp://www.math.ubc.ca/~bennett/BeGa-data.html,which documents the results of a month-long run on a MacBookPro. Re-alistically, with sufficient perseverance and suitably many machines, oneshould be able to readily extend the results described here to something likeK = 1010. In the remainder of this section, we will briefly summarize ourdata.4.5.1 Number of solutionsIn what follows, we tabulate the number of curves encountered for whichequation (4.1.1) has a given number Nk of integer solutions, with |k| ? 107.We include results for all values of k, and also those obtained by restrictingto 6-th power free k (the two most natural restrictions here are, in ouropinion, this one or the restriction to solutions with gcd(X,Y ) = 1). In therange under consideration, the maximum number of solutions encounteredfor positive values of k was 58, corresponding to the case k = 3470400. Fornegative values of k, the largest number of solutions we found was 66, fork = ?9754975.Regarding the largest value of Nk known (where, to avoid trivialities,we can, for example, consider only 6-th power free k), Noam Elkies kindly994.5. Numerical resultsprovided the following example, found in October of 2009 :k = 509142596247656696242225 = 52 ? 73 ? 112 ? 192 ? 149 ? 587 ? 15541336441.This value of k corresponds to an elliptic curve of rank (at least) 12, with(at least) 125 pairs of integral points (i.e. Nk ? 250), with X-coordinatesranging from ?79822305 to 801153865351455.Table 4.1 Number of Mordell curves with Nk integral points for positivevalues of k with 0 < k ? 107Nk # of curves Nk # of curves Nk # of curves0 8667066 12 3890 32 331 108 14 2186 34 182 1103303 16 1187 36 283 34 17 1 38 174 145142 18 589 40 115 33 20 347 42 66 55518 22 197 44 37 8 24 148 46 58 13595 26 91 48 29 6 28 63 56 110 6308 30 55 58 1Table 4.2 Number of Mordell curves with Nk integral points for negativevalues of k with |k| ? 107Nk # of curves Nk # of curves Nk # of curves0 9165396 11 4 32 81 167 12 1351 34 82 729968 14 655 36 23 23 16 340 38 14 67639 18 238 40 15 10 20 160 42 16 23531 22 107 44 27 3 24 71 46 28 7318 26 37 48 19 6 28 20 50 210 2912 30 15 66 11004.5. Numerical resultsTable 4.3 Number of Mordell curves with Nk integral points for positivek ? 107 6-th power freeNk # of curves Nk # of curves Nk # of curves0 8545578 12 3575 32 291 79 14 1998 34 182 1067023 16 1055 36 223 24 17 1 38 154 139090 18 506 40 95 10 20 278 42 66 51721 22 161 44 17 2 24 112 46 58 12271 26 76 48 29 1 28 58 56 110 5756 30 44Table 4.4 Number of Mordell curves with Nk integral points for negativek, |k| ? 107 6-th power freeNk # of curves Nk # of curves Nk # of curves0 9026739 11 1 32 81 109 12 1174 34 82 705268 14 562 36 23 12 16 291 38 14 63685 18 197 42 15 5 20 138 44 26 21883 22 96 46 27 3 24 68 48 18 6644 26 31 50 19 3 28 17 66 110 2561 30 134.5.2 Number of solutions by rankFrom a result of Gross and Silverman [34], if k is a 6-th power free integer,then the number of integral solutions Nk to equation (4.1.1) is boundedabove by a constant N(r) that depends only on the Mordell-Weil rank over Qof the corresponding elliptic curve. It is easy to show that we have N(0) = 6,corresponding to k = 1. For larger ranks, we have that N(1) ? 12 (wherethe only example of a rank 1 curve we know with Nk = 12 corresponds to1014.5. Numerical resultsk = 100), N(2) ? 26 (where N225 = 26), N(3) ? 46 (with N1334025 = 46),N(4) ? 56 (with N5472225 = 56), N(5) ? 50 (with N?9257031 = 50) andN(6) ? 66 (where N?9754975 = 66). The techniques of Ingram [46] mightenable one to prove that indeed one has N(1) = 12.4.5.3 Hall?s conjecture and large integral pointsSharp upper bounds for the heights of integer solutions to equation (4.1.1)are intimately connected to the ABC-conjecture of Masser and Oesterle. Inthis particular context, we have the following conjecture of Marshall Hall :Conjecture 2. (Hall) Given > 0, there exists a positive constant C sothat, if k is a nonzero integer, then the inequality|X| < C|k|2+holds for all solutions in integers (X,Y ) to equation (4.1.1).The original statement of this conjecture, in [35], actually predicts thata like inequality holds for = 0. The current thinking is that such a resultis unlikely to be true (though it has not been disproved).We next list all the Mordell curves encountered with Hall MeasureX1/2/|k|exceeding 1; in each case we round this measure to the second decimal place.In the range under consideration, we found no new examples to supplementthose previously known and recorded in Elkies [26] (note that the case withk = ?852135 was omitted from this paper due to a transcription error) andin work of Jime?nez Calvo, Herranz and Sa?ez [47].Table 4.5 Hall?s conjecture extrema for |k| ? 107k X X1/2/|k| k X X1/2/|k|-1641843 5853886516781223 46.60 1 2 1.411090 28187351 4.87 14668 384242766 1.3417 5234 4.26 14857 390620082 1.33225 720114 3.77 -2767769 12438517260105 1.2724 8158 3.76 8569 110781386 1.23-307 939787 3.16 -5190544 35495694227489 1.15-207 367806 2.93 -852135 952764389446 1.1528024 3790689201 2.20 11492 154319269 1.08117073 65589428378 2.19 618 421351 1.054401169 53197086958290 1.66 297 93844 1.031024.5. Numerical resultsOur final table lists all the values of k in the range under considerationfor which equation (4.1.1) has a sufficiently large solution :Table 4.6 Solutions to (4.1.1) with X > 1012 for |k| ? 107k Nk X k Nk X-1641843 6 5853886516781223 -1923767 2 24348907386264401169 6 53197086958290 -2860984 4 2115366915022-5190544 4 35495694227489 2383593 10 1854521158546-4090263 2 16544006443618 2381192 10 1852119707102-4203905 2 15972973971249 -4024909 4 1569699004069-2767769 4 12438517260105 -9218431 6 11838580501757008155 2 9137950007869 5066001 8 1026067837540-9698283 2 7067107221619 5059001 8 10242453374602214289 4 4608439927403 3537071 6 1007988055117In particular, by way of example, the equationY 2 = X3 ? 4090263thus has the feature that its smallest (indeed only) solution in positive in-tegers is given by(X,Y ) = (16544006443618, 67291628068556097113).103Chapter 5Bombieri method5.1 IntroductionIn 1996, Bombieri, van der Poorten and Vaaller [14] applied Bombieri?smethod to algebraic number of degree 3, and under a certain assumption,obtained an effective irrationality measure for some set of algebraic num-bers. Their method originated in the method introduced in section 2.1.3,using Dyson?s lemma in an essential way. They also introduce new technicaldevices. In this chapter we will closely follow [14] , and explicitly find theconstants in section 9 of [14] to solve inequality??x3 + pxy2 + qy3?? ? k, (5.1.1)under some restrictions on p and q.The outline of this method is as follows. Let K be a cubic extension ofk and v some valuation on k. Identify K with an embedding of K in kv.Fix an embedding of the Galois closure L of K/k, then determine a faithfulprojective representation ? ? P? of G = Gal(L/k) into PGL(2,Q). Nextstep is to define a subset ?(p) ? P1(L) byDefinition 5.1.?(P ) = {? ? P1(L) : ??1 = P?? for all ? in G}.For the points ? in ?(P ), we will show a new version of the Thue-Siegelprinciple, by introducing new metric ?v on P1(k?v). We see that the point[01]in P1(L), cannot have two excellent approximations ?1 and ?2 in ?(P ).for which both H(?1) and H(?2) are large. This will be expressed as an in-equality in which all the constants are explicitly given. It will be an effectiveversion of Thue-Siegel?s principal. We use this to find effective irrationalitymeasure for all generators of the cubic extension K/k. Finally, for the theirreducible polynomial x3 +px+q. we use the roots of f to construct a point?1 in ?(P ), which is a good approximation for the point[01]. According to1045.2. Preliminiariesthe Thue-Siegel?s principle we find an effective irrationality measure for thesmallest root of f and solve the Thue inequality (5.1.1).5.2 PreliminiariesLet k be a number field ,and d = [K : Q] . For every place v of K definedv = [Kv : Qv]. We will use two absolute values | |v and ? ?v; in particular,we have |x|v = ?x?dv/dv . The absolute value | |v is normalized as? if v|p then|p|v = p?[kv :Qv ]k:Q ,? if v|? and v is real then|x|v = |x|1/d,? if v|? and v is complex then|x|v = |x|2/d.Let Mk denote the set of places of K. In view of normalization, we havethe product formula?v?Mk|x|v = 1.We also define the absolute height of an algebraic number x ? K byH(x) =?v?Mkmax(1, |x|v),so thath(x) = logH(x) =?v?Mklog+ |x|v,where log+ x = max(log(x), 0). These absolute values have a unique exten-sion to ?v , the completion of an algebraic closure kv. We extend | |v to anorm on finite dimensional vector spaces over ?v . IfX =????x1x2? ? ?xN????1055.2. Preliminiariesis a column vector in ?Nv , we have?x?v ={max{?xn?v : 1 ? n ? N} if v -?{?Nn=1 ?x?2v}1/2if v | ?(5.2.1)and |x|v = ?x?dv/dv in both cases.We identify elements of Hom(?Nv ,?Mv)with M ?N matrices over ?v .Then we extend | |v to such matrices A by setting|A|v = sup{|Ax| : x ? ?Nv , |x| ? 