Integral Points on Curves Defined by the Equation Y 2 = X 3 + ax 2 + bx + c

Transcription

1 MSc Mathematics Master Thesis Integral Points on Curves Defined by the Equation Y 2 = X 3 + ax 2 + bx + c Author: Vadim J. Sharshov Supervisor: Dr. S.R. Dahmen Examination date: Thursday 28 th July, 2016 Korteweg-de Vries Institute for Mathematics VU Amsterdam, Department of Mathematics

2 Abstract In this project I will investigate, explain and compare different methods for finding all integral points on curves that are given by the equation Y 2 = X 3 + ax 2 + bx + c (with a, b, c being integers). The theory is based on the book of Nigel P. Smart "The Algorithmic Resolution of Diophantine Equations", most proofs of theorems that I use in my project can be found there. I have applied this knowledge, worked out some of the examples and wrote programs that use these techniques to determine all integral points on the curves defined by the given equations. Title: Integral Points on Curves Defined by the Equation Y 2 = X 3 + ax 2 + bx + c Author: Vadim J. Sharshov, wadishar@hotmail.com, Supervisor: Dr. S.R. Dahmen Second Examiner: Dr. M. Shen Examination date: Thursday 28 th July, 2016 Korteweg-de Vries Institute for Mathematics University of Amsterdam Science Park , 1098 XG Amsterdam Department of Mathematics VU Amsterdam De Boelelaan 1081a, 1081 HV Amsterdam 2

3 Contents Introduction 5 1. Skolem s method Introduction of Strassmann s theorem Theory of the method Examples of Skolem s method Improvement of Skolem s method Example of the improved method Thue equations Solving Thue equations Determining the value of A Determining the value of A Determining the value of A Determining the value of A LLL-algorithm Example of solving a Thue equation Method of Bilu and Hanrot Program Finding integral points using Thue equations Theory of the method Example of the method Program based on Thue equation method Results of running program One Mordell-Weil group Introduction of Mordell-Weil group Rules to compute the coordinates of point P + Q Determining the Mordell-Weil group Direct method Examples of the direct method Indirect method Birch-Swinnerton-Dyer algorithm Determining the generators of the Mordell-Weil group Theory for finding the generators Generators of the curve defined by Y 2 = X

4 5. Finding integral points using Mordell-Weil group Theory of the method Example for the curve Y 2 = X X Program based on Mordell-Weil group method Overview results program Two Discussion Comparison of two working programs Alternative methods Method using S-unit equations Alan Baker s theorem Method for special cases using an upper bound Further generalization Not only integral points Polynomials of higher degree Popular summary 65 Bibliography 65 A. Program One 68 B. Program Two 92 4

5 Introduction At the start of the previous century, the mathematician David Hilbert [9] published a list of twenty-three at that time unsolved mathematical problems. One of those was about Diophantine equations. A Diophantine equation is a polynomial equation with integer coefficients in which only integral solutions, solutions that only have integer coordinates, are searched. Hilbert s problem asked for a universal algorithm for determining whether a Diophantine equation had at least one integral solution. In 1970 the Russian mathematician Yuri V. Matiyasevich [11] showed the impossibility of this. However, under certain conditions it is possible not only to determine that there is at least one integral solution, but find all integral solutions of the Diophantine equation. In this thesis a specific kind of Diophantine equations is studied, namely those of the form Y 2 = X 3 + ax 2 + bx + c (0.1) where a, b, c N, to be solved in integers X and Y. Another restriction is that we consider only triplets (a, b, c) for which the cubic polynomial X 3 + ax 2 + bx + c has three different roots, in other words the discriminant of X 3 + ax 2 + bx + c should be nonzero. It is known that on this kind of equations there are finitely many integral solutions. We will study some of the methods for finding those solutions. For two of the methods I have implemented the theory in programs written by me. For this I used the mathematics software system SageMath, since my thesis is in the area of Algebraic Number Theory. SageMath is a free open-source mathematics software system, that has many build-in commands from this field. The codes of these programs can be found in Appendix A and B. In Chapter 1 we discuss the method, known as Skolem s method which makes use of p-adic numbers. We will explain more about these numbers in the next chapter. This method will illustrate the basic approach for finding integral solutions on Diophantine equations. We will illustrate Skolem s method with examples and introduce some important theorems for the theory of finding integral solutions to equations, namely Strassmann s theorem and Hensel s theorem. In Chapter 2 we give theory about how to find all integral solutions on Thue equations. A Thue equation is a Diophantine equation of the form: F (X, Y ) = m with nonzero m Z and F (X, Y ) Z[X, Y ] whose terms are of same degree 3. 5

6 This theory will be used in Chapter 3 where we discuss the process of finding all integral solutions for our equation 0.1. The program that is based on the theory explained in Chapter 2, is called programfindboundthueequation. It is an essential part of the program programone, which is the program that finds all integral solution of the equation 0.1. The codes of the programs programfindboundthueequation and programone can both be found in the Appendix A. They have been written by me and tested for various tuples of integers (a, b, c). In Chapter 4 we consider the Mordell-Weil group of the curve defined by the equation 0.1. This is the group of all rational points on our curve. We state the Mordell-Weil theorem, which tells us that the there are finitely many generators of the Mordell-Weil group and then consider different methods of how these generators can actually be determined. We do not implement these methods in our program, since SageMath has already a build-in function that finds the generators of the Mordell-Weil group based on this theory. Then in Chapter 5 we explain how you can use the found generators of the Mordell-Weil group to determine all integral solutions for our equation. Based on this method I have wrote program programtwo that actually finds all integral points on our curve. The code of this program can be found in Appendix B. We finish with Chapter 6, where we discuss differences between the method which uses Thue equations and the method that uses Mordell-Weil group generators. Also we consider some alternative methods for solving this problem. Besides that we think about the possibilities for further generalization of our problem and we propose ideas for further research. 6

