Diophantine Equations with Fibonacci and Pell Numbers By
Mahadi Ddamulira (
[email protected])
June 2015
A RESEARCH PROJECT PRESENTED TO AIMS-GHANA IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE AWARD OF MASTER OF SCIENCE IN MATHEMATICAL SCIENCES
Declaration This work was carried out at AIMS-Ghana in partial fulfilment of the requirements for a Master of Science Degree in Mathematical Sciences. I hereby declare that except where due acknowledgement is made, this work has never been presented wholly or in part for the award of a degree at AIMS-Ghana or any other University.
Student: Mahadi Ddamulira
Supervisor: Prof. Florian Luca
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Acknowledgements I would like to thank the Almighty Allah for guiding and giving me natural endowment, skills, healthy mind and a strong body to work on this research project. It is with immense gratitude that I acknowledge my supervisor, Prof. Florian Luca for his immeasurable supports, inspirations, criticisms, comments and suggestions rendered through out the process of preparing this research project. May the God bless you abundantly. Special thanks to Dr. Betty K. Nannyonga, Dr. David Ssevviiri and Mr. Alex B. Tumwesigye in the department of Mathematics at Makerere University, for it is by their efforts that I came to know about AIMS, in addition to a firm mathematical foundation they laid for me at the undergraduate level. May God bless you all. I can not forget the patience, love and care from my family members, I really appreciate it. Finally, the AIMS-Ghana 2014/2015 tutors: Tomiwa Ajagbonna, Seth Kurankyi Asante, Grace Omollo Misereh, Frantisek Hajnovic, Amsalework Ejigu and all those who are not mentioned but whose help are truly contributing in adding some significant information will not be forgotten, I thank you all and may the Almighty reward you.
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Dedication To my parents and relatives - the family of Sheikh Ali Muwoomya.
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Abstract There are many Diophantine equations whose aim is to determine the intersection of two binary recurrent sequences (Fibonacci numbers, Pell numbers, etc.). In this research project, we find all Fibonacci numbers which are products of two Pell numbers and all Pell numbers which are products of two Fibonacci numbers. Key words: Diophantine equations, Continuity fractions, Binary recurrent sequences, Fibonacci numbers, Pell numbers, Lower bounds for linear forms in logarithms.
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Contents Declaration
i
Acknowledgements
ii
Dedication
iii
Abstract
iv
1 Introduction
1
1.1
Background Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Literature Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.3
Layout of the research project . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
2 Algebraic Numbers and Fields
4
2.1
A field and examples of fields . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
2.2
Algebraic numbers and algebraic integers . . . . . . . . . . . . . . . . . . . . .
5
2.3
Field extensions and algebraic number fields
6
. . . . . . . . . . . . . . . . . . .
3 Continued Fractions
8
3.1
Finite continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
3.2
Some properties of continued fractions . . . . . . . . . . . . . . . . . . . . . .
9
3.3
Infinite continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
3.4
Convergents of continued fractions . . . . . . . . . . . . . . . . . . . . . . . .
11
4 Binary Recurrent Sequences
14
4.1
Basic definitions and properties . . . . . . . . . . . . . . . . . . . . . . . . . .
14
4.2
Examples of binary recurrent sequences . . . . . . . . . . . . . . . . . . . . . .
15
4.3
Some important inequalities involving Fibonacci and Pell sequences . . . . . . .
19
5 Lower Bounds for Linear Forms in Logarithms 5.1
Basic definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
21 21
5.2
Effective approximation the lower bound . . . . . . . . . . . . . . . . . . . . .
22
5.3
Reducing the bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
6 Proof of Theorem 1.1.1
24
6.1
Approximation of lower bounds on k, m and n . . . . . . . . . . . . . . . . . .
24
6.2
Reducing the bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
7 Comments and Conclusion
32
Appendix A
33
A.1 Sage source code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References
33 35
vi
1. Introduction 1.1
Background Overview
We define the sequences {An }n≥0 and {Bn }n≥0 for all positive integers n ∈ Z+ by ( An+2 = aAn+1 + An , A0 = 0, A1 = 1, Bn+2 = aBn+1 + Bn , B0 = 2, B1 = a.