1}.Lemma 5.1. ([84]) Let A = (amn) be an M ?N matrix over ?v If v - ?Then|A|v = max {|amn| : 1 ? m ?M, 1 ? n ? N} .If v | ? let A? denote the complex conjugate transpose of A and0 ? ?1 ? ?2 ? ? ? ? ? ?Ndenote the eigenvalues of the positive semi-definite matrix A?A , then|A|v = ?dv/2dN .Proof. Assume v -? and ~x ? ?Nv such that |~x| ? 1, then|A~x|v = max1?m?M{?????N?n=1amnxn?????v}? max1?m?M{max1?n?N|amn|v}.By choosing a suitable ~x we have equality in this inequality.If v | ? then A?A is a positive definite hermitian matrix. Thus there existsan orthogonal basis ~?1, ~?2, ? ? ? , ~?n for ?Nv such that ~?i is an eigenvector ofA?A with eigenvalues ?n and???~?i??? = 1 Let ~x ? ?Nv . we have~x =N?i=1?i~?i, ?i ? ?v,Then1065.2. Preliminiaries?~x?2v =?N?i=1?i~?i?2v=N?i=1??i~?i?2v=N?i=1??i?2v.And since the vectors ~?1, ~?2, ? ? ? , ~?n form an orthogonal basis we have?A~x?2v =~x?A?A~x=(N?i=1?i~?i)?( N?n=1?iA?A~?i)=(N?i=1?i~?i)?( N?i=1?i?i~?i)=N?i=1?i??i?2v??NN?i=1??i?2v=?N?~x?2v.Hence|A~x| ? ?dv/2dN |~x|v . (5.2.2)And by choosing ~x = ~?n we get the equality|A|v = ?dv/2dNas desired.Lemma 5.2. Let A be an N ? N matrix over,?v , rank(A) = N , and let~x ? ?Nv , then ??A?1???1v |~x|v ? |A~x|v ? |A|v |~x|v [84].1075.2. PreliminiariesProof. If v | ? then by (5.2.2) we have|A~x|v ??dv/2dN |~x|v= |A|v |~x|v .If v | ? , let A = (amn) , 1 ? m,n ? N and ~x =?????x1x2...xN?????, then|A~x|v = max1?m?N{|N?n=1amnxn|v}? max1?m?N{max1?n?N{|amn|v|~x|v}}? |A|v |~x|v .This completes the proof of the right hand side inequality; To prove theother inequality for any place v we have|~x|v =??A?1A~x??v???A?1??v |A~x|v .And so we have ??A?1???1v |~x|v ? |A~x|v .The following corollary is the immediate conclusion of this lemma:Corollary 5.1. Let A,Bbe N ?N nonsingular matrices over ?v . Then|AB|v ? |A|v |B|v .Define the map?v : GL(N,?v)? [1,?)by ?v(A) = |A|v|A?1|v. Since ?v(aA) = ?v(A) for all a 6= 0 in ?v , ?v is welldefined as a map?v : PGL(N,?v)? [1,?).1085.2. PreliminiariesMoreover let A ? GL(N,?v) for any ~x ? ?Nv by corollary 5.1 we have|~x| =??AA?1~x??v? |A|v??A?1~x??v? |A|v??A?1??v |~x|v .By dividing both sides by |~x|v we can conclude that |A|v??A?1??v ? 1 andso we have ?v(A) ? 1 for all A ? GL(N,?v).In the case of N = 2 we have a closed simple formula for ?v(A).Lemma 5.3.?v(A) = |detA|?1v |A|2v [84].Proof. Let A =(a bc d)Then A?1 = 1detA(d ?b?c a)Therefore??A?1??v =????1detA(d ?b?c a)????v=1|detA|v|A|v .So?v(A) = |A|v??A?1??v= |A|v|A|v|detA|v=|A|2v|detA|v.Let e1, e2, ? ? ? eN be the standard basis vectors in ?nv . For each subsetI ? {1, 2, ? ? ? , N}, let eI be the corresponding standard basis vector in theexterior algebra ?(?Nv ). This is the graded algebra?(?Nv ) =??n=0?n(?Nv )in which each subspace ?n(?Nv ) has dimension(Nn)and basis vector {eI :|I| = n}. By applying (5.2.1) to the basis {eI : I ? {1, 2, ? ? ?N} we canextend | | to ?(?Nv ) . We define a second map?v : ?Nv ? ?Nv ? [0, 1]1095.2. Preliminiariesby?v(x, y) =|x ? y|v|x|v |y|v.Since ?v(ax, by) = ?v(x, y) for all a 6= 0 and b 6= 0 in ?v, it follows that ?v iswell defined as map?v : PN?1 ? PN?1 ? [0, 1]One can show that ?v is metric on PN?1(?v)[71].Lemma 5.4. In the case N = 2 the maps ?v and ?v are related to eachother by inequality?v(A)?1?v(x, y) ? ?v(Ax,Ay) ? ?v(A)?v(x, y) (5.2.3)for all A in PGL(2,?v) and x, y in P1(?v).Proof. Note that?v(Ax,Ay) = |Ax|?1v |Ay|?1v |detA|v |x ? y|v? |A|?2v |detA|v |x|?1v |y|?1 |x ? y|v by corolarry 5.1=?v(A)?1?v(x, y) by lemma 5.3.This proves the left hand side inequality. The right hand side follows,since ?v(A?1) = ?v(A).Remark 5.1. Let ? =[?1]in P1(k) , ? =[?1]in P1(k) Since ?v(a, b) ?|?? ?|v . To find the lower bound for |?? ?|v we can actually find a lowerbound for ?v(a, b).If ? occurs in kn , We define its heights byH(?) =?v|?|v .Next we will prove Liouville?s lower bound for the projective metric. Supposethat ? =[?1]in homogenous coordinates with ? algebraic over k of degreer ? 2 . As K = k(?) is embedded in kv, there exists a place w? of K withw? | v , [Kw? : kv] = 1 , and |?|rw? = |?|v . It follows that?v(?, ?) = ?w?(?, ?)r ?{?w?w(?, ?)}r,1105.3. The Thue-Siegel inequalitySince ? 6= ?, we obtain?v(?, ?) ?{?w|?|?1w |?|?1w |? ? ?|w}r(5.2.4)= {H(?)H(?)}?rand this is the Liouville?s lower bound for the projective metric. If r = 2,then with N = 2 this result is sharp [15] .5.3 The Thue-Siegel inequalityLet ? ? kv be an algebraic number of degree 3 over k, so that K = k(?) ? kv. ??, ??? are conjugates of ? in k?v and L = K(?, ??, ???) ? kv denotes theGalois closure of the extension K/k. We denote the points ?, ??, ??? by thecorresponding points in P1(L) as? =[?1], ?? =[??1], ??? =[???1].For each ? in G = Gal(L/k) there exists a unique element Q? in PGL(2, L)such that?(?) = Q??, ?(??) = Q???, ?(???) = Q????.Hence we can obtain a faithful projective representation of G in PGL(2, L): ? ? Q?. We will define another representation of G in PGL(2,Q). Let ?be the unique element in PGL(2, L) such that?? =[01], ??? =[11], ???? =[10]. (5.3.1)Then the conjugate representation ?? ? ?Q???1 is a faithful projectiverepresentation of G in PGL(2,Q), which permutes the elements of the set{[01],[11],[10]}.Remark 5.2. If G is noncyclic thenp? ?{[1 00 1],[0 11 0],[?1 10 1],[0 1?1 1],[1 01 ?1],[1 ?11 0]}and if G is cyclic thenp? ?{[1 00 1],[0 1?1 1],[1 ?11 0]}.1115.3. The Thue-Siegel inequalityIn the next lemma we will find the set ?(p?) of definition 5.1 for thisrepresentation .Lemma 5.5. Let ? belong to P1(L). Then the identity??1(?) = P?(?)holds for all ? in G if and only if ??1? occurs in P1(k) [14].Proof. For any ? in G we have??1(?)? = ??1(?)??1(?(?)) = ??1 {??(?)}= ??1{?Q???1??}= ??1 {P???} .Since P??? occurs in the set{[01],[11],[10]}, it is fixed by ??1 and there-fore ??1(?)? = P???. By similar argument we have ??1(?)?? = P???? and??1(?)??? = P?????. It means that ??1(?) = P??. as elements of PGL(2, L)for each ? in G. Assume ??1(?) = P?? for all ? in G; then we have??1(??1?) = (??1(?))?1??1(?) = (P??)?1P?? = ??1?.Since ??1? is fixed by all elements of Gal(L/k), it belongs to P1(k) and? ? ?P1(k). Conversely assume ??1? occurs in P1(k); then ? = ?? forsome ? in P1(k). Then we have??1(?) = ??1(?)? = P??? = P??,for all ? in G. This completes the proof and it means that?(P ) = {? ? P1(L) : ??1 = P?? for all ? in G} ={?? : ? ? P1(k)}.The only points in P1(k) fixed by each element of P? are[?61]and[?561]where ?6 is primitive sixth root of unity. These are fixed points only if Gis cyclic therefore ?(P ) does not intersect P1(Q) [14]. The main problem isgiving an effective lower bound for ?v([01], ?). For this matter, we defineexponent approximation ev(?) for each ? in ?(P ) byev(?) =? log ?v([01], ?)log {8H(?)},1125.3. The Thue-Siegel inequalityso that?v([01], ?)= 8H(?)?ev(?).We also set?w(P ) = max {?w(P?) : ? ? G}for each place w of L and?(P ) =?w?w(P ).By remark 5.2 we can see that ?(P ) = 12(3 + 51/2). If [L : k] = 6 and??, ??? occurs in kv; then by Liouville?s bound for projective metric we haveev(?) < 6. Moreover in [14] it has been shown thatTheorem 5.1. For each point ? in ?(P ) we have 0 ? ev(?) < 3.For the points ?1 and ?2 in ?(P ) definer(?1, ?2) = min{log {8H(?1)}log {8H(?2)},log {8H(?2)}log {8H(?1)}}ands(?1, ?2) = (log {8H(?1)})?1 + (log {8H(?2)})?1 .By this definition, following is the Thue-Siegel principle for cubic exten-sionTheorem 5.2. If ?1 and ?2 are points in ?(P ) thenev(?1)ev(?2) ? 6 + 19s(?1, ?2)1/3 + 24r(?1, ?2)1/2[14].So in the case that both r(?1, ?2) and s(?1, ?2) are small , we will geta much better inequality for ev(?1)ev(?2) than theorem 5.1. Assuming theThue-Siegel inequality (theorem5.2), we can find an effective irrationalitymeasure for generators of the cubic extension . Let??v(?1) = ev(?1)?1{6 + 19 (log {8H(?1)})?1/3}. (5.3.2)Assume we can find a point ?1 in ?(P ) such that ??v(?1) is less than 3. Then for all generators ? of the cubic extension K = k(?) ? kv,we havethe following theorem [14].1135.3. The Thue-Siegel inequalityTheorem 5.3. Assume that ?1 in ?(P ) satisfies ??v(?) < 3 . Then? log ?v(?, ?) ? ??v(?1) log {8H(?)}+O{(log 8H(?))1/2}for all ? in P1(k) . The implied constant depends on ?1 and ? .Proof. First note that if ev(?1) < 2 since ev(?2) < 3 theorems 5.2 and 5.3follow immediately. So we assume 2 ? eV (?) ? 3. By mean value theoremwe have(x?1 + y?1)1/3? x?1/3 ?13x2/3y?1, (5.3.3)where x and y are positive real numbers. Let x = log{8H(?1)} andy = log{8H(?2)} where ?2 belongs to ?(P ) . Therefore from the aboveinequality, since 2 ? ev(?1), we have:ev(?1)?1{6 + 19s(?1, ?2)1/3}???v(?1) ?4 (log {8H(?1)})2/3 (log {8H(?2)})?1(5.3.4)and from theorem 5.2 we obtain that:ev(?2) ???v(?1) + 4 (log {8H(?1)})2/3 (log {8H(?2)})?1+ 12 (log {8H(?1)})1/2 (log {8H(?2)})?1/2 .(5.3.5)Now let ? be an element of PGL(2, L) as defined in 5.3.1 . From lemma5.3 we have?(?) =?w?w(?) =?wdet(?)?1|?|2w =?w|?