7 Notation My thesis considers a problem from the area of Algebraic Number Theory, therefore we will quickly refresh some basis notation from this field of mathematics. For more information we refer to the following article of F. Beukers [2] or to Chapter 4 from [6]. Now follows a list of definitions that we will use throughout this thesis. Q(θ) will denote the number field, often for the same number field we will use K instead of Q(θ). A number field is a field extension of the rational numbers (Q) with an element θ that is a root of an irreducible polynomial with rational coefficients. (θ) (i) will stand for the i-th conjugate of the element θ. O K will denote the ring of integers of a number field K. g will denote the absolute value of the norm of an element g of O K. η i will denote (in context of number fields) the i-th fundamental unit, so a generator for the unit group of the ring of integers O K of a number field K. ɛ will denote (in context of number fields) a unit from the unit group, so an element of O K whose absolute value of the norm is one. Q p will denote the field of p-adic numbers, with p a prime number. The norm of a p-adic number will be denoted by p. log(α) will denote a discontinuous function from C to C defined in the following way: C \ R 0 we take the analytically extended logarithm with log(1) = 0, defined for C \ R 0, so with the branch cut being the negative real axis, R 0 we take the analytically extended logarithm with log(1) = 0, defined for C \ ir 0, so the branch cut along the negative imaginary axis. We will work with Thue equations. During this approach we often use the following principle, that of their factorization. So when we consider a polynomial of the form X m ay m (a Z), we set θ = m a and we use the fact that X θy divides X m ay m : m 1 X m ay m = (X θy )( X i (θy ) m 1 i ). i=0 7

8 1. Skolem s method To illustrate the development of various methods used to find integral points, we will show a method which was the main method in the first half of the twentieth century. This method for finding solutions of Diophantine equations dates back to Thoralf Skolem [16]. This chapter is based on Chapter III of [17], we use same examples only work them out in more detail. In the following chapters of this thesis we will see how the idea behind this method is still used in other methods Introduction of Strassmann s theorem Strassmann theorem uses p-adic numbers and also p-adic convergence, so before we introduce the theorem we recall what the p-adic numbers are. For any prime number p the set of p-adic numbers, noted as Q p, is a completion of Q, i.e. it consists of Q together with the limit points of every convergent sequence. The difference with R, the usual completion of Q is the norm. For more about completion of Q and more theory about p-adic fields we refer to Chapter 4 of [5]. Certain points that are considered far apart in R are close by each other in Q p. We will now introduce this p-adic norm. Every nonzero p-adic number n can be uniquely written as n = p k ( n i p i ), where k Z and n i {0,..., p 1} i=0 for i > 0 and n 0 {1,..., p 1}. From this notation the p-adic order and the p-adic norm of a p-adic number are defined as follows: ord p (n) = k (p-adic order), n p = p k (p-adic norm). So that rational number with "high" powers of p in the numerator have low p-adic norm and a positive order. Thus we see that for the following series of integers p i, i=0 the p-adic norm of each term is less then the p-adic norm of the previous term and goes to zero. It is not difficult to prove that p-adic norm is an ultrametric norm. And lemma from [5] tells us that in an ultrametric space a series converges if and only if its terms in norm tend to 0. Thus it would mean that the series i N pi converges, we will often in this chapter make use of this property. 8

9 Theory of the method Skolem s method makes use of Strassmann s theorem, introduced by Reinhold Strassmann [19] in This is a very useful theorem that gives a solution to converging power series. Theorem 1.1 (Strassmann s theorem). Let (a i ) i N 0 be a sequence of p-adic numbers, and let f(x) = i 0 a i X i denote a power series which converges for all x Z p. When N is defined such that a N p = max i 0 a i p, and a i p < a N p for all i > N, then there are at most N elements in α Z p such that f(α) = 0. The proof of this theorem can be found in Theorem of [5] Examples of Skolem s method The applience of this theorem will be illustrated by finding the integral solutions of the following curves: X 4 2Y 4 = ±1, X 3 + 6Y 3 = ±1 and X 3 + 2Y 3 = ±1. Al lot of this calculations for these examples can be done with the help of SageMath. Equation X 3 + 6Y 3 = ±1 We will consider the number field K = Q(θ), where θ is defined as a solution of the equation θ = 0. We can calculate that the discriminant of this number field is 972 = For the definition of the discriminant we refer to proposition of [6]. The rank of of OK is one, thus for the basis we can choose one fundamental unit, let it be 1 + 6θ + 3θ 2. When we take X, Y Z such that they are solutions of our equation and therefore X θy OK and because we have taken 1 + 6θ + 3θ2 as the basis element of OK, we get that there exists a k such that: X θy = ±(1 + 6θ + 3θ 2 ) k. We can now expand the brackets and write the right hand side of this equation as a series 9

10 in k. ( ( ) k X θy = ± )1 i (6θ) j (3θ 2 ) k i j i, j, k i j 0 i,j k ( ( ( ) ( ) k k = ± θ + )θ 2 + k 1, 1, 0 k 1, 0, 1 (( ) ( ) ( ) k + 9 (2θ) 2 k + (2θ 3 k ) + )θ 4 k 2, 2, 0 k 2, 1, 1 k 2, 0, 2 ) = ± = ± + 27 (...) +... ( ( ( ( ( (...) +... ( ) ( ) k k θ + )θ 2 + k 1, 1, 0 k 1, 0, 1 ) ( ) ( k θ 2 k + ( 12) 6 k 2, 2, 0 k 2, 1, 1 ) ( 2kθ + kθ 2) ( 2k(k 1)θ 2) + 27 ( 4k(k 1) k(k 1)θ +...) +... ) k k 2, 0, 2 ) ) θ In other words there are constants c 1, c 2, c 3 Z[k], such that X θy = c 1 + c 2 θ + c 3 θ 2. By construction we see that c i have ascending powers of 3 in the numerator, thus they converge in 3-adic norm, which is a necessary requirement if we want to use Strassmann s Theorem. Since it is impossible to make a non-trivial factorization in a linear and quadratic term over Q 3 of the polynomial of our initial equation, X 3 + 6, it is irreducible over Q 3. Also we use a theorem from [6], that states if a prime number (in our case we take p = 3) divides the discriminant of the number field K, the ideal (3) is ramified. Using the fact that our equation is an Eisenstein polynomial, we get from Proposition of [5], that the extension Q 3 (θ)/q 3 is a ramified extension. Therefore, we can equate the coefficients of {1, θ, θ 2 } and conclude that c 3 = 0: 0 = 3(k) + 9(2k(k 1)) We have constructed the coefficients such that in 3-adic norm they converge to 0, therefore we get that the series converge in 3-adic norm. So now we can apply Strassmann s theorem and we find that there the unique solution to this 3-adic equation is k = 0. With this value for k we obtain the solution (X, Y ) = (±1, 0). 10