(1.1.1)
For a = 1 we write {An } = {Fn } and {Bn } = {Ln }, which are the Fibonacci and Lucas sequences, respectively, given by ( Fn+2 = Fn+1 + Fn , F0 = 0, F1 = 1, (1.1.2) Ln+2 = Ln+1 + Ln , L0 = 2, L1 = 1. The Binet formulas for the general terms of the Fibonacci and Lucas sequences are obtained using standard techniques for solving linear recurrent sequences that we discuss in Chapter 4, are given by Fn =
αn − β n α−β
and Ln = αn + β n ,
(1.1.3)
where α and β are the roots of the equation x2 − x − 1 = 0. For a = 2, we write {An } = {Pn } and {Bn } = {Qn }, which are the Pell and Pell-Lucas sequences, respectively, given by ( Pn+2 = 2Pn+1 + Pn , P0 = 0, P1 = 1, (1.1.4) Qn+2 = 2Qn+1 + Qn , Q0 = 2, Q1 = 2. The Binet formulas for the general terms of these sequences are given by Pn =
γ n − δn γ−δ
and Qn = γ n + δ n ,
(1.1.5)
where γ and δ are the roots of the equation x2 − 2x − 1 = 0. Further details about the Fibonacci, Lucas and Pell sequences are discussed in Chapter 4. In this research project, we study the Diophantine equations involving the well-known Fibonacci and Pell sequences, given below: Fk = Pm Pn
(1.1.6)
P k = Fm Fn .
(1.1.7)
and
Our results are summarised in the following theorem. 1
Section 1.2. Literature Review
Page 2
Theorem 1.1.1. Considering the two Diophantine equations above: (i) the only positive integer solutions (k, m, n) of equation (1.1.6) have k = 1, 2, 5, 12. (ii) the only positive integer solutions (k, m, n) of equation (1.1.7) have k = 1, 2, 3, 7. It is known that 144 = 122 and 169 = 132 are the largest squares in the Fibonacci and Pell sequences, respectively, and 12 and 13 are Pell and Fibonacci numbers respectively. The two results were proved by J. H. E. Cohn in 1964 in [1] and in 1996 in [2], respectively. Therefore, Theorem 1.1.1 tells us that there are no larger Fibonacci or Pell numbers which are products of two numbers from the other sequence. The detailed proof of Theorem 1.1.1 is given in Chapter 6. In the process of proving the theorem we also use the properties of linear forms in logarithms and the Dujella-Peth˝o [3] reduction procedure which are discussed in Chapter 5 and Chapter 3, respectively, to solve the Diophantine equations (1.1.6) and (1.1.7).