|2w.Where the last equation follows from product rule. So for ? in P1(k) wehaveH(??) ??w|?|w|?|w = ?(?)1/2H(?). (5.3.6)From inequality (5.2.3) we conclude that1145.4. Dyson?s lemma, proof of Thue-Siegel?s principle?v(?, ?) =?v(??1[01],??1??)? ?v(?)?1?v([01],??)? ?(?)?1?v([01],??).(5.3.7)By lemma 5.5 ?? occurs in ?(P ) so the inequalities (5.3.5) and (5.3.7)can be combined to let us conclude that? log ?v(?, ?) ???v log {8H(??)}+ (log {8H(?1)})1/2 (log {8H(??)})1/24 (log {8H(?1)})2/3 + log ?(?).(5.3.8)Finally using the upper bound (5.3.6) for H(??) completes the proof.5.4 Dyson?s lemma, proof of Thue-Siegel?sprincipleIn this section we will will closely follow [14] to show how we can constructthe auxiliary polynomials using a projective version of Dyson?s lemma andhow we can use it to prove Thue-Siegel? s principle (Theorem 5.2).5.4.1 PreliminariesAssume that M and N are nonnegative integers, w is a place of the numberfield L and ?w is the completion of an algebraic closure L?w. The vector spaceEw = ?(M+1)(N+1)w can be identified with bihomogeneous polynomialsF (x, y) =M?m=0N?n=0fm,nxm0 xM?m1 yn0 yN?n1 .We define two norms on Ew. The first one is|F |w = sup{|F (x, y)|w : x ? ?2w, y ? ?2w, |x|w ? 1 and |y|w ? 1}.1155.4. Dyson?s lemma, proof of Thue-Siegel?s principleWe define a representation (A,B) ? ?(A,B) of GL(2,?w) ? GL(2,?w) inGL(Ew) by(?(A,B)F )(x, y) = F (A?1x,B?1y),so that ?(A,B)o?(C,D) = ?(AC,BD). To define the second norm, first we setw ={0 if w -?[Lw : Qw] [L : Q]?1 if w | ?.(5.4.1)Definition 5.2. If w -?, we define[F ]w = max {|fm,n|w : 0 ? m ?M and 0 ? n ? N} ,and if w | ? we define[F ]w ={M?m=0N?n=0(Mm)?1(Nn)?1||fm,n||2w}w/2.Suppose N = 0; then F is simply a homogenous polynomial in ?w[x] sowe can writeF (x) =M?m=0fmxm0 xM?m1 . (5.4.2)If F is not identically zero then there exist nonzero vectors?1, ?2, ? ? ? , ?M in ?2w such that,F (x) =M?m=1(x ? ?m). (5.4.3)The next lemma relates both of the norms we defined for this case:Lemma 5.6. Let F (x) in Ew be given by (5.4.2) and (5.4.3) . If w | ?thenlog |f |w = log[f ]w =M?m=1log |?m|w (5.4.4)and if w | ? thenlog |F |w ? log[F ]w ?M?m=1log |?m|w?w2{M ? log(M + 1)}+ log[f ]w?w2M + log |F |w[14].(5.4.5)1165.4. Dyson?s lemma, proof of Thue-Siegel?s principleWe also define? =?(M,N, ?, ?)={(m,n) ? Z2 : 0 ? m ?M, 0 ? n ? N, and 1 ?m?M+n?N}.Also setYw = Yw(?, ?) ? Ewto be the set of polynomials F (x, y) in Ew such thatfm,n = 0 wheneverm?M+n?N< 1.Let F in Ew and (A,B) ? GL(2,?w)?GL(2,?w) . Then ?(A,B) acts onboth F and the vector of coefficients (m,n)? fm,n(?(A,B)F ))(x, y) =F (A?1x,B?1y),M?m=0N?n=0(?(A,B)f)m,nxm0 xM?m1 yn0 yN?n1 ,So that (m,n)? (?(A,B)f)m,n is the vector of the coefficients of ?(A,B)F .Then we defineIndex(F,A,B, ?M, ?N) = min{ m?M+n?N: (?(A,B)F )m,n 6= 0}.If F is identically zero we set the index equal to ?The index function has the following simple properties:Proposition 5.1. 1. F occurs in Yw if and only ifIndex (F, I2, I2, ?M, ?N) ? 1Where I2 is the identity matrix, I2 =[1 00 1].2. For ? > 0 we have??1 Index (F ;A,B, ?M, ?N) = Index (F ;A,B, ??M, ??N).3. For the representation ?Index(?(C,D)F ;A,B, ?M, ?N) = Index (F ;AC,BD, ?M, ?N).1175.4. Dyson?s lemma, proof of Thue-Siegel?s principle4. For nonzero elements a and b in ?w we haveIndex (F ; aA, bB, ??M, ??N) = Index (F ;A,B, ??M, ??N).From the last property, we can see that the map(A,B)? Index (F ;A,B, ??M, ??N)is well defined for (A,B) in PGL(2,?w)? PGL(2,?w). The next theoremgives a local estimate for the norm of F [14].Theorem 5.4. Let F be a polynomial in Ew , x ? P1(?w) and y ? P1(?w).Assume that1 ? Index (F ;A,B, ?M, ?N)for some pair (A,B) in PGL(2,?w)? PGL(2,?w); then we have|xw|?M |y|?Nw |F (x, y)|w ?{(M + 1)(N + 1)2M+N}/2?w(A)m?w(B)n|F |w max{?w([01], x)?M, ?w([01], y)?N}.Now we try to find a global estimate for the norm. As previously,we assume L/k is a Galois extension of an algebraic number field, withG = Gal(L/k) and ? ? ?? is a faithful projective representation of G inPGL(2, k) at each place v of k. As previously, we assume L/k is a Galoisextension of an algebraic number field, with G = Gal(L/k) and ? ? ?? isa faithful projective representation of G in PGL(2, k). At each place v of kwe define?v(?) = max {?v(??) : ? ? G}And with the normalization, we define?v(?) =?w|v(max {?w(??) : ? ? G}) .We also define?(?) =?v?v(?).Since ?v(?) = 1 for almost all places of k, it is therefore well defined. Nextwe define?(?) ={? ? P1(L) : ??1(?) = ??? for all ? in G}.1185.4. Dyson?s lemma, proof of Thue-Siegel?s principleFor any v fixed place of k, the group G acts on places w of L such that w | v. If ? is in G then ?w is the place of L determined by|x|?w =????1(x)??w for all x in Lfor all ? in G.Let E = L(M+1)(N+1) denote the L-vector space of bihomgenous polynomialsF (x, y) =M?m=0N?n=0fm,nxm0 xM?m1 yn0 yN?n1in L(x, y) having bidegree (M,N).We can apply theorem 5.4 at each place w of L. Taking the product toall of the places, we can obtain a global estimate for the normTheorem 5.5. Let F be a polynomial in E and let ?1 and ?2 be points in?(?). Assume that F (?1, ?2) 6= 0 and that1 ? Index (F ;??, ??, ?M, ?N)for all ? in G . Then for any place v of k and any embedding of L into K?vwe have0 ?M log{21/2?(?)H(?1)}+N log{21/2?(?)H(?2)}+12log {(M + 1)(N + 1)}+?wlog |F |w+ max{?M log ?v([01], x)?M, ?N log ?v([01], y)?N}[14].5.4.2 Dyson?s lemmaAssume ? is an algebraically closed file of characteristic zero and, like inthe previous sections, F (x, y) is a bihomogeneous polynomial of bidegree(M,N) with coefficients in ?. First we try to generalize the Index definitionfor points in P1(?)? P1(?) .We will define stabilizer of[01]in PGL(2,?)stab{[01]}={U ? PGL(2,?) : U[01]=[01]}{[u11 u12u21 u22]? PGL(2,?) : u12 = 0}.1195.4. Dyson?s lemma, proof of Thue-Siegel?s principleFor any elements U and V in stab{[01]}we haveIndex (F ;A,B, ?M, ?N) = Index (F ;UA, V B, ?M, ?N)so the map(A,B)? Index (F ;A,B, ?M, ?N)is constant on all right cosets of stab{[01]}? stab{[01]}. For the point(?, ?) in P1(?)?P1(?) we select (A,B) so that A?1[01]= ? and B?1[01]=? then we setIndex (F ;?, ?, ?M, ?N) = Index (F ;A,B, ?M, ?N).According to the previous argument, it is well defined. We now describethe projective form of Dyson?s lemma introduced in the section 2.1.3 . Forthis let ?1, ?2, ? ? ? , ?J be distinct points in P1(?) and let ?1, ?2, ? ? ? , ?j bedistinct points in P1(?) and also 0 < ?j < 1, 0 < ?j < 1 for j = 1, 2, ? ? ? , JThe projective formulation of lemma 2.5 will be as following [14] :Lemma 5.7. Let F be a bihomogeneous polynomial of bidegree (M,N) in?[x, y] which is not identically zero. IfIndex (F ;?j , ?j , ?jM, ?jN) ? 1for each j = 1, 2, ? ? ? , J then12J?j=1?j?j ? 1 +(J ? 22)min{MN,NM}.Corollary 5.2. Let F 6= 0 be a bihomgeneous polynomial of bidegree (M,N)in ?[x, y] then12J?j=1{?j?j,?j?j, ?j?j (Index (F ;?j , ?j , ?jM, ?jN))2}1 +(J ? 22)min{MN,NM}[14].1205.4. Dyson?s lemma, proof of Thue-Siegel?s principleProof. Let ?j , j = 1, 2 ? ? ? , J be chosen such that0 < ?j < min{??1j , ??1j , Index (F ;?j , ?j , ?jM, ?jN)}Therefore we have 0 < ?j?j < 1, 0 < ?j?j < 1 and1 ? Index (F ;?j , ?j , ?jM, ?jN) = Index (F ;?j , ?j , ?j?jM,?j?jN).From Dyson?s lemma12J?j=1?j?j?2j ? 1 +(J ? 22)min{MN,NM}By taking supremum on the left hand side over all values of K and consid-ering the way we choose ?, the corollary follows.To apply Dyson?s lemma, it is important to know what happens to theindex when we apply a differential operator to F . For any bihomogeneouspolynomial of bidegree (R,S) in ?[x, y] there exists a corresponding linearpartial differential operatorT (D) =R?r=1S?s=1tr,s1r!(??x0)r 1(R? r)!(??x1)R?r 1s!(??y0)s 1(S ? s)!(??y1)S?s.(5.4.6)We haveIndex (T (D)F ;?, ?, ?M, ?N) ? Index (F ;?, ?, ?M, ?N)+Index (T ; I2?, I2?, ?M, ?N)?(R?M+S?N),(5.4.7)where I2 =[0 ?11 0].Let A?1[01]= ? and B?1[01]= ? . We call the pair of integers (R,S)critical with respect to the index at (?, ?) ifIndex (F ;?, ?, ?M, ?N) = min{ m?M+n?N: (?(A,B)f)m,n 6= 0}=R?M+S?N.(5.4.8)1215.4. Dyson?s lemma, proof of Thue-Siegel?s principleRemark 5.3. Assume that (R,S) is critical with respect to (?, ?), then by5.4.7 and 5.4.8 we haveIndex (T (D)F ;?, ?, ?M, ?N)?? Index (T ; I2?, I2?, ?M, ?N).And{T (D)F} (?, ?) 6= 0?? T (I2?, I2?) 6= 05.4.3 The auxiliary polynomialsDuring this section we assume that 0 < ? < 1, 0 < ? < 1 and also that?? < 23 . We define?? = ??(M,N, ?, ?){(m,n)Z2 : 0 ? m ?M, 0 ? n ? N, and 1 ?m?M+n?N?1?+1?? 1}and??? =???(M,N, ?, ?){(m,n)Z2 : 0 ? m ?M, 0 ? n ? N, andm?M+n?N< 1}.Let M ?? N ?? in such a way that0 < limM/N <? (5.4.9)thenlimM?1N?1?????? = 1? ??limM?1N?1??????? =12??.ThereforelimM?1N?1(??????????????)= 1?32??.So since ?? < 2/3 we have particularly |??| ? |???