11 Equation X 3 + 2Y 3 = ±1 For this equation we again consider the number field K = Q(θ) but this time with θ being a solution of the equation θ = 0. we find that the discriminant of this field is 108 = We see that the rank of OK is again one, so we can choose the fundamental unit 1 θ as the basis of OK. So for some k Z we can write X θy = ±( 1 θ) k If we now consider the prime ideal lying above (3), this completely ramifies, because 3 divides the discriminant of K. For Strassmann s theorem we want again convergence in 3- adic norm, therefore we do the following quick calculation to get ( 1 θ) 3 = 1 3θ(1+θ). Therefore it makes sense to consider the equation for k = 3s + k, where k {0, 1, 2}, which leads to three different series. First, the case where k = 0. This gives the following: X θy = ±( 1 θ) k = ±(1 3θ(1 + θ)) s ( ) ( ) s s = ±(1 + 3 ( θ(1 + θ)) + 9 ( θ(1 + θ)) ) 1 2 ( ) ( ) s s = ±(1 + 3 ( θ θ 2 ) + 9 (θ 2 + 2θ 3 + θ 4 )) +...) 1 2 ( ) ( ) s s = ±(1 + 3 ( θ θ 2 ) + 9 (θ ( 2) 2θ)) +...) 1 2 = ±(1 + 3s( θ θ 2 s(s 1) ) + 9 (θ 2 4 2θ)) +...) 2 Using the fact that the ideal (3) is ramified, we can equate the coefficients of θ 2, which gives us s(s 1) 0 = 3s Strassmann s theorem tells us that this equation has at most one solution. This solution is found when s = 0, and thus k = 0. This gives the solution of the initial equation (X, Y ) = (±1, 0). 11

12 The case k = 1 gives the following: X θy = ±( 1 θ) k = ±(( 1 θ) (1 3θ(1 + θ)) s ( ) ( ) s s = ±(( 1 θ) (1 + 3 ( θ(1 + θ)) + 9 ( θ(1 + θ)) ) 1 2 ( ) s = ±( 1 θ + 3(sθ(1 + θ) 2 ) + 9( ( θ 2 (1 + θ) 3 ) +...) 2 ( ) s = ±( 1 θ + 3(sθ(1 + 2θ + θ 2 )) + 9 ( θ 2 (1 3θ(1 + θ)) +...) 2 ( ) s = ±( 1 θ + 3(s(θ + 2θ 2 + θ 3 )) + 9 ( (θ 2 3θ 3 3θ 4 )) +...) 2 ( ) s = ±( 1 θ + 3(s(θ + 2θ 2 2)) + 9 ( (θ θ)) +...) 2 From equating the coefficients of θ 2 at both sides follows that: s(s 1) 0 = 6s Therefore, by Strassmann s theorem, there is at most one solution for this equation, which is when s = 0, and thus k = 1. This leads to another solution of the initial equation, namely (X, Y ) = ( 1, ±1). Finally we consider the case that k = 2. Doing the same procedure gives us: X θy = ±( 1 θ) k = ±(( 1 θ) 2 (1 3θ(1 + θ)) s ( ) ( ) s s = ±((1 + 2θ + θ 2 ) (1 + 3 ( θ(1 + θ)) + 9 ( θ(1 + θ)) ) 1 2 ( ) s = ±(1 + 2θ + θ 2 3(sθ(1 + θ) 3 ) + 9( (θ 2 (1 + θ) 4 ) +...) 2 ( ) s = ±(1 + 2θ + θ 2 3s(θ 3θ 2 + 6) + 9 (θ 2 2θ 3 6θ 4 3θ 5 ) +...) 2 ( ) s = ±(1 + 2θ + θ 2 3s(θ 3θ 2 + 6) + 9 (θ θ + 6θ 2 ) +...) 2 ( ) s = ±(1 + 2θ + θ 2 3s(θ 3θ 2 + 6) + 9 (7θ θ) +...) 2 Equating the coefficients of θ 2 from both sides gives that s(s 1) 0 = 1 + 9s This equation has no solutions for any s, thus, for k = 2, we get no solutions for the initial equation. Hence, equation X 3 + 2Y 3 = ±1, has the following integral solutions: ( 1, ±1) and (±1, 0). 12

13 Equation X 4 2Y 4 = ±1 This time we will set θ as a solution of the equation θ 4 2 = 0 and use its corresponding number field Q(θ). We compute that the discriminant of this number field is 2048 = In this case the rank of OK is 2, we can take the following two fundamental elements as the basis of OK, namely 1 + θ2 and 1 + θ. We see that the smallest prime number that stays prime in Q(θ) is 5. We consider the image of the elements 1 + θ 2 and 1 + θ in O K /(5)O K and find that their multiplicative orders are respectively 12 and 32. Moreover we can rewrite the series as following: Now we consider: (1 + θ 2 ) 12 = θ (...) +... and (1 + θ) 312 = (4θ 2 + 3θ 3 ) (...) X θ Y = ±(1 + θ 2 ) δ 1 (1 + θ) δ 2 (1 + ((1 + θ 2 ) 12 1)) k 1 (1 + ((1 + θ) 312 1)) k 2 After doing the same steps as in the previous examples we can equate the coefficients of θ 2 and θ 3 to get for each of them a series in two variables on the right hand side. Substituting the relation from the one in the other will eventually give a series in one variable on which we can apply Strassmann s theorem. But we have to repeat this procedure for δ 1 ranging from 0 to 11 and δ 2 ranging from 0 to 311. This requires a lot of calculations, in the next part we explain a way how this number can be decreased Improvement of Skolem s method In the previous section we have seen how we can find integral solutions of equations by working in an appropriate number field Q(θ) to reduce the problem to an easier equation on which we could apply Strassmann s theorem to get a bound on the number of possible solutions. However, with some equations it was needed to repeat this procedure multiple times. This number of repetitions can be reduced when we use Hensel s theorem together with Strassmann s theorem. This theorem is named after Kurt Hensel. He was a German mathematician who described p-adic numbers in 1897 in his publication [8], This theorem is also known as Hensel s lifting theorem and broadly speaking it states that a solution of a polynomial modulo a prime number uniquely corresponds to solutions modulo a higher power of that prime under certain conditions. We will state the Multi-Dimensional Hensel theorem later on. When we combine Skolem s Method with Multi-Dimensional Hensel Theorem, we can find a smaller bound for the maximum exponent. With the consequence that it reduces the number of needed computations. We will now introduce the Multi-Dimensional Hensel theorem and we will illustrate it with an example. In this theorem we will use the 13