1.2
Literature Review
Diophantine equations are polynomial equations which contain two or more unknowns, and only the integer values of these unknowns are to be found. Some forms of Diophantine equations include a linear Diophantine equation, which is an equation between two sums of monomials of degree zero or one, and an exponential Diophantine equation, which is one in which exponents on terms can be unknowns. Diophantine equations were named after the Hellenistic mathematician of the 3rd century, Diophantus of Alexandria who studied such equations and was also the first to introduce the usage of symbols in algebra [4]. While Diophantine equations are mostly studied as a form of theoretical mathematics, they do have a few applications which include cryptography and also in the field of chemistry, where Diophantine equations are used to balance out chemical equations and determine the molecular formulas of compounds. Some famous examples of Diophantine equations include the Pell equation which is in the form x2 + dy 2 = ±1 as well as the Erd˝os-Straus conjecture which states that for every integer n ≥ 2 there exist three positive integers x, y, z such that 4/n = 1/x+1/y+1/z. This is discussed in [5]. When m = 1 in equation (1.1.6) or k = 1 in equation (1.1.7), the resulting Diophantine equation is of the form Un = Vm for some m, n ≥ 0, (1.2.1) where {Un }n≥0 and {Vm }m≥0 are the Fibonacci and Pell sequences, respectively. More generally, there is a lot of literature on how to solve equations like (1.2.1) in case {Un }n≥0 and {Vm }m≥0 are two non degenerate linearly recurrent sequences with dominant roots. See, for example, [6], [7] and [8]. The theory of linear forms in logarithms “´a la Baker” gives that, under reasonable conditions (say, the dominant roots of {Un }n≥0 and {Vm }m≥0 are multiplicatively independent), equation (1.2.1) has only finitely many solutions which are effectively computable. In fact, a
Section 1.3. Layout of the research project
Page 3
straightforward linear form in logarithms gives some very large bounds on max{m, n}, which then are reduced in practice either by using the LLL algorithm or by using a procedure originally discovered by Baker and Davenport [9] and perfected by Dujella and Peth˝o [3]. In this research project, after identifying this gap in literature, we concentrate on solving the diophantine equations of the forms Uk = Vm Vn
and Vk = Um Un
for some k, m, n ≥ 0,
(1.2.2)
where {Uk }k≥0 and {Vk }k≥0 are the Fibonacci and Pell sequences, respectively, as before. These forms represent the equations (1.1.6) and (1.1.7). We also use linear forms in logarithms and the Dujella-Peth˝o’s reduction procedure to solve equations (1.1.6) and (1.1.7).
1.3
Layout of the research project
In Chapter 2, we give an introduction to algebraic numbers and number fields and some of the basic properties of these, that will later on be used in the subsequent chapters of this research project. In Chapter 3, we introduce all the aspects of the continued fractions that will be useful in the subsequent chapters. We shall also insist on the property of convergents of an irrational number α as being “the best” approximations of α by rationals. In Chapter 4, we take a brief tour of the theory of linearly recurrent sequences of order 2. We also define the Lucas sequences and observe that the sequence of Fibonacci numbers is an example of such a sequence. Various examples are given on expressing these sequences in terms of their Binet forms. In Chapter 5, we give various specific statements of which belong to what is known as the theory of lower bounds for linear forms in logarithms of algebraic numbers. We give various examples on determining the logarithmic height of an algebraic number, which are later used in Chapter 6. It is here, where we give the statement of the Matveev theorem. In Chapter 6, we give the detailed proof of Theorem 1.1.1, which is our main result, by applying the theorem of Matveev [10] to determine the effective lower bounds on k, m, n and also apply the Dujella-Peth˝o reduction procedure to reduce these bounds. There are certain little computations along the way, in the subsequent sections of the research project, which we do using sage but, they can also be done with either Maple or Mathematica.
2. Algebraic Numbers and Fields In this chapter, we give an introduction to algebraic numbers and number fields and some of the basic properties of these, that will later on be used in this project. We assume familiarity with the definition and basic properties of vector spaces.
2.1
A field and examples of fields
Definition 2.1.1 (Field). A field is a set F with two operations, called addition (+) and multiplication (×) such that for any x, y, z ∈ F : • x + y and x × y (= xy) are uniquely defined elements of F , • x + (y + z) = (x + y) + z, • x + y = y + x, • there is an element 0 ∈ F such that 0 + x = x, • for any x ∈ F there is an element −x ∈ F such that (−x) + x = 0, • x(yz) = (xy)z, • xy = yx, • there is an element 1 ∈ F \{0} such that 1 × x = x, • for any x 6= 0 ∈ F there is an x−1 ∈ F such that xx−1 = 1, • x(y + z) = xy + xz. A field is just a set of elements that you can add, subtract, multiply and divide (non-zero elements) so that the usual rules of algebra are satisfied. The familiar examples of fields are Q, R and C; without the division (so if we don’t ask of every non-zero element to have an inverse) then the structure is called ring. If additionally the multiplication is not commutative it is non-commutative ring for instance, matrices form a non-commutative ring as opposed to integers or polynomials which form commutative rings. Definition 2.1.2 (Subfield). A subfield of a field F is subset that also forms a field under the same addition (+) and multiplication (×). Thus, Q is a subfield of R which is in turn a subfield of C an so on.