| > 0.Now we plan to construct an auxiliary polynomial F (x, y) such that Fhas a large index (bigger than 1) with respect to the points p1 =([01],[01]),p2 =([10],[10])and p3 =([11],[11]), and h(P ) is not too large . By1225.4. Dyson?s lemma, proof of Thue-Siegel?s principledefinition of ??, it is necessary for the coefficients fm,n of F to be supportedon ?? so that F has an index bigger than 1 in p1 and p2. Moreover F hasan index bigger than 1 in P3 if?(m,n)???(mi)(nj)fm,n = 0for all pairs (i, j) in ???. Therefore what we need is basically a nontrivialsolution in integers to |???| homogeneous linear equations in |??| variablesand we choose M,N such that |??| ? |???| > 0. Siegel?s Box principle lemmaguarantees existence of such a polynomial with bounded height. We willsummarize our argument in the next theorem.Theorem 5.6. Let M and N be positive integers that satisfy (5.4.9) and|??| ? |???| > 0. Then there exists a bihomogeneous polynomialF (x, y) =M?m=0N?n=0fm,nxm0 xM?m1 yn0 yN?n1of bidegree (M,N) in Z[x, y] such that1. F is not identically zero.2. 1 ? Index(F ;[01],[01], ?M, ?N).3. 1 ? Index(F ;[10],[10], ?M, ?N).4. 1 ? Index(F ;[10],[10], ?M, ?N).5. For the coefficient fm,n of F we havelog |fm,n| ?14(1?32??)?1(?M + ?N) + o(M) + o(N).6. eitherF (x, y) = F((0 11 0)x,(0 11 0)y)OrF (x, y) = F((0 11 0)x,(0 11 0)y).[14]1235.4. Dyson?s lemma, proof of Thue-Siegel?s principleProof of Thue-Siegel?s inequalityIn this section we will prove the Thue-Siegel inequlityev(?1)ev(?2) ? 6(1 + Css(?1, ?2)1/3 + Crr(?1, ?2)1/2),where Cs = 2?337/2 + 2?334/3 = 3.1522... and Cr = (3/4)71/2 + 2?1/271/2= 3.8551... ; thus it is slightly stronger than theorem ( 5.2) First note that,since for elements of ?(P ) we have ev(?) < 3 if ev(?1) or ev(?2) is less than2, then the theorem follows immediately. So we may assume2 < ev(?1) < 3, 3 < ev(?2) < 3and by similar argument we assumeCss(?1, ?2)1/3 + Crr(?1, ?2)1/2 < 1/2. (5.4.10)Let G be the subgroup in the remark 5.2; then G acts on P1(k?v). Every orbithas six elements with only three exceptions:{[01],[11],[10]},{[?11],[12],[21]}, and{[?61],[?561]}.For all of these points we have ev(?) < 2 . Hence we assume that the orbitof ?1 and ?2 under the action of G has six points.Let 0 < ? < 1 and 12 < ?? <23 . We chose positive integers M and N suchthat12r(?1, ?2) ? min{MN,NM}? 2r(?1, ?2).Therefore by (5.4.10)1 +72min{MN,NM}<92??. (5.4.11)Let F be a bihomogenous polynomial of bidegree (M,N) in Z[x] thatsatisfies the hypothesis of theorem 5.6. Assume G = Gal(L/k) is noncyclic.Then G = P? : ? ? G therefore P??i = ??1?i and it means that for anyA ? G the index at (?1, ?2) is the same as the index at (A?1, A?2). If G iscyclic then Let I2 =[0 11 0],G = {p? : ? ? G} ? {I2p? : ? ? G}Then we have1245.4. Dyson?s lemma, proof of Thue-Siegel?s principleIndex (F ; I2p??1, I2p??2, ?M, ?N)=Index ((?I2,I2)F ; p??1, p??2, ?M, ?N) , By (3) of proposition 5.1=Index (F ; p??1, p??2, ?M, ?N) , By (6) of theorem 5.6=Index(F ;??1?1, ??1?2, ?M, ?N), Since ?1 ? ?(p)=Index (F ;?1, ?2, ?M, ?N) Since F (x, y) ? Z[x] .In any case, for any A ? G the mapA? Index (F ;?1, ?2, ?M, ?N)is constant. So let us set? = Index (F ;?1, ?2, ?M, ?N) .Assume ? ? 1. Then by applying Dyson?s lemma 5.7 at the nine points{(A?1, A?2) : A ? G} ?{([01],[01]),([11],[11]),([10],[10])}in P1(k?v)? P1(k?v), we obtain that92?? ? 1 +72min{MN,NM},Which contradicts (5.4.11) . Therefore 0 ? ? < 1 and F vanishes at (?, ?2)with low order . Next use the corollary 5.2 of Dyson?s lemma to get theinequality32?? + 3???2 ? 1 +72min{MN,NM}.Now we suppose that the pair of nonnegative integers (R,S) is criticalwith respect to the index at (?1, ?2) . We chose a bihomogeneous polynomialT (x, y) in Z[x, y] of bidegree (R,S) such that T (I2?1, I2?2) 6= 0, for exampleT (x, y) = xr0xR?r1 ys0yS?s1 , where 0 ? r ? R and 0 ? s ? S. By the remark5.3 we have{T (D)F} (?1, ?2) 6= 0And from equation (5.4.8)1? ? ? Index(T (D)F ;[01],[01], ?M, ?N)1255.4. Dyson?s lemma, proof of Thue-Siegel?s principleand similarly at([11],[11])and([10],[10]). So we may apply globalestimate theorem 5.5 to {T (D)F} (x, y) with (M,N) replaced by (M ?R,N ? S), ? replaced by (1 ? ?)(M ? R)?1?M , and ? replaced by (1 ??)(N ? S)?1?N so we obtain:0 ?(M ?R) log{21/2?(p)H(?1)}+ (N ? S) log{21/2?(p)H(?2)}+12log {(M + 1)(N + 1)}+?wlog |T (D)F |w+ (1? ?) max{?M log ?v([01], x)?M, ?N log ?v([01], y)?N}.(5.4.12)Without loss of generality, assume T (D)F has relatively prime integercoefficients. Considering the expansion of T (D)F and estimate (5) in theo-rem 5.6, for a coefficient of T (D)F we obtainlog????fm+r,n+s(m+ rr)(M ?m? rR? r)(n+ ss)(N ? n? sS ? s)????? |fm+r,n+s|+ (m+ r)?(rm+ r)+ (M ?m? r)?(R? rM ?m? r)+ (n+ s)?(sn+ s)+ (N ? n? s)?(S ? sN ? n? s)14(1?32??)?1(?M + ?N) + o(M) + o(N) +M?(RM)+N?( sn).Therefore?wlog |T (D)F |w ??wlog [T (D)F ]w14(1?32??)?1(?M + ?N) + (M +N) log 2 + o(M) + o(N).(5.4.13)By 5.4.12 and 5.4.13 we have:1265.4. Dyson?s lemma, proof of Thue-Siegel?s principle0 ?M log {8H(?1)}+N log {8H(?2)}+14(1?32??)?1(?M + ?N)+ (1? ?) max{?M log ?v([01], x)?M, ?N log ?v([01], y)?N}+ o(M) + o(N).(5.4.14)Now let 0 < t < 14 ; we can choose ? and ?? ={2(1? t)ev(?2)3ev(?1)}1/2, ? ={2(1? t)ev(?1)3ev(?2)}1/2.Then 0 < ? < 1 and 0 < ? < 1 and since ?? = 23(1 ? t) we have12?? <23 .And we let M ?? and N ?? in such a way thatMN?log {8H(?1)}log {8H(?2)}.By compactness we can restrict (M,N) to a suitable subsequence along with?? ?? with(1? t) + 2(1? t)(??)2 ? 1 +72r(?1, ?2)With our choice of ?, ? , M and N and from inequality (5.4.14)(1? ??){23(1? t)ev(?1)ev(?2)}1/2? 2 + (4t)?1s(?1, ?2)and from these two latter inequalities we obtain{ev(?1)ev?2} ?61/2{1 + (8t)?1s(?1, ?2)(1? t)1/2 ?(12 t+74r(?1, ?2))1/2}?61/2{1 + (8t)?1s(?1, ?2)1?(32 t+74r(?1, ?2))1/2}.(5.4.15)On the square 0 ? x ? b < 1, and 0 ? y ? b < 1 we have the simpleinequality1275.5. The Thue inequality??x3 + pxy2 + qy3?? < k(1 + x1? y)2? 1 + 2(1? b)?2(x+ y). (5.4.16)Let t = (24)?1/3s(?1, ?2)2/3 , take x = 14 (3s(?1, ?2))1/3 andy ={14 (3s(?1, ?2))2/3 + 74r(?1, ?2)}1/2so by inequality (5.4.10) we can takeb = 3 ? 23/2 and all conditions of inequality (5.4.16) is satisfied. Thus wecan obtainev(?1)ev(?2)? 6{1 + (1? b)?2{14(3s(?1, ?2))1/3 +{14(3s(?1, ?2))2/3 +74r(?1, ?2)}1/2}}? 6{1 + (1? b)?2{32(3s(?1, ?2))1/3 + (7r(?1, ?2))1/2}}= 6{1 + cs(?1, ?2)1/3 + crr(?1, ?2)1/2}.5.5 The Thue inequality??x3 + pxy2 + qy3?? < kWakabayashi [87], studied the Thue inequality??x3 + pxy2 + qy3?? < k byusing Pade? approximation method , and proved under some condition thatthis inequality only has nontrivial solutions. More precisely, he proved thefamily of cubic Thue inequality??x3 + pxy2 + qy3?? ? k,when p is positive and greater than 360q4 ,and k = p+ |q|+ 1 has onlytrivial solutions, namely(0, 0), (?1, 0), (?1, 1), (?q/d, p/d) where d = gcd(p, q)In this section we will use the machinery introduced in previous sections(Bombieri method) to solve the same Thue inequality??x3 + pxy2 + qy3?? < k (5.5.1)under some stronger restrictions on p and q. Since the proof is related toheights of the Thue polynomials, the restriction that we assumed is strongerthan the one assumed by Wakabayashi.1285.5. The Thue inequality??x3 + pxy2 + qy3?? < kLet f(x) = x3 + px+ q be an irreducible polynomial in Q[x]. And let Lbe the splitting field of f wheref(x) = (x? ?)(x? ??)(x? ???)in L[x]. Assume |?| < |??| < |???| . The unique element of PGL(2, L), definedin (5.3.1) is? =[(?? ? ?)?1 ??(?? ? ?)?1?(??? ? ??)?1 ???(??? ? ??)?1]. (5.5.2)We have the following theorems:Theorem 5.7. Assume p ? 360q6; then for any rational number ? = xy Wehave????? ?xy???? >1(8My)? . (8My)9.01?log(p+15/12)?2?log 8My .p1.73?log p/2+15/12 .e2.13?log p+15/12 .M2,where ?(?) =?w?w(?) and the product is over all the valuations of L.? ?(3 log p+ 15) + 5.76(3 log p+ 15)2/332 log p? log q ? 0.19andM =(?p2 + 5 |q|+ 5 +?8 |q|+ 1)?2.Let (x, y) be a solution for (5.5.1); then the transformation (x, y) ?(?x,?y) gives another solution of (5.5.1), so without lose of generality wecan assume for any solution we have y ? 0.Theorem 5.8. Assume the hypothesis of theorem 5.7 is satisfied and log p >8012 . For any solutions of Thue inequality (5.5.1) we have|y| < p518k.Theorem 5.9. With the same hypothesis as theorem 5.7, the only primitivesolutions with y ? 0, of the Thue inequality??x3 + px2y + qy3?? ? p+ |q|+ 1are (0, 0), (?1, 0), (?1, 1), (?q/d, p/d), where d = gcd(p, q).1295.5. The Thue inequality??x3 + pxy2 + qy3?? < k5.5.1 Irrationality measureFor the proof of theorem 5.7, we use (5.3.8) to find an effective irrationalitymeasure of ?. Set? = (? ? ??)(?? ? ???)(??? ? ?).The discriminant of F is ?2 = ?4p3 ? 27q2. As previously, we define? =[?1], ?? =[??1], ??? =[???1]be points in P1(L). To apply Theorem 5.3 we need to find a point in ?(p)that satisfies the hypothesis of theorem 5.3. We find that?[3q?2p]=[?????]= ?1. (5.5.3)Therefore ? belongs to ?(p). Moreover we have to show that ? is actuallya good approximation for the point[01]. The first step toward this goal isto find an upper bound for logH(?1) in terms of p and q. We introducepolynomialsr1(x) =(x+???)(x+?????)(x+?????)