14 symbol (J f ( a)) to denote the determinant of the Jacobian matrix, which is the matrix with entries f i a j evaluated in a. Theorem 1.2 (Multi-Dimensional Hensel). Let p be a prime, n in Z 0 f be an n-vector of power series in n-variables with coefficients coming from Z p. Suppose there is a vector a Z n p such that f( a) = 0 mod p 2δ+1, where δ = ord p (J f ( a)) < Then there is a unique zero a Z n p of f, α, such that α a (mod p δ+1 ). This is the Multi-Dimensional Hensel theorem, which states how to find a solution to a vector of power series. This theorem has different names and many equivalent formulations. For a prove of this theorem we refer to Section of [5]. Now we will illustrate Skolem s method and also the application of Multi-Dimansional Hensel theorem for the curve X 4 2Y 4 = ± Example of the improved method Let us take the equation: X 4 2Y 4 = ±1. To find all integral solutions, we will first study the quartic number field K(θ), with θ, being the solution of the equation θ 4 2 = 0. The unit rank of the ring of integers is two, so we need to take two fundamental units, so we take the following elements: η 1 = 1 + θ 2 and η 2 = 1 + θ. We now need to find all possible powers a 1, a 2, such that X θy = ±η a 1 1 ηa 2 2. (1.1) The smallest prime number that is still prime in K is 5. In the residue field O K /(5)O K the image of η 1 has multiplicative order 12 and in the residue field of the image of η 2 has multiplicative order 312. When we calculate, η1 12 and η2 312, we rewrite them as: When fill this in the equation 1.1: η 12 1 =1 + 5 (2θ 2 ) (...) +... η =1 + 5 (4θ 2 + 3θ 3 ) (...) +... X θy = η β 1 1 ηβ 2 2 (1 + (η12 1 1)) k 1 (1 + (η )) k 2. Now we equate the coefficients of θ 2 and θ 3 from both sides and get two power series in two variables k 1 and k 2. But since we need to check it for all values of 0 β 1 11 and 0 β 2 311, we want to lower this bounds to reduce the number of calculations. 14

15 We get four equations from the four roots of θ 4 2 = 0, they are of the form X θ i Y = β i. Now we eliminate the X and the Y from the equation. We are get two equations, that are also called Siegel s identities: (θ 3 θ 2 )β 1 + (θ 1 θ 3 )β 2 + (θ 2 θ 1 )β 3 = 0 (θ 4 θ 2 )β 1 + (θ 1 θ 4 )β 2 + (θ 2 θ 1 )β 4 = 0 Now we will consider the polynomial modulo 7, because the prime 7 decomposes in the field K as a product of three prime ideals, one of degree 2 and two of degree 1. We see that: x 4 2 (x + 2)(x + 5)(x 2 + 4) (mod 7). Now because 7 is not an index divisor, we can write θ 1 and θ 2 as 7-adic roots of x 4 2, given by θ 1 = (...) and θ 2 = (...). Now we take θ 3 = Ω and θ 4 = Ω to be the roots of the polynomial g(x) = In the localization of K the x 4 2 (x θ 1 )(x θ 2 ). elements η i satisfy η1 6 = 1 mod 7 and η48 2 = 1 mod 7. So we write a 1 = b 1 + 6k 1 and a 2 = b k 2, we need to find what values of 0 b 1 5 and 0 b 2 47 solve the equation defined above. But this are only 6 48 = 288 possibilities. But of these possibilities only 6 are good, namely: (b 1, b 2 ) = (0, 0); (0, 1); (2, 23); (3, 24); (3, 25); (5, 47). We need too expand in all the six cases the 7-adic power series. Every cases gives at most two possibilities, so we find in total at most 12 possibilities, but there are only 6 possibilities okay. We get the following identities: (θ 3 θ 2 )η (1)b 1 1 η (1)b (θ 1 θ 3 )η (2)b 1 1 η (2)b (θ 2 θ 1 )η (3)b 1 1 η (3)b (mod 7), (θ 4 θ 2 )η (1)b 1 1 η (2)b (θ 1 θ 4 )η (2)b 1 1 η (2)b (θ 2 θ 1 )η (4)b 1 1 η (4)b (mod 7). Now we look through all the possible pairs of (b 1, b 2 ), there are 6 48 = 288 possibilities, we look whether these pairs fulfill the requirements of Siegel s identities. As a result we get the following 6 pairs: (b 1, b 2 ) = (0, 0); (0, 1); (2, 23); (3, 24); (3, 25); (5, 47). Now for all these individual pairs we have to expand our power series in two variables k 1, k 2, we get the following: (1) The case that (b 1, b 2 ) = (0, 0). (2) The case that (b 1, b 2 ) = (0, 1). f 1 = 5k 1 + k 2 + 6Ωk 2 + 7(...) f 2 = 5k 1 + k 2 + Ωk 2 + 7(...) f 1 = 5k 1 + 6k 2 + 5Ωk 1 + 7(...) f 2 = 5k 1 + 6k 2 + 2Ωk 1 + 7(...) 15

16 (3) The case that (b 1, b 2 ) = (2, 23). (4) The case that (b 1, b 2 ) = (3, 24). (5) The case that (b 1, b 2 ) = (3, 25). (6) The case that (b 1, b 2 ) = (5, 47). f 1 = 4 + 5k 1 + 3k 2 + Ω(2k 1 + 5k 2 ) + 7(...) f 2 = 4 + 5k 1 + 3k 2 + Ω(5k 1 + 2k 2 ) + 7(...) f 1 = 4 + 2k 1 + 6k 2 + Ω(4 + k 2 ) + 7(...) f 2 = 4 + 2k 1 + 6k 2 + Ω(3 + 6k 2 ) + 7(...) f 1 = 5 + 2k 1 + k 2 + Ω(1 + 2k 1 ) + 7(...) f 2 = 5 + 2k 1 + k 2 + Ω(6 + 5k 1 ) + 7(...) f 1 = 6 + 2k 1 + 4k 2 + Ω(5k 1 + 2k 2 ) + 7(...) f 2 = 6 + 2k 1 + 4k 2 + Ω(2k 1 + 5k 2 ) + 7(...) For all cases we apply Multi-Dimensional Hensel Theorem from [17]. After using this theorem we see that in all our six cases we have only one unique solution in Z 2 7. And every such solution corresponds to two solutions of our Thue equation, X 4 2Y 4 = ±1, so we get that an upper bound on the number of solution is 12. After testing we get the following solutions: b 1 b 2 X Y Table 1.1.: All solutions for X 4 2Y 4 = ±1 We have found 6 solutions, since we found out that the bound is 12, there is a possibility that 6 other do exist, so we can repeat this method and use it on an other prime to find out if there are other solutions. I have tried to do this also for p = 11, but unfortunately did not find any new solutions. We partially did solve the problem, but not completely. We will proceed to the next chapter, where we will learn another method for finding all integral solutions to a Thue equation. 16