4
Section 2.2. Algebraic numbers and algebraic integers
2.2
Page 5
Algebraic numbers and algebraic integers
Definition 2.2.1 (Algebraic number). An algebraic number is a root of a monic polynomial with rational coefficients. That is α is an algebraic number if and only if there exists n ≥ 1 and a1 , a2 , . . . , an ∈ Q such that αn + a1 αn−1 + a2 αn−2 + · · · + an−1 α + an = 0.
(2.2.1)
Definition 2.2.2 (Algebraic integer). An algebraic integer is a root of a monic polynomial with integer coefficients. That is β is an algebraic integer if and only if there exists n ≥ 1 and b1 , b2 , . . . , bn ∈ Z such that β n + b1 β n−1 + b2 β n−2 + · · · + bn−1 β + an = 0.
(2.2.2)
From the above definitions, we can clearly see that, every algebraic integer is an algebraic number. We also think of algebraic integers and algebraic numbers as generalisations of integers and rational numbers, respectively. Thus, all integers are algebraic integers and all rational numbers are algebraic numbers [11]. Some examples of algebraic integers include: √
√ √ 1 3 (1 + 5), 2 cos 2, i, 5, 2
2π 9
,....
The examples of algebraic numbers include: r √ 1 1 1 2π , (1 + i), (1 + 3i), 2 sin ,.... 5 2 2 7 Remark 2.2.3. The numbers e and π are neither algebraic numbers nor algebraic integers, they are called transcendental numbers. Definition 2.2.4 (Minimum polynomial). Let α be an algebraic number. Then among all polynomials with rational coefficients with α as a root, there exists a unique polynomial f (X) of the lowest degree, which is monic (with leading coefficient equal to one) and also irreducible (can not be factorised further non-trivially). This polynomial is called the minimum polynomial of the algebraic number α. Definition 2.2.5. The degree of the algebraic number α is the degree of its minimum polynomial f (X). Remark 2.2.6. The existence of minimum polynomials of any degree n implies the existence of algebraic numbers of degree n. For example, all rational numbers are algebraic numbers of degree one. The complex number i is an algebraic number of √ degree two since it is a root of the irreducible polynomial x2 + 1 over the field Q. The number n 3, where n is any positive integer, is an algebraic number of degree n since it is a root of the irreducible polynomial xn − 3 over Q [11].
Section 2.3. Field extensions and algebraic number fields
2.3
Page 6
Field extensions and algebraic number fields
Definition 2.3.1 (Field extension). Let F and K be fields with F a subfield of K. Then we call K a field extension of F . This is denoted by K/F . Definition 2.3.2 (Degree of a finite field extension). A field extension K/F is finite if K is a finite dimensional vector space over F . The degree [K : F ] of a finite extension K/F is the dimension of K as a vector space over F . Definition 2.3.3. An element α ∈ K is algebraic over a field F if α is the root of some nonzero polynomial f ∈ F [x] (This is the minimum polynomial of α over F ). A field extension K/F is algebraic if every element α of K is algebraic over F . Proposition 2.3.4. Suppose that K/F is a field extension and that α is an element of K. Suppose that α is algebraic over F , with minimum polynomial f ∈ F [x] of degree d. Then the extension F (α)/F is finite and has degree d. Proof. See [12]. Proposition 2.3.5 (Tower law). Suppose K/E and E/F are field extensions. Then K/F is a finite field extension if and only if K/E and E/F are finite and in this case the degree [K : F ] = [K : E][E : F ].