=(x3 ?32x2 ?32x+ 1)+?2q(x2 ? x)r2(x) =(x+???)(x+?????)(x+????)=(x3 ?32x2 ?32x+ 1)??2q(x2 ? x)and their homogenizationR1(x) = x31r1(x0x1), R2(x) = x31r2(x0x1).Lemma 5.8. Let w be a place of L. If w -? thenlog[R1]w = log[R2]w =12log+????p3q2????wIf w | ? then1305.5. The Thue inequality??x3 + pxy2 + qy3?? < klog[R1]w = log[R2]w=w2log{72+23????p3q2+274????w}Where w is as (5.4.1). Moreover, the polynomials r1(x) and r2(x) areirreducible in k(?)[x][14].Proof. If w -?. Then obviously we have [R1]w = [R2]w and we havelog[R1]w =12log[R1R2]wWe find thatR1(x)R2(x) =x60 ? 3x50x1 +(p3q2+ 6)x40x21=?(2p3q2+ 7)x30x31 +(p3q2+ 6)x20x41 ? 3x0x51 + x61.By using definition (5.2), the proof for this case is completed.If w | ? then by definition (5.2)[R1]w ={1 +13?????2q?32????2w+13?????2q+32????2w+ 1}w/2={72+23?????2q????2w}w/2={72+23????p3q2+274????2w}w/2.Similar calculations hold for R2.The subgroup Gal(L/k(?)) ? Gal(L/k) is cyclic of order 3. Assume r1 hasa root in k(?) then since Gal(L/k(?)) acts transitively on roots of f ,it actstransitively on roots of r1 . Therefore r1 has a root of multiplicity 3, butthe discriminant of r1 is p6q?4 6= 0. This contradiction leads us to concludethat r1(x) and r2(x) are irreducible in k(?)[x].1315.5. The Thue inequality??x3 + pxy2 + qy3?? < kLemma 5.9. Let ?1 in ?(P ) be defined by (5.5.3); then16?u-?log+????p3q2????+?u|?du6dlog{72+23????p3q2+274????u}? logH(?1)?12+16?u|?log+????p3q2????u+?u|?du6dlog{78+16????p3q2+274????u}[14].Proof. r1(x) is an irreducible polynomial over k(?)(x), thus it is the minimalpolynomial of ????? . Since the action of Gal(L/k(?)) is transitive, we findthat the points(?????),(?????), and(????)have the same height. So wecan apply (5.4.4) , (5.4.5) and sum over all places w of L. We obtain theinequality?wlog[R1]w ? 3 logH(?1) ?(32? log 2)+?wlog[R1]w.The lemma follows by applying the values of log[R1]w from lemma 5.8 .Let v be a place of k, and assume that f has a root in kv, and we identifyL with an embedding of L in k?v, such that|?| ??????? ???????? . (5.5.4)Lemma 5.10. Let ? in ?(P ) be defined as (5.5.3), then if v -? we have? log ?v([01], ?1)=12log+????p3q2????and if v | ? thendvdlog{78+16????p3q2+274????}?? log ?v([01], ?1)?3dvd+dvdlog{78+16????p3q2+274????}[14].1325.5. The Thue inequality??x3 + pxy2 + qy3?? < kProof. Our embedding of L into k?v determines a place w? of L such thatw? | v and | |w? = | |v on k?v. If v | ? then?v([01], ?1)=|?|v????[01]????v????[?????]????v=|?|v|???|v.On the other hand, from lemma 5.8 and equation (5.4.4) we have12log+????p3q2????w?= log[R2]w? = log+???????????w?+ log+?????????????w?+ log+????????????w?.Normalized with respect to v and using (5.5.4) we have:12log+????p3q2????v= log+????????????v= ? log ?v([01], ?1).If v | ? then?v([01], ?1)=????????????v????[01]????v????[?????]????v?????????dv/2d={???v??2 + ???2?v}dv/2d.On the other hand, by lemma 5.8 and inequality(5.4.5) we have:w?2log{72+23????p3q2+274????w?}= log[R2]w??w?2log{1 + ?????2w?}+w?2log{1 + ???????2w?}+w?2log{1 + ??????2w?}?3w?2+w?2log{78+16????p3q2+274????}.In the above equation by (5.5.4) we have 0 ? log{1 + ? ??? ?2w?}? 2 and0 ? log{1 + ? ????? ?2w?}? 2 . Therefore the result follows immediately.1335.5. The Thue inequality??x3 + pxy2 + qy3?? < kSo to summarize, lemma 5.9 and lemma 5.10 can provide us with anupper bound for irrationality measure??v(?1) =(?v([01], ?1))?1 (6 log {8H(?1)}+ 19 log {8H(?1)}2/3)(5.5.5)in terms of p and q. In the next section we will prove theorem 5.7 usingprecise bounds of lemmas 5.9 and 5.10.5.5.2 Proof of theorems 5.7, 5.8During this section we assume v is the usual absolute value on Q. k is thefield of rational numbers. Then kv = R, and k?v = C. We also assume thatp ? 360q6 and log p ? 1000.Lemma 5.11. Let p 6= 0 and q be integers and f(x) = x3 + px + q beirreducible . Assume |p| > e1000. Moreover,p ? 360q6. (5.5.6)Then F has a unique real root ?. The effective irrationality measure for ?is ?. where? ?(3 log p+ 15) + 5.76(3 log p+ 15)2/332 log p? log q ? 0.19.Proof. The discriminant of f is ?2 = ?4p3 ? 27q2. When 0 ? p then thediscriminant is negative so f has a unique real root, say ?. So ? ? kv. Theother roots of f are complex conjugates . ?? and ??? f(x) = (x ? ?)(x ???)(x? ???) we have? =? 2a,?? =a? bi,??? =a+ bi,(5.5.7)where a, b are real numbers. Moreover?3a2 + b2 = p,2a(a2 + b2) = q.(5.5.8)1345.5. The Thue inequality??x3 + pxy2 + qy3?? < kIdentify L with an embedding of L in k?v = C, then |?|v = 2a ,|??|v =|???|v =?a2 + b2 =?4a2 + p and we have|?|v ???????v ????????v .So all the conditions of lemmas 5.9 and 5.10 are satisfied. From lemma 5.9we have6 log {8H(?1)}? 6 log 8 + 3 +?u-?log+????p3q2????u+?u|?dudlog{78+16????p3q2+274????u}? 6 log 8 + 3 +?u-?log+????p3q2????u+?u|?dudlog{78+16????p3q2+274????d/duu}6 log 8 + 3 +?u-?log+????p3q2????u+?u|?ddu{15????p3q2????d/duu}? h(p3q2)+ 15.(5.5.9)We use the argument in the proof of lemma 5.10 to obtain a betterlowerbound for ? log ?v([01], ?1). Note that? log ?v([01], ?1)=dvdlog{1 +?????????2v?????}.Also from equations (5.5.7) and (5.5.8), we have??????????? = 1 and?? ????? <1/1000. Therefore by the same argument as the proof of lemma 5.10 wehavew?2log{74+13????p3q2+274????w?}+11000< ? log ?v([01], ?1)hence12hv(p3q2)? 0.19 < ? log ?v([01], ?1), (5.5.10)and the lemma is followed immediately by (5.5.5), (5.5.9) and (5.5.10).1355.5. The Thue inequality??x3 + pxy2 + qy3?? < kRecall thatev(?) =? log ?v([01], ?1)log 8H(?1).By (5.5.9), and (5.5.10) we haveev(?1) >12hv(p3q2)? 0.1916h(p3q2)+ 52=32 log p? log q ? 0.1912 log p+52>83 log p+log 3606 ? 0.1912 log p+52by (5.5.6)>2.666 ( since log p > 6) .(5.5.11)Lemma 5.12. Let ? be in P1(Q), then we have? log ?v(?, ?)? ??v log {8H(??)}+ 9.01 (log {8H(?1)})1/2 (log {8H(??)})?1/2 ?2.38 (log {8H(?1)})2/3 + log ?(?).Proof. Let ?2 be in ?(p). By the theorem 5.2 we haveev(?2) ? ev(?)?1(6 + 19s(?, ?2)1/3 + 24r(?, ?2)).From (5.3.3) , by letting x = log {8H(?)} and y = log {8H(?2)} , we canobtain thats(?, ?2) ?13log {8H(?)}2/3 log {8H(?2)}?1 + log {8H(?)}1/3 .Henceev(?2) ? ev(?)?1{6 + 19 log {8H(?)}1/3 +193log {8H(?)}2/3 log {8H(?2)}?1 + 24r(?, ?2)}.By the definition of (5.3.2) , we have1365.5. The Thue inequality??x3 + pxy2 + qy3?? < kev(?2) ??+ ev(?)?1(193log {8H(?)}2/3 log {8H(?2)}?1 + 24(log {8H(?)}log {8H(?2)})1/2).From (5.5.11) we have ev(?1) > 2.666, thereforeev(?2) ??+(2.38 log {8H(?)}2/3 log {8H(?2)}?1 + 9.01(log {8H(?)}log {8H(?2)})1/2).(5.5.12)Also from (5.3.7) we have?v(?, ?) ? ?(?)?1?v([01],??). (5.5.13)Since ?? belongs to ?(p) we can combine (5.5.13) and (5.5.12) to obtainthe bound:? log |?? ?|v ???v log {8H(??)}+ 9.01 (log {8H(?1)})1/2 (log {8H(??)})1/22.38 (log {8H(?1)})2/3 + log ?(?),(5.5.14)as desired.From equation 5.3 we have ?w(A) = |detA|?1w |A|2w so?(?) =?w?w(?) =?w|det(?)|?1w |?|2w .And by product rule?(?) =?w|?|2w .AlsoH(??) ??w|?|w |?|w = ?(?)1/2H(?).1375.5. The Thue inequality??x3 + pxy2 + qy3?? < kTherefore to find the effective irrationality measure for ? we only needto find an upper for ?(?) . ? is an element of PGL(2, L) so we rewrite? =[(?? ? ?)?1 ??(?? ? ?)?1?(??? ? ??)?1 ???(??? ? ??)?1]as?? =[(??? ? ??) ??(??? ? ??)?(?? ? ?) ???(?? ? ?)].Consider ??, If w - ? , since all entries of ??, are algebraic integers ,|?|w < 1, therefore |?(??)| < ??(??)?2 <?2??(??)?1.Using equations (5.5.7) we will write ? in terms of a, b?? =[2bi 4abi?3a+ bi 3a2 + b2 + 2abi].Hence from matrix column norm and using equations 5.5.8, we obtain?(?) < M2. whereM =(?p2 + 5 |q|+ 5 +?8 |q|+ 1)?2.Also ?v(?, ?) ? |?? ?|v so using lemma 5.12 and (5.5.9) we obtain:????? ?xy???? >1(8My)? . (8My)9.01?log(p+15/12)?2?log 8My .p1.73?log p/2+15/12 .e2.13?log p+15/12 .M2.This completes the proof of theorem 5.7.Lemma 5.13. Let (x, y) be a solution of (5.5.1) then we have????? ?xy???? <kpy3.Proof. ([87]) We have????? ?xy???? ?k???(?? ? xy)(??? ? xy)??? |y|3.Note that ?????? = ?q , ?2 + p = ?q/? and from equations (5.5.8)????(?? ?xy)(??? ?xy)???? ??? Im ????2 =??????2 ?(Re ??)2=|q|???24=?2 + p??24> p..1385.6. Proof of theorem 5.9Proof of theorem 5.8Proof. Let (x, y) be a solution of (5.5.1) with y 6= 0 , theorem 5.7 provide alower bound for???? ? xy??? and lemma 5.13 give an upper bound for???? ? xy??? .Let C = ?