17 2. Thue equations In the next chapter we will search for integral points on curves that are defined by Y 2 = X 3 + ax 2 + bx + c, for a, b, c Z. Thue equation play an important part in that method. For if we want to be able to find all integral points on the defined curve, we will need to be able to find all integral solutions of certain Thue equation of degree 4. In this chapter we follow the theory of Chapter VII of [17]. We will explain how the method for finding all integral solutions of a Thue equation works. We start by explaining what a Thue equation is Solving Thue equations A Diophantine equation F (X, Y ) = m (2.1) where F (X, Y ) Z[X, Y ] a homogenous polynomial of degree at least 3 and m a given nonzero integer is called a Thue equation if F (X, 1) is an irreducible polynomial. We are interested only in the solutions (X, Y ) with X and Y both being integers. The polynomial F (X, Y ) of degree n can in general be written as: F (X, Y ) = n a i X i Y n i with a i Z. i=0 If a n 1, in other words if F (X, 1) is non monic, it can be transformed into equivalent Thue equation (F (X, Y ) = m ) where F (X, 1) is a monic polynomial. To get these F and m we have to multiply the Thue equation 2.1 with a n 1 n and substitute X/a n for X. The integral points on our new Thue equation will correspond with integral points on the equation 2.1. Therefore we will from now on assume that our Thue equation F (X, Y ) = m has a monic polynomial F (X, 1). The easiest way to find all the solutions of a Thue equation, is to find an upper bound on X or Y and simply check all possibilities. If the polynomial F (X, 1) has no real roots it is easy to find such a bound. Lemma 2.1. Let F (X, Y ) = m be a Thue equation where F (X, 1) has no real roots and let θ be one of those roots. If (X, Y ) is a solution, then Y m min 1 i n I(θ (i) ). 17

18 Proof. Let X, Y Z such that (X, Y ) is a solution of the Thue equation. By the definition of θ, we know that F (X, 1) = n i=1 (X θ(i) ). We get that n i=1 (X Y θ(i) ) = m, since (X, Y ) is a solution. This means that there is an i such that X Y θ (i) m, and since X has no imaginary part, Y I(θ (i) ) X Y θ (i) m. From this it follows that Y m I(θ (i) ) m min 1 i n I(θ (i) ). If F (X, 1) does have a real root, it is harder to find the solutions of the Thue equations. In this case, let θ be a root of F (X, 1) (with the conjugates of θ ordered in the standard way, that is θ (i) R if 1 i s, θ (i) = θ (i+t) if s + 1 i s + t). We will work over the number field K = Q(θ). Obviously the norm of an element g of O K stays the same after it is multiplied by a unit ɛ (i). By Dirichlet s Unit theorem we now that the unit group OK is finitely generated. We can define a group action of the unit group (OK ) of O K on an element g, by defining OK g := {eg, e O K }. We want to define a complete set of representatives of O K of a certain norm. We define this set as following. First we define the set M m = {x O K x = m}, this is the set of all elements that have norm m. Now we can divide this set though our action of U. We get M m = M /U. Then we embed this set M m canonically into M m and now M m is exactly the complete set of representatives of O K of norm m, in order words U M m = M. Now we formulate an important lemma. Lemma 2.2. Given any number field K. Take M m as being the complete set of representatives of O K of norm m as defined earlier. Then M m is finite. Proof. The norm of an element g O K is the same as the norm of the ideal (g). In a Dedekind domain, there are only a finite number of ways to factorize any ideal into prime ideals. Moreover there are only finitely many prime ideals of certain norm, we refer to Theorem 5.17c) of [18]. Thus there are only a finite number of combinations. This implies that the are only finitely many elements of certain norm modulo the group action of the O K. So M m is a finite set. For any solution (X, Y ) of the Thue equation we set: β (i) = X θ (i) Y. factorization of the ideal (X θy ), we get that By unique β (i) = µ (i) ɛ (i), where ɛ (i) is a unit and µ (i) M m (i) as defined earlier, from Lemma 2.2 we now that there are only finitely many possibilities for µ (i). 18

19 During the procedure of the first method it will be necessary to solve Thue equations of degree 4. Consider the case where F (X, 1) has at least s 1 real roots and pairs of t complex conjugate roots. If deg(f (X, Y )) = 4, this means that s + 2t = 4 The set of fundamental units in O K is {η 1,..., η r }, with r = s + t 1. The formulas hold for all conjugates, but we will formulate them with the coefficients corresponding with the identity conjugate θ and use β and µ and η i. With this we get the following β = µ for a set of integers {a 1,..., a r }. Hence to find all possible β we need to find an upper bound on A = max i ( a i ). We will use linear forms to find this A. Define for a given i ( ) r Λ k,j η (k) i i := log( α) + a i log i=1 η (j) + a 0 2π 1, with α := µ(k) (θ (i) θ (j) ) µ (j) (θ (k) θ (i) ) i for all combinations of k, j different from i. Because we have defined a discontinuous log we get a 0 times 2π 1, but they will not play a role. We want to know how small the linear forms can become, because this gives us this bound on the largest value that coordinates of an integral point on the curve can take. To find the bound for the linear forms we will introduce an auxiliary lemma, that will help us to relate the size of the logarithm to the maximum exponent of the chosen fundamental unit. Lemma 2.3 (Lemma for the bound on the maximum exponent). Let K be a number field with r fundamental units η i K and let a i Z for 1 i r. Set A := max( a i ) and ɛ = r i=1 ηa i i. Let I = {i 1,... i r } be an arbitrary set of r different indices from the set {1,..., r + 1} then the matrix U I = is invertible. Let t I such that it then holds that log( ɛ (t) ) r i=1 η a i i log( η (i 1) 1 ) log( η (i 1) r )..... log( η (ir) 1 ) log( η r (ir) ) log( ɛ (t) ) = max 1 i r+1 log( ɛ(i) ), A U 1 I. Proof. For the proof we refer to the proof of Lemma VII.2 of [17]. When we use this lemma for our case, we get different cases in all of them we will calculate different bounds A i. In the end our bound on all of the a i will be the maximum of {A 1, A 2, A 3, A 4 }. Now we will explain how in all the cases these A i can be determined as follows. 19