(2.3.1)
Proof. Suppose [K : E] = n and [E : F ] = m. We assume B1 = {α1 , α2 , . . . , αn } is a basis for K over E and B2 = {β1 , β2 , . . . , βm } is a basis for E over F . Then every element α ∈ K can be written as a linear combination n X α = a1 α1 + a2 α2 + · · · + an αn = ai αi , ai ∈ E. i=1
By writing each element ai ∈ E as a linear combination of the elements of the basis B2 , that is to say, ai = bi1 β1 + bi2 β2 + · · · + bim βm =
m X
bij βj ,
j=1
we get α=
n X
ai αi =
i=1
n m X X i=1
j=1
! bij βj
αi =
n X m X
bij αi βj ,
bij ∈ F.
i=1 j=1
Then to prove that αi βj ∈ K,where 1 ≤ i ≤ n and 1 ≤ j ≤ m are linearly independent, we set α = 0. Then we have ! n m m X X X bij βj αi = 0 which implies bij βj = 0, i=1
j=1
j=1
because B1 is a linearly independent set of K over E. We also have that, B2 is a linearly independent set of E over F , this also implies that bij = 0 for all i and j. Thus it immediately follows that [K : F ] = nm. A slightly similar proof is given in [12].
Section 2.3. Field extensions and algebraic number fields
Page 7
√ √ Example 2.3.6. We consider the field extension Q( 2, 5)/Q which is nothing other than the sequence of two simple field extensions √ √ √ Q ⊂ Q( 2) ⊂ Q( 2, 5). We use Proposition 2.3.4 to find the degrees of the individual√field √extensions and then apply the Tower law to determine the degree of the filed extension Q( 2, 5)/Q. √ √ Firstly, we consider the field extension Q( 2)/Q. The minimum polynomial of 2 over Q is √ is x2 − 2 because x2 − 2 is a monic polynomial in Q[x] for which √ 2 is a root. √ This polynomial √ also irreducible over Q since any factorisation of x2 − 2 is (x − 2)(x + 2) for which 2 ∈ / Q. Thus by Proposition 2.3.4, the degree of the first field extension is given by h √ i Q : Q( 2) = deg(x2 − 2) = 2. √ √ Secondly, we let F = Q(√ 2), so that the second extension is written as F ( 5)/F where the minimum polynomial of 5 over F is x2 − 5 for the same reasons as before. Then, the degree of the second field extension is given by h √ √ √ i Q( 2) : Q( 2, 5) = deg(x2 − 5) = 2. √ √ Finally, by the Tower law, the degree of the field extension Q( 2, 5)/Q is given by h √ √ i h √ ih √ √ √ i Q : Q( 2, 5) = Q : Q( 2) Q( 2) : Q( 2, 5) = 2 × 2 = 4. Definition 2.3.7 (Algebraic number field). An algebraic number field (or simply number field) is a finite degree field extension of the field of rational numbers. Here its dimension as a vector space over Q is simply called its degree. Example 2.3.8. We consider the following examples of number fields which are considered in the subsequent chapters. √ (i) For any square-free integer d, the quadratic field Q( d), is a number field obtained by adjoining the square √ root of√d to the √ field of rational numbers. For instance, these are quadratic fields Q( 2), Q( 3), Q( 5), . . .. The degree of this number field is 2. √ √ (ii) The number field, L = Q( 2, 5) is obtained by adjoining the square roots of 2 and 5 to the field of√rational √ numbers. The degree of this number field is the degree of the field extension Q( 2, 5)/Q, which is 4.
3. Continued Fractions In this chapter, we recall those aspects of the continued fractions that will be useful later on in the course of the project.
3.1
Finite continued fractions
A continued fraction is an expression obtained from an iterative process of representing a number as a sum of its integer part and the reciprocal of another number, then writing this other number as a sum of its integer part and another reciprocal, and so on. In a finite continued fraction, the iteration is terminated after finitely many steps by using an integer in lieu of another continued fraction. On the other hand, an infinite continued fraction is an infinite expression [13]. Definition 3.1.1.