+9.01?log(p+15/12)?2?log 8My. Comparing these bounds, we havey3?C ?M2+C .8C .e2.13?log p+5/4 p1.73?log p+5/4pk (5.5.15)Assume ? < 2.7. If p > 360q6 it happens whenever log p > 8012 and ifq = 1 then it happens when log p > 1528. Under this condition, from theinequality (5.5.15) we obtain thaty < p518.k.This completes the proof of theorem 5.8.5.6 Proof of theorem 5.9To prove theorem 5.9 , we fix the value k in theorem 5.8 to be k = p+ |q|+1.By theorem 5.8, there is no solution for the inequality (5.5.1) with y > p520.So to complete the proof of theorem 5.9, we have to show that there isno relatively small solution of the inequality 5.5.1 other than the trivialsolutions(0, 0), (?1, 0), (?1, 1), (?q/d, p/d).It is easy to see that for any solution (x, y) of inequality (5.5.1), x/yis a principal convergent of ? . So if we can find the continued factionconvergence to ? we can find the solutions for (5.5.1) but since we need tofind this convergence to high order there will be some technical difficultiesin finding the continued fractions with integer partial quotients, so instead,following Wakabayashi [87], we introduce the continued fractions with partialquotients.5.6.1 Continued fractions and Legendre?s theoremLet ? be a real number. For ? we chose a rational number k0 satisfyingk0 < ? < k0 + 11395.6. Proof of theorem 5.9and define ?1 by?1 =1? ? k0and then let k1 be a rational number such thatk1 < ?1 < k1 + 1.Since ?1 > 1, we have k1 > 0 and then define ?2 by?2 =1?1 ? k1,and continue the same process for each i ? 1; namely, choose ki in eachstep, satisfyingki < ?i < ki + 1and define ?i+1 by?i+1 =1?i ? ki.For each i, since ?i > 1 , we have ki > 0; then we have? = [k0, k1, k2, ? ? ? , kn, ?n+1] = k0 +1k1 + k2 + ? ? ?+ kn + ?n+1.We call this a continued fraction expansion with rational partial quotientsfor ?. From the way we define this convergence it is clear that the expansionis not unique. We define convergence pn/qn by{p0 = 1, p1 = k0, pn+1 = knpn + pn?1 (n ? 1)q0 = 1, q1 = 1, qn+1 = knqn + qn?1 (n ? 1) .By this definition, for n ? 1 we havepnqn?1 ? qnpn?1 = (?1)n, (5.6.1)? =pn?n + pn?1qn?n + qn?1and????? ?pnqn???? =1qn (qn?n + qn?1).We also choose positive rational number dn, so that both dnpn, dnqn ? Z..Call dn the common denominator of pn and qn. Note that dn is not unique.1405.6. Proof of theorem 5.9Theorem 5.10. (Generalized Legendre Theorem [87]) Let ? be a real num-ber, and let pn/qn (n = 1, 2, ...) be the convergence defined by a continuedfraction expansion with partial quotients for ?. For a fixed n ? 1 assumethat qn/dn?1 < qn+1/dn . If integers p and q satisfyqndn?1? q ?qn+1dn(5.6.2)and ????? ?pq???? <1dn (dn?1 + dn+1) q2, (5.6.3)thenpq=pnqn.Proof. Let us assume p/q 6= pn/qn. Let (x, y) is the solution for this linearsystem{pnx+ pn+1y = pqnx+ qn+1y = q(5.6.4)By (5.6.1) the unique solution is given by x = (?1)n+1 (pn+1q ? qn+1p), y = (?1)n+1 (qnp? pnq) . Since p and q are integers we can see thatdny ? Z. If x = 0 then q = qn+1y so y > 0 and since dny ? N wehave q = qn+1y ? qn+1/dn which contradicts (5.6.2) . Also if y = 0 thenp/q = pn/qn which contradicts our assumption, therefore xy 6= 0. Now wewill show that xy < 0. Assume xy > 0; then since qn, qn+1 are positive, ifboth x, y are negative, from the second equation of (5.6.4) we would haveq be negative, which contradicts (5.6.2). As well, if x > 0 and y > 0 thenq > qn+1y ? qn+1/dn which again contradicts (5.6.2), then x and y havedifferent signs . On the other hand ? ? pn/qn and ? ? pn+1/qn+1 havedifferent signs. Hence x(??pn/qn) and y(??pn+1/qn+1) have the same signand therefore|?q ? p| = |x (?qn ? pn) + y (?qn+1 ? pn+1)|= |x (?qn ? pn)|+ |y (?qn+1 ? pn+1)|? |x (?qn ? pn)| ? |(?qn ? pn)| /dn+1.Hence????? ?pnqn???? ?dn+1qqn????? ?pq???? . (5.6.5)1415.6. Proof of theorem 5.9Then from (5.6.5) , (5.6.3) and (5.6.2) we have1dnqnq?????pq?pnqn???? ?????? ?pq????+????? ?pnqn???? ?????? ?pq????+dn+1qqn????? ?pq????<qn + dn+1qqndn (dn?1 + dn+1) q2?dn?1q + dn+1qqndn (dn?1 + dn+1) q2=1dnqnq.This contradiction completes the proof.A solution of a Thue inequality often satisfies the assumptions of Gener-alized Legendre theorem. Therefore their quotient x/y is one of convergencepn/qn. So one way to show that a Thue inequality has no non trivial solutionis to verify that the convergents do not give a solution. But since the valuespn and qn become large for large value of n and since we want to find themto a high order, the direct verification may be very time consuming. Toovercome this difficulty we use an alternative method to verify that conver-gents do not give solutions. It is originally due to Petho? [68] and formulatedfor partial rational quotients by Wakabayashi [87] .Theorem 5.11. Let F (x, y) be a homogenous polynomial of degree d ?3 with integer coefficients , and let k be a positive integer . Let ? =?(1), ?(2), ? ? ? , ?(d) be the solutions of F (x, 1) = 0 and assume ? is a sim-ple root. Let (x, y) be a solution of the Thue inequality|F (x, y)| ? k (5.6.6)Suppose that for a positive number A we have?????d?i=2(?i ?xy)?????? A. (5.6.7)Choose a continued fraction expansion with rational partial quotientskn (n ? 0) for ? , and let pn/qn be its convergents and let dn be a com-mon denominator of pn and qn . Moreover, assume for a fixed n ? 1 thefollowing three conditions (5.6.8) (5.6.9) and (5.6.10) are satisfiedqndn?1<qn + 1dn, (5.6.8)(kdn(dn?1 + dn)A)1/(d?2)dn?1 < qn (5.6.9)1425.6. Proof of theorem 5.9and?????(k(kn+1+1/kn?1)A)1/(d?2)dd/(d?2)n?1 ? qn if n ? 2(k(k1+1)A)1/(d?2)dd/(d?2)0 ? q1 if n = 1 .(5.6.10)Theny /?[qndn?1,qn+1dn].Proof. Assumeqndn?1? y ?qn+1dn. (5.6.11)From (5.6.6) , (5.6.7) , (5.6.11) and (5.6.9) we have????? ?xy???? ?k????di=2(?(i) ? x/y??? yd?2y2?kA(qn/dn?1)d?2y2<1dn(dn?1 + dn+1)y2.So by the generalized Legendre theorem we have x/y = pn/qn. Letn ? 2; then since kn < ?n < kn + 1 we have????? ?xy???? =????? ?pnqn???? =1qn(qn?n + qn?1)>1qn (qn(kn + 1) + qn?1)=1q2n (kn + 1 + qn?1/qn)=1q2n (kn + 1 + qn/(kn?1qn?1 + qn?2))?1q2n (kn + 1 + 1/kn?1).Then by (5.6.6),(5.6.7) and (5.6.11) we have1q2n (kn + 1 + 1/kn?1)<????? ?xy???? ?kAyd?kA (qn/dn?1)d ,contradicts (5.6.10). For case n = 1 since qn?1 = q0 = 0 we have a similarcontradiction.1435.6. Proof of theorem 5.95.6.2 Applying Legendre?s theorm for small values of yAs usual we will take k = p+ |q|+1. By using GL2 transformation (x, y)?(x,?y) we can assume y ? 0. Moreover, instead of (5.5.1) we will consider??x3 + px? q?? ? p+ q + 1. (5.6.12)Now let ?1 be the real zero of (5.6.12) then ?1 = ?? theorems 5.7 and5.8 hold with same value of ?. Then we assume without loss of generalitythat q > 0. It helps us to find only positive continued fractions and avoidthe difficulties caused by the minus sign during our computations. For thereal number ?1 we calculated the continued fraction expansion with therational quotients up to the order 450, by using the software Maple. Thenwe obtained?1 = [k0, k1, k2, ? ? ? , k450, ....],withk0 = 0, k1 = p/q, k2 = p2/q, k3 = p/2q ....The convergents pn/qn are given byp0 = 1, p1 = 0, p2 = 1, p3 = p2/q, ....and common denominators dn of pn and qn are given byd0 = 1, d1 = 1, d2 = q, d3 = q2, d4 = 2q3.....The full output is available at the weblinkhttp://www.math.ubc.ca/~amir/fracdata.htm.Lemma 5.14. Suppose q > 0 andp ? (q2 + 1)2Then the only primitive solutions (x, y) of (5.6.12), with 0 ? y < q3/d2 =p3/q3 + 1/q are the trivial ones [87].Proof. If y = 1 then since p ? (q2 + 1)2, we can see that x = 0 or x = ?1.Hence we may assume 2 ? y < q3/d2 and x 6= 0. First note that x > 0since otherwise we would have??x3 + px2y ? qy3?? ? 1 + 4p+ 8q > p+ q + 1contradicts our choice of value k in the inequality (5.6.12). Next we willconsider two different cases1445.6. Proof of theorem 5.9Case (i) 2 ? y ? q + 1. In this case we have??x3 + pxy2 ? qy3??? (p+ q + 1) ? 1 + py2 ? qy3 ? (p+ q + 1)= p(y2 ? 1)? q(y3 + 1) ? 3(p2 + 1)2 ? q((q + 1)3 + 1)> 0.Which contradicts the inequality (5.6.12).Case (ii) q + 2 ? y < p/q . In this case we apply Legendre theorem5.10 for n=1. For n = 1. we have 1 = q1/do < q2/d1 = p/q and thereforeq1/do ? y ? q2/d1 by lemma 5.13 and assumption for y we have?????1 ?xy???? <p+ q + 1py.y2?p+ q + 1p(q + 2)y2=(p+ q + 1)(q + 1)p(q + 2)d1(d0 + d2)y2.Moreover we have p(q + 2) ? (p + q + 10(q = 1) = p ? (q + 1)2 ? 0 so????1 ? xy??? < 1/d1(d0 + d2)y2. Hence all the hypothesis of theorem 5.10 aresatisfied for n = 1 so we have x/y = p1/q1 = 0, and x = 0, a contradiction.Case(iii) p/q ? y < q3/d2. For this case we apply Legendre theorem 5.10for n=2. By similar argument we have????1 ? xy??? < 1/d2(d1 +d3)y2 Thereforefrom Theorem 5.10 we have x/y = p2/q2 = p/q = (p/d)/(q/d) so x = p/dand y = q/d and it completes the proof .Lemma 5.15. 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(1? z)B#3r,g = z2r+1Kr(z)Also define A?r , B?r by the formulasA?r = xrA#r,gandB?r = xrB#r,g.If ?r,0 = 0 for some r , setting u = ?31 and v = ?31 ? ?31 ,we have?3?3=(u? v)(B?r,0(u, v))3u(A?r(u, v)3)Using the same argument as proof of lemma 6.3 Bennett [9]. Let ar bethe integral ideal in M = Q(???) generated by (u ? v)(B?r,0(u, v))3andu(A?r(u, v)3). Moreover let N(ar) be its absolute norm, thenN(ar)1/2 |v|?3r?1 |z1|r 3??|Kr(z1)|? 1And therefore|?1|3r ?N(ar)1/2 |v|?3r?1(3??)r+1|Kr(z1)|(A.0.1)153r=6To find an upper bound for N(ar)1/2 |v|?3r?1. Define M1 to be an extensionof M where the ideal genrated by u and v is principal, say generated by w.Set u1 = uw , v1 =vw and let the extension of ar to M1 be br. Further if wedefine?r = (A?r(u1, v1), B?r (u1, v1)),then by [6] , formula (6.10)br ? w3r+1B?r (0, 1)?3r . (A.0.2)We use this to find N(ar) , and find an upper bound for |?1|. We will showthat if ?r,0 = 0 for any 6 ? r ? 