20 Determining the value of A 1 Before we can determine the value of A 1 we have to calculate some important constants. We took θ to be the root of F (X, 1), we have set θ (k) to be the conjugates of θ in 2 the order we have explained earlier. Then we can define c 1 = min i j ( θ (i) θ (j) ) and c 2 = max i j k i ( θ(i) θ (j) ) and c θ (j) θ (k) 3 = c 1 c 2. Now to determine c 4 we have to use the following matrix defined in Lemma 2.3. So we r is the number of fundamental units that we have, i.e. r = s+t 1. We remind that I = {i 1,..., i r } is an r-subset of {1,..., s+t}, and the matrix U I is defined in the following way: log( η (i 1) 1 ) log( η (i 1) r )..... log( η (ir) 1 ) log( η r (ir) ) We calculate the U 1 I, where Ui 1 is the inverse of the just defined matrix U I. Now we c 4 = max( U 1 J ) for all possible r-subsets (J) of {1,..., s + t}. Now we take c 5 to be a positive real number smaller then c 4 n 1, for example you can take c c 5 = 4 n Recalling that µ were the elements of O K of norm m, we set c 6 = max 1 i n µ (i) 1. Now we are ready to calculate A 1. This is the case that β (i) > e c5 A (index i is chosen such that β (i) is the smallest of all the conjugates of β) and the case that ɛ (t) e c 4 A 1 according to Lemma 2.3. We choose c 6 to be the maximum of all. So now we get µ (t) e c4 A ɛ (t) = β(t) µ (t) c 6 β (t). Now we get an upper bound on the β (t) as follows: β (t) < m l t β (l) 1 < m β (i) (n 1) < m e c 5 (n 1) A. Combing everything we get the following inequality: e c 4 A < c 6 m e c 5 (n 1) A. We take the logarithm at both sides to get: c 4 A < log(c 6 m ) + c 5 (n 1) A. This leads to the following bound on A: So the value of A 1 is equal to A < log(c 6 m ) c 4 (n 1)c 5. log(c 6 m ) c 4 c 5 (n 1) Determining the value of A 2 This will be the other case that β (i) > e c 5 A and the case that ɛ (t) e c 4 A. In this case all the c i, which we have determined earlier are the same, we define only c 7, which is equal to min 1 i n µ (i) 1. Now we get the following inequalities: e c 4 A ɛ (t) = β(t) µ (t) c 7 β (t) c 7 β (i) > c 7 e c 5 A. 20

21 Again after we take on both sides the logarithm we get c 4 A log(c 7 ) c 5 A. We get the following bound on A: A log(c 7) c 4 c 5. Therefore we take A 2 to be equal to log(c 7) c 4 c Determining the value of A 3 This is the case that β (i) e c5 A. We again have two cases either A > log(2c 3) c 5 or not. So if this is not the case then we find that the value of the third bound A 3 is log(2c 3) c Determining the value of A 4 This is the case that β (i) e c 5 A and A > log(2c 3 )/c 5. In this case we get that Λ 2c 3 e c 5 A. We use the theory of linear forms in logarithms from Chapter V of [17]. To define the bound A 4 we will make use of height of functions called absolute logarithmic Weil height h( ) and the modified height h m ( ), that are defined for all algebraic numbers. For the definition of the absolute logarithmic height we refer to [3] and the h m is defined as follows: ( h m (a) = max h(a), log(a), 1 d d Now when A > 3 in order to define a bound on A we need to introduce the following constants, first we define c 8, its value is very large number ). 18 (n + 1)!n n+1 (32 D) n+2 log(2nd) i h m (α i ). Here the integer D is the degree of the extension of the splitting field S, in other words D = [S : Q] and for α i such that S = Q(α 1,..., α n ). Using the Theorem A.1 of Appendix A of [17] we a bound on our A, which we define as A 4 : A 4 := 2 c 5 (log(2c 3 ) + c 8 log( s + t 2 ) + c 8 log( c 8 c 5 )). But since c 8 is a "big" constant, we have that A 4 becomes an even larger number. With the help of the LLL-algorithm we can try to lower the bound of A 4 to get a "smaller" bound that we define as A 4. The construction algorithm of A 4 from A 4 we will explain with an example in the next section. As the result for our bound on A we get the maximum of all previously defined A i and 3, in other words: A max(3, A 1, A 2, A 3, A 4 ). 21

22 2.2. LLL-algorithm We have come across linear forms in this section we will explain how we will use the linear forms. Given a linear form Λ = d 1 a 1 + d 2 a 2... d r a r, where a i are variable and d i C are coefficients and let A = max 1 i n ( a i ), our goal is to find a reasonably small bound on A. The LLL-algorithm takes as basis the columns of a matrix and transforms them into a smaller basis. This algorithm was invented by Arjen Lenstra, Hendrik Lenstra and László Lovász in 1982 and is now being used by many programs and engines, also by SageMath. To read more about this algorithm we refer to their publication [10] and for more recent information to Chapter 2 from [6]. We will use the LLL-algorithm to reduce the bound of our A 4. We take the bound of A 4 and we use it as an initial bound. We take a C that is larger than A r 4 (r is the number of variables that we have in our linear form). For the linear form inequality: Λ = d 1 a 1 + d 2 a 2... d r a r f 1 e f 2 A q, for some constants f 1, f 2 and q Z. we construct the r r matrix that is defined by the following columns (R defines real part of a complex number and I its imaginary part) C R(d 1 )... C R(d r 2 ) C R(d r 1 ) C R(d r ) C I(d 1 )... C I(d r 2 ) C I(d r 1 ) C I(d r ) Using the columns of this matrix as a basis we do an LLL-algorithm to reduce this basis. Now to calculate the new bound on A, we first define some constants S = (r 2) C 2 and T = (1+r C) 2 and C special = k1 1 b 1 (where b 1 is the first column of the reduced LLLbasis and k 1 = max 1 i n ( b 1 b ), b i is the orthonormal reduced basis of the new reduced i matrix). Then A new the new bound on A is given by the formula A new := q 1 f 2 (log(c f 1 )) log( C special S T ). We can repeat this process a couple of times until the bound on A stops getting smaller or if we get a small enough bound to run through all possible cases, this last bound we will define as A Example of solving a Thue equation We will illustrate now the explained theory on an example of a Thue equation. Before we start we will introduce two height functions defined on an element of the number field 22