(i) A finite continued fraction is an expression of the form 1
a0 +
,
1
a1 +
a2 + · · · +
1 an−1 +
(3.1.1) 1 an
where a0 ∈ R and ai ∈ R+ for all 1 ≤ i ≤ n. For the purpose of simplicity, we shall use the notation [a0 , . . . , an ] for the above expression. (ii) The continued fraction [a0 , . . . , an ] is called simple if a0 , . . . , an ∈ Z. (iii) The continued fraction Ck = [a0 , . . . , ak ] with 0 ≤ k ≤ n is called the k−th convergent of [a0 , . . . , an ]. Example 3.1.2. We can write 13 1 1 1 1 47 =2+ =2+ =2+ =2+ =2+ = [2, 1, 3, 4]. 17 4 1 1 17 17 1+ 1+ 1+ 13 1 13 13 3+ 4 4 It is clear that every simple continued fraction is a rational number. Conversely, by the Euclidean algorithm, every rational number can be represented as a simple continued fraction. In fact, if p and q > 0 are relatively prime and we write p = r0 a0 + r1 , r0 = r1 a1 + r2 , ... = ..., rn−1 = rn an ,
1 ≤ r1 < r0 := q; 1 ≤ r2 < r1 ;
(3.1.2)
where n ≥ 0 is maximal such that rn ≥ 1, then we can easily see that the expression in (3.1.1) is p/q through a series of backward substitutions. 8
Section 3.2. Some properties of continued fractions
3.2
Page 9
Some properties of continued fractions
For each continued fraction [a0 , . . . , an ] we define p0 , . . . , pn and q0 , . . . , qn via p 0 = a0 , q0 = 1; p1 = a0 a1 + 1, q 1 = a1 ; pk = ak pk−1 + pk−2 , qk = ak qk−1 + qk−2 for all k = 2, . . . , n. Proposition 3.2.1. With the previous notation, we have: (i) Ck = pk /qk ; (ii) pk qk−1 − pk−1 qk = (−1)k−1 for all k ≥ 1; (iii) Ck − Ck−1 =
(−1)k−1 qk qk−1
for 1 ≤ k ≤ n;
(iv) Ck − Ck−2 =
(−1)k ak qk qk−2
for 2 ≤ k ≤ n.
Proof.
(i) We use induction on k. If k = 0, then C0 = [a0 ] = p0 /q0 . If k = 1, then C1 = [a0 , a1 ] = a0 +
a0 a1 + 1 p1 1 = = . a1 a1 q1
If k = 2, then C2 = [a0 , a1 , a2 ] = a0 +
1
= a0 +
a2 a1 a2 + 1
1 a2 a2 p 1 + p 0 a2 (a0 a1 + 1) + a0 p2 = = = . a2 a1 + 1 a2 q 1 + q 0 q2 a1 +
Now, we let k ≥ 2 and assume that Ci = pi /qi for all i = 0, . . . , k. We also observe that pk−2 , qk−2 , pk−1 and qk−1 depend only on a0 , . . . , ak−1 . Thus we have, (ak + 1/ak+1 )pk−1 + pk−2 1 = Ck+1 = a0 , a1 , . . . , ak−1 , ak + ak+1 (ak + 1/ak+1 )qk + qk−1 ak+1 (ak pk−1 + pk−2 ) + pk−1 ak+1 pk + pk−1 pk+1 = = = . ak+1 (ak qk−1 + qk−2 ) + pk−1 ak+1 qk + qk−1 qk+1 (ii) If k = 1, then we have p1 q0 − p0 q1 = (a0 a1 + 1)1 − a0 a1 = 1. By induction, if k ≥ 2, then pk qk−1 − pk−1 qk = (ak pk−1 + pk−2 )qk−1 − pk−1 (ak qk−1 + qk−2 ) = −(pk−1 qk−2 − pk−2 qk−1 ) = −(−1)k−2 = (−1)k−1 .