15 then |?1| < ?2/3 , but we assumed|?1| > ?2/3. This contradiction completes the proof of nonvanishing of ?r,0,for each r.r=6A#6,0(z) = 988z6 ? 31122z5 + 266760z4 ? 960336z3 + 1662120z2 ? 1371249z + 433026B#6,0(z) = 187z6 ? 11781z5 + 141372z4 ? 636174z3 + 1301265z2 ? 1226907z + 433026K6,0(z) = 6539203z6 ? 278018298z5 + 2790451188z4?? 11179031292z3 + 20974929426z2 ? 18462496536z + 6154165512Setting:F6(x, y) = 79202404017x5 ? 159949407645yx4 + 109186695090y2x3?? 29296172862y3x2 + 2680421535y4x? 43464971y5G6(x, y) = 79202404017x5 ? 186350208984yx4 + 153702897192y2x3? 52808618772y3x2 + 6928019514y4x? 229643804y5We can see thatF6(x, y)A?6(x, y)?G6(x, y)B?6 (x, y) = ?106562117376x5y6B?5 (x, y)A?6(x, y)?A?5(x, y)B?6 (x, y) = 1144y11From this two equation we conclude that?6 ? (106562117376u51v61 , 1144v111 ) ? 106562117376(v1)6since B6(0, 1) = 187 and K6(z1) > 6154165492 formulas (A.0.2) and (A.0.1)|?1| < 72.16?7/36this contradiction proves non-vanishing of ?(6,0).154r=7r=7A#7,0(z) = ?494z7 + 20748z6 ? 240084z5 + 1200420z4 ? 3047220z3+ 4113747z2 ? 2814669z + 767637B#7,0(z) = ?85z7 + 7140z6 ? 115668z5 + 722925z4 ? 2168775z3+ 3346110z2 ? 2558790z + 767637K7,0(z) = ?614125z7 + 34819841z6 ? 471884616z5 + 2626058475z4? 7227728820z3 + 10412142975z2 ? 7521051447z + 2148871842SettingF7(x, y) = 218954190636x6 ? 547175054385yx5 + 500968274391y2x4? 206106707898y3x3 + 37428532188y4x2 ? 2465893775y5x+ 29807035y6G7(x, y) = 218954190636x6 ? 620159784597yx5 + 659031715782y2x4? 325814595210y3x3 + 75233403960y4x2 ? 7055472502y5x+ 173231474y6We can see thatF7(x, y)A?7(x, y)?G7(x, y)B?7 (x, y) = ?40351037922y7x6B?6 (x, y)A?7(x, y)?A?6(x, y)B?7 (x, y) = ?8398y13From this two equation we conclude that?7 ? (40351037922u61v71 , 8398v131 ) ? 40351037922(v1)6since B7(0, 1) = ?85 and K7(z1) > 2148871830 formulas (A.0.2) and (A.0.1)|?1| < 33.74?4/21this contradiction proves non-vanishing of ?(7,0).r=8A#8,0(z) = 2470z8 ? 133380z7 + 2000700z6 ? 13204620z5 + 45708300z4?? 89131185z3 + 98513415z2 ? 57572775z + 13817466B#8,0(z) = 391z8 ? 42228z7 + 886788z6 ? 7316001z5 + 29929095z4?? 66699126z3 + 82393038z2 ? 52966953z + 13817466K#8,0(z) = 59776471z8 ? 4358130075z7 + 76570832520z6 ? 562540646655z5+ 2111528671200z4 ? 4394226699135z3 + 5127936702525z2?? 3139880973840z + 784970243460155r=9SettingF8(x, y) = 257842779464439x7 ? 769043822230242yx6 + 886262102158212y2x5?? 497428263302319y3x4 + 140750028719955y4x3 ? 18693133706664y5x2+ 934070121522y6x? 8748416597y7G8(x, y) = 257842779464439x7 ? 854991415385055yx6 + 1113960845183355y2x5?? 723315871395180y3x4 + 244946909822940y4x3 ? 41180909742180y5x2++ 2916341162280y6x? 55264933490y7We can see thatF8(x, y)A?8(x, y)?G8(x, y)B?8 (x, y) = ?64363009411584y8x7B?7 (x, y)A?8(x, y)?A?7(x, y)B?8 (x, y) = 16796y15From this two equation we conclude that?8 ??64363009411584u71v81 , 16796v151???64363009411584v71?.Since B8(0, 1) = 391 and K8(z1) > 784970240000 formulas (A.0.2) and (A.0.1) imply:|?1| < 54.13?316this contradiction proves non-vanishing of ?(8,0).r=9A#9,0(z) = ?2660z9 + 179550z8 ? 3385800z7 + 28440720z6 ? 127983240z5+335956005z4 ? 530456850z3 + 496011600z2 ? 252965916z + 54206982B#9,0(z) = 391z9 + 52785z8 ? 1393524z7 + 14632002z6 ? 77815647z5+233446941z4 ? 411965190z3 + 423735624z2 ? 234896922z + 54206982K#9,0(z) = ?59776471z9 + 5448151226z8 ? 120346110612z7 + 1125360861408z6? 5491758685140z5 + 15385952831760z4 ? 25651868175360z3+25132846361040z2 ? 13352805876060z + 2967290194680156r=10SettingF9(x, y) = 6766871023502262x8 ? 23475154665503151yx7+ 32675890549211586y2x6 ? 23369579565193512y3x5+ 9132870725085555y4x4 ? 1913581724600637y5x3+ 195321674212056y6x2 ? 7678452802046y7x+ 57387225227y8G9(x, y) = 6766871023502262x8 ? 25730778340003905yx7+ 39749067546212385y2x6 ? 32070900651078645y3x5+ 14462687637429120y4x4 ? 3616439390830530y5x3+ 465907291477830y6x2 ? 25884419856610y7x+ 390409256020y8We can see thatF9(x, y)A?9(x, y)?G9(x, y)B?9 (x, y) = ?488514352385880y9x8B?8 (x, y)A?9(x, y)?A?8(x, y)B?9 (x, y) = ?74290y17From this two equation we conclude that?9 ??488514352385880u81v91 , 74290v171???488514352385880v91?.Since B9(0, 1) = 391 and K9(z1) > 2967290180000 formulas (A.0.2) and (A.0.1) imply:|?1| < 43.15?527this contradiction, proves non-vanishing of ?(9,0).r=10A#10,0(z) = 82460z10 ? 6802950z9 + 157439700z8 ? 1637372880z7+ 9257454360z6 ? 31243908465z5 + 65776649400z4? 87132704400z3 + 70577490564z2 ? 31927912398z + 6179595948B#10,0(z) = 11339z10 ? 1870935z9 + 60618294z8 ? 788037822z7+ 5265525447z6 ? 20309883867z5 + 47787962040z4? 69633887544z3 + 61308096642z2 ? 29868047082z + 6179595948K#10,0(z) = 1457888351219z10 ? 162413351268624z9 + 4403229639855552z8? 50981179991962968z7 + 312597353265293124z6? 1126073476171223496z5 + 2503363190796325872z4? 3474864128904362640z3 + 2932290444707346060z2? 1375716274987503600z + 275143254997500720157r=10SettingF10(x, y) = 75405019023099125130258x9 ? 298443893232215402955342yx8+ 487458919611631529227188y2x7 ? 424970558615713360076442y3x6+ 213642344179905359118183y4x5 ? 62434602120851944637913y5x4+ 10158885120735920047122y6x3 ? 825196853165322203388y7x2+ 26232672827487996507y8x? 160127062812539113y9G10(x, y) = 75405019023099125130258x9 ? 323578899573248444665428yx8+ 578561881908692316308940y2x7 ? 558951063375461610547890y3x6+ 316456470215341452410715y4x5 ? 106513778241440912510280y5x4+ 20641721853824927604000y6x3 ? 2112005918316317639400y7x2+ 94700573189355078330y8x? 1164483428831640820y9we can see thatF10(x, y)A?10(x, y)?G10(x, y)B?10(x, y) = ?45002510860392218972160y10x9B?9 (x, y)A?10(x, y)?A?9(x, y)B?10(x, y) = ?2080120y19From this two equation we conclude that?10 ??45002510860392218972160u91v101 , 2080120v191???45002510860392218972160v101?.Since B10(0, 1) = 11339 and K10(z1) > 275143253000000000 formulas (A.0.2) and (A.0.1)imply:|?1| < 183.83?1160this contradiction, proves non-vanishing of ?(10,0).158r=11r=11A#11,0(z) = ?20615z11 + 2040885z10 ? 56853225z9 + 716350635z8? 4959350550z7 + 20829272310z6 ? 55910151990z5+ 98024292450z4 ? 111747693393z3 + 79819780995z2? 32442878727z + 5725213893B#11,0(z) = ?2668z11 + 528264z10 ? 20602296z9 + 324486162z8? 2654886780z7 + 12743456544z6 ? 38230369632z5+ 73729998576z4 ? 91361085192z3 + 70277757840z2? 30534474096z + 5725213893K#11,0(z) = ?18991421632z11 + 2538969762665z10 ? 82860204771729z9+ 1162359271452456z8 ? 8727488325684576z7 + 39125898864421488z6? 110904610688488764z5 + 203758154517092436z4? 242003539343672064z3 + 179278783381721142z2? 75300131449876410z + 13690932990886620settingF11,0(x, y) = 10086380874316141564545x10 ? 44867702308185202410144yx9+ 84177949522001285566548y2x8 ? 86728758817629067960410y3x7+ 53544447564487777981062y4x6 ? 20286382993552760112756y5x5+ 4642061682309232093164y6x4 ? 606506715319471273458y7x3+ 40240887892938973440y8x2 ? 1056968121552891712y9x+ 5369611493366676y10G11,0(x, y) = 10086380874316141564545x10 ? 48229829266290582931659yx9+ 98013141305361226196091y2x8 ? 110425391171850655570635y3x7+ 75455462185942660347330y4x6 ? 32152155740359388970390y5x5+ 8461604539337689022850y6x4 ? 1314812676605125272030y7x3+ 109725035932977136335y8x2 ? 4059459622628640965y9x+ 41489707996909305y10.We can see thatF11(x, y)A?11(x, y)?G11(x, y)B?11(x, y) = ?399028707182191791915y11x10B?10(x, y)A?11(x, y)?A?10(x, y)B?11(x, y) = ?13750205y21Since B11(0, 1) = 2668 and K11(z1) > 13690932900000000 formulas (A.0.2) and (A.0.1)imply:|?1| < 73.93?211this contradiction, proves non-vanishing of ?(11,0).159r=12r=12A#12,0(z) = 108965z12 ? 12748905z11 + 420713865z10? 6310707975z9 + 52427420100z8 ? 267379842510z7+ 886575267270z6 ? 1968887931210z5 + 2953331896815z4? 2953331896815z3 + 1886321663127z2 ? 696022431849z+ 112868502462B#12,0(z) = 13340z12 ? 3121560z11 + 144216072z10? 2704051350z9 + 26548867800z8 ? 154741972320z7+ 573455544480z6 ? 1400869972944z5 + 2284027129800z4? 2459721524400z3 + 1679396075280z2 ? 658399597695z+ 112868502462K#12,0(z) = 2373927704000z12 ? 375089280379875z11 + 14501580864313725z10? 242183646588189150z9 + 2182172532842687580z8? 11879523104989091400z7 + 41597624578535753868z6? 96807735644023943748z5 + 151292328990025212000z4? 156915475662561817410z3 + 103571640742394696850z2? 39375570332663090928z + 6562595055443848488,160r=12settingF12(x, y) = 16564533422227754585497501869x11? 81831530392603072324167751920yx10+ 173483945186867693764217267616y2x9? 206475946310970890176350078648y3x8+ 151546120738868355567191593488y4x7? 70954048020900642058780095024y5x6+ 21189749516162281303370832600y6x5? 3921491602361190913545847716y7x4+ 421810845344399723947616178y8x3? 23329066630579789398301528y9x2+ 515169318819791836231200y10x? 2212379957408092237460y11G12(x, y) = 16564533422227754585497501869x11? 87353041533345657186000252543yx10+ 198920618270821189584995684715y2x9? 256234037837182480962076547385y3x8+ 205465103260720137096648512850y4x7? 