23 K of degree d that we will use in this example: absolute logarithmic Weil height h(a) for its definition we refer to [3] and the modified height (h m ). The modified height will be defined as ( h m (a) = max h(a), log(a), 1 ). d d We will find all the solutions of the equation X 4 2Y 4 = ±1. We set θ to be the root of the equation θ 4 2 = 0 and we take two fundamental units to be: η 1 = 1+θ 2, η 2 = 1+θ. Hence we find all possible exponents a i such that X θy = β = ±η a 1. We label now 1 ηa 2 2 the roots of X 4 2 = 0, so that θ (1) = 4 2, θ (2) = 4 2, θ (3) = θ (4) = We can calculate c 1, c 2, c 3 and c 4. Now we take c 5 = 0.65, so that it satisfies the inequality c 5 < c 4 n 1 = c 4 3. Also we can take the constants c 6 and c 7 equal to 1. So we get the trivial upper bounds of A 1 = A 2 = 0. We need to consider two cases for when Y Y 1 = 1, there are two cases (where i = 1 or i = 2), for the index i, such that β (i) = min 1 i 4 β(i) Let us first consider the case i = 1. We set j = 3 and k = 2. We consider the linear form in logarithms and use the LLL-algorithm: Λ = log( α 3 ) + r i=1 We notice that α 2 = ±( θ(1) θ (3) θ (2) θ (1) ( (2) ) η 1 η (3) = 1 ( 1 + θ 2 a i log( η(2) i η (3) i ) + a 0 2π 1, with a 0 Z = ± we also have 1 θ 2 ) ( (2) ) ( ) η 2 1 θ, η (3) = θ 2 The first has a minimum polynomial: X 2 + 6X + 1 and the second has a minimum polynomial: X 8 + 8X X 6 136X X 4 136X X 2 + 8X + 1. Using SageMath we compute the heights: ( (2) ) ( (2) ) η η h(α 2 ) = , h = , h = η (3) 1 2 η (3) 2 Now we consider modified heights, using appendix A1 of [17]. h m (α 2 ) = max{0.3465, , 0.125} = ( (2) ) η 1 h m η (3) = max{ , , 0.125} = ( (2) ) η 2 h m η (3) = max{ , , 0.125} =

24 Using the theory of the appendix A1 of [17], so we can give a lower bound to our Λ : ( ) 3A log Λ c 8 log, 2 ( (2) η where c 8 = 18 5! (log 64)h m ( 1)h m (α 2 )h m η (3) 1 = ( (2) ) η 2 )h m η (3) 2 Now we can see that if A A 3 = 1.866, then A A 4 = Now we can use the LLL-algorithm to try to get a smaller bound on A. Using LLL -algorithm with C = we get that A new = 49, after this we repeat the algorithm using C = 10 5, which gives us eventually A 4 = 8. This bound is reasonable small and we can enumerate all possibilities to find all solutions. Let us now consider the case i = 2. We choose again ( j and k, ) now j = 3 and k = 1. θ We consider the values of Λ. We see that α 2 = ± (2) θ(3) = ± 1+ 1 θ (1) θ (3) 2, we also get ( (1) ) η 1 η (3) = 1 + ( θ2 (1) ) η 1 θ 2, 2 1 η (3) = 1 + θ 1 + 1θ 2 These two numbers have the minimal polynomials: X 2 + 6X + 1 and X 8 + 8X X 6 136X X 4 136X X 2 + 8X + 1. We compute the heights and then also the modified heights and we get that c 8 = So we get that A A 4 = Using the LLL-algorithm with C = 10 40, we get that A new = 49 and after using the same algorithm with C = 10 4, we get that A 4 = 7. So in both cases A 8. So we need to check all the solution below this bound. And then we find that the only integral solutions below this bound to the equation X 4 2Y 4 = ±1 are (X, Y ) = {±(1, 0), ±(1, 1), ±(1, 1)}. This means that in the previous hapter we did determine all possible solutions to the Thue equation Y 2 X 4 = Method of Bilu and Hanrot This is an alternative method for determining the upper bound of A for the Thue equation. We let U I to be the matrix as defined in Lemma 2.3 and we take an i {1,..., s + t} \ I and we let (u i,j ) = U 1 I, with I being the same as before, so that β (i) = min 1 l r+1 β(l) 24

25 From the following equation: U i a 1. a r = log x θ(i 1 ) y µ (i 1 ). log x θ(ir) y µ (ir) we deduce for k = 1,..., r that a k = = = r u k,j log x θ (ij) y µ (i j) j=1 r u k,j log y + j=1 r u k,j log y + j=1 = δ k log y + λ k + r u k,j log j=1 x y θ(i j) µ (i j) r u k,j log θ (i) θ (i j) + j=1 µ (i j) r x u k,j log y θ(i j) θ (i) θ (i j) j=1 r x u k,j log y θ(i j) θ (i) θ (i j) j=1 We have defined δ k = r j=1 u k,j and λ k = r j=1 u k,j log x large enough, we get that y θ(i j ) θ (i) θ (i j ) n+2 n+1. Using this fact we can estimate the term r j=1 u k,j log r x u k,j log y θ(i j) θ (i) θ (i j) j=1 1 n r j=1 u k,j log n + 2 n + 1 r u k,j j=1 θ(i) θ (i j ) µ (i j ) x y θ(i j ) θ (i) θ (i j ) r j=1. When y becomes, we get that u k,j log n + 2 n + 1 Then it follows that ( A c 10 log y + ) c 11, where c 10 = max 1 k r δ k and 1 c 11 = max r 1 k r n j=1 u k,j + λ k. Rewriting the previous inequality we get y 1 c 12 e c13a, where c 12 = exp( c 11 c 10 ) and c 13 = c Now when y is sufficiently large, we see 25

26 that: x log y θ(i j) x θ (i) θ (i j) log(1 + y θ(i) θ (i) θ (i j) ) x y 2 θ(i) θ (i) θ (i j) x c 1 y θ(i) c 1 c 9 y n c 1 c 9 c n 12e nc 13A = c 14 e c 15A Using the fact we now see that we get the following inequality: r x δ k log y a k + λ k u k,j log y θ(i j) θ j=1 (i) θ (i j) r u k,j c 14 e c 15A where we define j=1 c 16 e c 15A, c 16 := c 14 max 1 k r r j=1 u k,j = c 14 U 1 1. Now we write down the Bilu and Hanrot equation, first we define index h to be the one for which holds: δ h = max 1 k r δ k. From the others we define g to be an other index than h. Then we define δ = δg δ h and λ = δgλ h δ h λ g δ h then we get the following: a g δa h + λ = a g + δa h δ gλ h δ h λ g δ h = δ g log y a g + λ g δ g log y + δa h δλ h = (δ g log y a g + λ g ) δ(δ h log y a h + λ h ) δ g log y a g + λ g + δ δ h log y a h + λ h (1 + δ )c 16 e c 15A 2c 16 e c 15A. We can use this inequality to reduce the bound on A using LLL reduction of two dimensional lattices. However it is more efficient to use continued fractions. We can use this inequality to search for the small solutions to the equation when we can reduce the bound on A. This would mean that we would need to check (2A 6 + 1) r possibilities. But we can reduce the sieving method on the exponents. Rewriting the right hand sides depending on y in stead of A, in the cases when y becomes big we get the following: a g 1 δ h (δ g a h δ g λ h + δ h λ g ) <