106368872500017696858627681270y5x6+ 35689548164123863423957803390y6x5? 7588152966949903270265545350y7x4+ 969656379093261326061310275y8x3? 67381836937780814496984625y9x2+ 2093703367638590949649005y10x? 18071362972936489554335y11We can see thatF12(x, y)A?12(x, y)?G12(x, y)B?12(x, y) = ?914029782676539674755645440y12x11B?11(x, y)A?12(x, y)?A?11(x, y)B?12(x, y) = ?15714520y23Since B12(0, 1) = 13340 and K12(z1) > 6562595010000000000 formulas (A.0.2) and (A.0.1)imply:|?1| < 173.83?1372this contradiction, proves non-vanishing of ?(12,0).161r=13r=13A#13,0(z) = ?11470z13 + 1565655z12 ? 60389550z11+ 1062856080z10 ? 10424165400z9 + 63326804805z8? 253307219220z7 + 690837870600z6 ? 1305683575434z5+ 1709823729735z4 ? 1522294675506z3 + 879186229704z2? 297022374900z + 44553356235B#13,0(z) = ?1334z13 + 364182z12 ? 19665828z11+ 432648216z10 ? 5014786140z9 + 34816943772z8? 155652219216z7 + 466956657648z6 ? 959291394516z5+ 1352846838420z4 ? 1287536991048z3 + 790079517234z2? 282171256155z + 44553356235K#13,0(z) = ?2373927704z13 + 437617194280z12 ? 19776004312428z11+ 387525903467742z10 ? 4122328258070217z9+ 26732502858952926z8 ? 112925788901103168z7+ 322751842740529812z6 ? 635560579106005113z5+ 863236915928990064z4 ? 794254560542886960z3+ 472640666946852420z2 ? 164115373424072850z+ 25248518988318900,162r=13settingF13(x, y) = 4557135961912237412250339x12? 24759238796653830786822816yx11+ 58549045762250078628504636y2x10? 79104811966199986251402576y3x9+ 67395028434242449136138460y4x8? 37705683843096490607145552y5x7+ 13991355218250609959166624y6x6? 3399084437079641632899984y7x5+ 521173737973418190859476y8x4? 47057879496466425465540y9x3+ 2205314743618990186008y10x2? 41535728476524984068y11x+ 152777968424470498y12G13(x, y) = 4557135961912237412250339x12? 26278284117291243257572929yx11+ 66295776920922218067195537y2x10? 96151438709662070608803459y3x9+ 88598661175033115189940198y4x8? 54114036417612847385561970y5x7+ 22185394491287605337006670y6x6? 6050547455194282058396010y7x5+ 1064941408953812674487655y8x4? 114138346260455273626785y9x3+ 6715243088313676806705y10x2? 177823999604293260655y11x+ 1313615665538738090y12.We can see thatF13(x, y)A?13(x, y)?G13(x, y)B?13(x, y) = ?6956492992094113593750y13x12B?12(x, y)A?13(x, y)?A?12(x, y)B?13(x, y) = ?7650490y25Since B13(0, 1) = 1334 and K13(z1) > 25248518824203527 formulas (A.0.2) and (A.0.1)imply:|?1| < 47?739this contradiction, proves non-vanishing of ?(13,0).163r=14r=14A#14,0 = 493210z14 ? 77680575z13 + 3462334200z12? 70631617680z11 + 806830401960z10 ? 5748666613965z9+ 27230526066150z8 ? 89118085307400z7 + 205862777060094z6? 338203133741583z5 + 392752026280548z4 ? 315041732310600z3+ 166035507569100z2 ? 51726446588835z + 7217643710070B#14,0 =54694z14 ? 17228610z13 + 1075065264z12? 27414164232z11 + 370091217132z10 ? 3013599910932z9+ 15954352469640z8 ? 57435668890704z7+ 144213472975572z6 ? 255146913726012z5+ 316734099797808z4 ? 269943835054950z3+ 150397279530615z2 ? 49320565352145z + 7217643710070K#14,0 =163613471287384z14 ? 34802000734704264z13 + 1817391854469497292z12? 41279776184184885378z11 + 511451389462905300249z10? 3889993184036806675896z9 + 19459875151016939648052z8? 66742985230004787962220z7 + 160639639553407956137505z6? 273729016682091111345750z5 + 328513030934711203617660z4? 271518908064701256378900z3 + 147079259156386264856850z2? 46996665144124333289400z + 6713809306303476184200164r=14SettingF14,0(x, y) = 16081960716183328596163062459067095x13? 95315491892056898184734815997043879x12y+ 248786320282590585982669697193813774x11y2? 376366639121744389317457341158614824x10y3+ 365472565693995112653695895678601188x9y4? 238356924732462027204149338865139588x8y5+ 106166025872804934840280286806051824x7y6? 32202194222781556839906771562778448x6y7+ 6515393037852325418599013184588012x5y8? 843345574480971712492022357472348x4y9+ 64927370960819614086248526357888x3y10? 2613328852222066663267892558100x2y11+ 42491140539608352987480589614xy12? 135373722566161155430179998y13G14,0(x, y) = 16081960716183328596163062459067095x13? 100676145464118007716789170150066244x12y+ 278771266389255848866896517808487612x11y2? 449697515963501801426720104029663126x10y3+ 468507050848111861213331099073579855x9y4? 330282656538347376302815645153153398x8y5+ 160503645414233690985067374601203120x7y6? 53754745220736691551385089149002620x6y7+ 12201653749896332062590242813939505x5y8? 1811811131106426461897599110642000x4y9+ 165466932299883388917261787036860x3y10? 8355918025805974118330433499530x2y11+ 190901137541784504341875653735xy12? 1220749510126455250479377570y13we can see thatF14(x, y)A?14(x, y)?G14(x, y)B?14(x, y)= ?276456962676998247747586007040000y14x13B?13(x, y)A?14(x, y)?A?13(x, y)B?14(x, y) = 30601960y27165r=15Since B14(0, 1) = 5469 and K14(z1) > 6713809259306811040076 ,formulas (A.0.2) and(A.0.1) imply:|?1| < 167.35?528this contradiction, proves non-vanishing of ?(14,0).r=15A#15,0(z) = ?246605z15 + 44388900z14 ? 2263833900z13+ 52973713260z12 ? 696808074420z11 + 5748666613965z10? 31768947077175z9 + 122537367297675z8 ? 338203133741583z7+ 676406267483166z6 ? 981880065701370z5 + 1023885630009450z4? 747159784060950z3 + 362085126121845z2 ? 104655833796015z+ 13650760929915B#15,0(z) = ?26158z15 + 9416880z14 ? 672365232z13+ 19666683036z12 ? 305727527196z11 + 2882573827848z10? 17804132466120z9 + 75540390606252z8 ? 226621171818756z7+ 488107139301936z6 ? 757407629951280z5 + 839173226366475z4? 647362203196995z3 + 330233350618710z2 ? 100105580152710z+ 13650760929915K#15,0(z) =17898375136312z15 + 4351100634458147z14 ? 259994842193246004z13+ 6774021286078915836z12 ? 96647473965326155740z11+ 851160111253722402660z10 ? 4967712033417551079585z9+ 20080985161211047238715z8 ? 57747913555709356686135z7+ 119795821794959862592755z6 ? 179717549112912333644550z5+ 193102695263085916168575z4 ? 144835773449177287631700z3+ 71991953475180418982925z2 ? 21304082910296402357625z+ 2840544388039520314350166r=15settingF15,0(x, y) = 10608841267802658448327686619891890x14? 68123327982300788729018622452576820x13y+ 194566233202521242858983803658624461x12y2? 325933705839696266676797826258367953x11y3+ 355586047898432309109030891440420208x10y4? 265270226827715760641875285025265246x9y5+ 138256478865942775048749452343172884x8y6? 50540942853850457834821171427766498x7y7+ 12822620466797800991335912972826790x6y8? 2200439216462721116843319664341402x5y9+ 244097842988546146854864994206186x4y10? 16231364736674936168446007289804x3y11+ 567528671203688119104456558732x2y12? 8048892794225901061430511712xy13+ 22427241016498991359805734y14G15,0(x, y) = 10608841267802658448327686619891890x14? 71659608404901674878461184659207450x13y+ 216095249055754543718842490407273191x12y2? 383874724987429350020909351317739910x11y3+ 446380712028489137463269836702702830x10y4? 357111042505136534668378445916613290x9y5+ 201089687590208498929678951602911649x8y6? 80163083272603204177118307682407900x7y7+ 22445013271842962717917311303531660x6y8? 4318928869456449471643930302604020x5y9+ 549333381838396230451655514623025x4y10? 43309263859855463266356355167060x3y11+ 1898941985903806873256333221110x2y12? 37823099960348889581878989020xy13+ 211433204789117431160061665y14We can see thatF15(x, y)A?15(x, y)?G15(x, y)B?15(x, y)= ?23808310652994993059767072566750y15x14B?14(x, y)A?15(x, y)?A?14(x, y)B?15(x, y) = ?586426690y29167r=15Since B15(0, 1) = 26158 and K15(z1) > 2840544366735437404054 formulas (A.0.2) and(A.0.1) imply:|?1| < 117.11?845this contradiction, proves non-vanishing of ?(15,0).168
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Thue equations Ghadermarzi, Amir 2014
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Title | Thue equations |
Creator |
Ghadermarzi, Amir |
Publisher | University of British Columbia |
Date Issued | 2014 |
Description | In this dissertation, we are mainly interested in effective methods to solve parametrized Thue equations. After briefly talking about the different effective methods, two parametrized families of cubic Thue equations are completely solved by using Pad é approximation and linear forms in logarithms. The Thue inequality ∣x³ + pxy² + qy³ ∣≤ k, is studied by using Bombieri's method. We find all solutions under some conditions on k, p and q. As an application of Thue equations, we find the integral points on the Mordell curves Y² = X³ + k for all nonzero integers k with ∣k∣ ≤ 10⁷ . Our approach uses a classical connection between these equations and cubic Thue equations. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2014-01-10 |
Provider | Vancouver : University of British Columbia Library |
Rights | Attribution-NonCommercial-NoDerivatives 4.0 International |
DOI | 10.14288/1.0166844 |
URI | http://hdl.handle.net/2429/45757 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
GraduationDate | 2014-05 |
Campus |
UBCV |
Scholarly Level | Graduate |
Rights URI | http://creativecommons.org/licenses/by-nc-nd/4.0/ |
AggregatedSourceRepository | DSpace |
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