27 And now we loop through all possible values of a k and then all possible values of a g for all g h are determined explicitly. Hence we need only check 2A exponent vectors. So what implies that the final search is only depending on A 6 and not on r and so not on the degree of the equation. But you still need to calculate the fundamental units and the complete set of representatives for the µ and then the calculating and solving the Thue equation is a fast process. For more information about this method we refer to Section VII.3 of [17] 2.5. Program I have written the program programfindboundthueequation, based on the theory of this chapter. It solves Thue equations, for the code of this program we refer to Appendix A. This program is essential part of programone which will be explained in the next chapter. I have tried to write program programfindboundthueequation so that it can solve the general case of a Thue equation, unfortunately in the end I had to make some restrictions. The program inputs a homogeneous polynomial Q(p, q) defined in variables p,q and an array of numbers m i, for which you want to solve the Thue equation Q(p, q) = m 2 i, m 2 i Z. Careful that here it is important that m 2 i Z, this construction was done for the smooth running of program programone. Therefore it is allowed to choose imaginary numbers and square roots m i as long as m 2 i stays Z. The beginning of program programfindboundthueequation is not dependent on the degree the polynomial. Only in the end there is a restriction on the number of fundamental units. Since programone produces always a degree four Thue equations that needs to be solved, I have restricted the end of program programfindboundthueequation to the cases that can occur only in degree four extensions of Q, namely 1 or 2 or 3 fundamental units. If you want to solve a higher degree Thue equation, you can use my program with a slight upgrade. Then you need to add other possible cases on the number of fundamental units. How program programfindboundthueequation works, is made more clear in the Appendix A and also in the last section of the next chapter, where we discuss program the working of program programone. 27

28 3. Finding integral points using Thue equations 3.1. Theory of the method In this chapter we explain how the theory of the Thue equations can be applied to find all integral point on a curve defined by the equation Y 2 = X 3 + ax 2 + bx + c, where a, b, c Z. We remind that we will consider only those a, b, c for which X 3 +ax 2 +bx +c does not have multiple roots, in other words the discriminant should be nonzero. We follow the principle explained in Section 2 of [4]. We denote A to be the algebra, with A = Q[X]/(X 3 + ax 2 + bx + c). Then there are three possibilities for A to be. If X 3 +ax 2 +bx +c has three different roots that are integers, then A = Q Q Q, If X 3 + ax 2 + bx + c has one root that is an integer and two quadratic roots θ = θ (1), θ (2), then A = Q Q(θ), If X 3 +ax 2 +bx +c has no roots that are Z, so X 3 +ax 2 +bx +c is irreducible, then there is a θ, such that θ 3 + aθ 2 + bθ + c = 0 and in this case A = Q(θ). We consider the last case, we start by writing A = r i=1 A i, where A i = Q(θ i ) (for i {1, 2, 3}) and θ i (with i {1, 2, 3}) being the corresponding root of cubic polynomial X 3 + ax 2 + bx + c = (X θ 1 )(X θ 2 )(X θ 3 ). Suppose (X, Y ) is an integral solution to our equation, then X θ i A i is an algebraic integer. We consider a prime ideal O Li that is a divisor of an odd degree of the ideal (X θ i ). From the unique factorization of ideals and the equation of ideals that we have (Y ) 2 = (X θ i )(X θ j )(X θ k ) we get the following information. Since divides the right hand side, it has to divide the left hand side, thus it should divide the ideal (Y ). Then since it divides (Y ), an even power of ideal divides the left hand side and because an odd power of divides the ideal (X θ i ), has to divide the ideal (X θ j ), for a j i. Now because divides both the ideals (X θ i ) and (X θ j ) it has to divide also their difference, so divides the ideal (θ j θ i ). From this follows that ideal divides the discriminant of X 3 + ax 2 + bx + c. So we are interested in the set of prime ideals that divide the discriminant, we will call this set S i. 28

29 We can write now X θ i = α i β 2 i, with the following three conditions that α i A i (S i, 2) (α i is square free, so we denote with A i (S i, 2) the set A i /(A i )2 with the added generators of the prime ideals that did divide the discriminant); β i = (xb 1 + yb 2 + zb 3 ) A i (where b 1, b 2, b 3 form a basis of A i ); r i=1 N A i /Q(α i ) is a rational square. In this situation three quadratic forms Q 1, Q 2, Q 3 with integral coefficients in three variables x, y, z can always be constructed, just by rewriting α i β 2 i to a form Q 1θ 2 1 +Q 2θ 1 +Q 3, using the equality that θ 3 1 = aθ2 1 bθ 1 c. After equating the coefficients of the equality X θ i = α i β 2 i we see that finding an integral point on our curve is equivalent to finding integers x, y, z to the system: Q 1 (x, y, z) = 0, Q 2 (x, y, z) = 1, Q 3 (x, y, z) = X. We illustrate this method with the following example Example of the method We consider the curve Y 2 = X 3 6X 14. We take θ, being a solution of θ 3 6θ 14 = 0. We determine that class number of Q(θ) is one, so one fundamental unit forms a basis of OK, we take for it η = (5 8θ+2θ2 ) 3. We find also that as the integral basis of Q(θ) we can take ω 1 = 1, ω 2 = θ and ω 3 = 1 θ+θ2 3. The discriminant of X 3 6X 14 is 4 1 ( 6) ( 14) 2 = 4428 = So we want to find the generators of the prime ideals that divide (2), (3) and (41), which split in O K into: (2) = 3 2, (3) = 3 ( 3) 2, (41) = 41 ( 41) 2. The generators of these prime ideals we can take the following: π 2 = 8 + θ θ2 3 π 3 = 5 + θ θ2 3 π θ + θ2 = θ 2θ2 π 41 = 3 π θ + θ2 = 3 for prime ideal 2, for prime ideal 3, for prime ideal 3, for prime ideal 41, for prime ideal