Section 3.3. Infinite continued fractions
Page 10
(iii) By (i) and (ii), we have that Ck − Ck−1 =
pk qk−1 − pk−1 qk (−1)k−1 pk pk−1 − = = . qk qk−1 qk qk−1 qk qk−1
(iv) Finally, we have pk pk−2 pk qk−2 − pk−2 qk − = qk qk−2 qk qk−2 (ak pk−1 + pk−2 )qk−2 − (ak qk−1 + qk−2 )pk−2 = qk qk−2 ak (pk−1 qk−2 − pk−2 qk−1 ) (−1)k−2 ak = = . qk qk−2 qk qk−2
Ck − Ck−2 =
∴ Ck − Ck−2
3.3
Infinite continued fractions
Proposition 3.2.1 shows that Ck < Ck−2 if k ≥ 3 is odd and Ck > Ck−2 if k ≥ 2 is even. Thus, we have, C1 > C3 > C5 > · · ·
and C2 < C4 < C6 < · · ·
Furthermore, C2m − C2m−1 =
(−1)2m−1 < 0, q2m q2m−1
therefore, C1 > C3 > C5 > · · · C6 > C4 > C2 . In particular, if {an }n≥0 is an infinite sequence of integers with an > 0 for all n ≥ 1, then setting Ck = [a0 , . . . , ak ], we infer that: (i) the sequence {C2n+1 }n≥0 is decreasing and bounded, in particular it is convergent. (ii) the sequence {C2n }n≥0 is increasing and bounded, in particular it is convergent. (iii) the sequence C2n − C2n+1 tends to zero. In particular the sequence {Cn }n≥0 is convergent. This fact will help us define correctly the infinite continued fractions.
Section 3.4. Convergents of continued fractions
Page 11
Definition 3.3.1. Let {an }n≥0 be an infinite sequence of integers with an > 0 for all n ≥ 1. We define the infinite continued fraction as the limit of the finite continued fraction [a0 , a1 , . . .] = lim Cn . n→∞
It is also easy to see that the infinite continued fractions represent irrational numbers. Conversely, every irrational number α can be represented as an infinite continued fraction [14]. Proposition 3.3.2. Given α = α0 ∈ R \ Q, let {an }n≥0 be defined as ak = bαk c,
αk+1 =
1 α k − ak
for all k ≥ 0.
Then α = [a0 , a1 , . . .]. Proof. Clearly, α = α0 = a0 +
1 = [a0 , α1 ] = a0 + α1
1 a1 +
1 α2
= [a0 , a1 , α2 ] = · · · = [a0 , a1 , . . . , ak , αk+1 ].
By (i) of Proposition 3.2.1, α=
αk+1 pk + pk−1 , αk+1 qk + qk−1
therefore, αk+1 pk + pk−1 pk −(pk qk−1 − qk pk−1 ) 1 1 = − = < 2. |α − Ck | = αk+1 qk + qk−1 qk (αk+1 qk + qk−1 )qk (αk+1 qk + qk−1 )qk qk Since qk ≥ k for k ≥ 1, we have that 1/qk2 → 0 as k → ∞, which completes the proof.
3.4
Convergents of continued fractions
In this section we show that the convergents pk /qk of an irrational number α give the best approximations of α by rationals. The following result is due to Legendre. Proposition 3.4.1. (i) Let α be an irrational number and let Ck = pk /qk for k ≥ 0 be the convergents of the continued fraction of α. If x, y ∈ Z with y > 0 and k is a positive integer such that |yα − x| < |qk α − pk |, then y ≥ qk+1 .
Section 3.4. Convergents of continued fractions
Page 12
(ii) If α is irrational and x/y is a rational number with y > 0 such that α − x < 1 , y 2y 2 then x/y is a convergent of α. Proof. See [14]. Remark 3.4.2. Part (ii) of Proposition 3.4.1 is known as Legendre’s criterion for a rational x/y to be a convergent of α. This result was generalised by several authors [14]. Here we give a couple of examples in what follows. Let Ck = pk /qk be a convergent of the continued fraction of α. For each n ≥ 0 we define pk,n = npk+1 + pk ,
qk,n = nqk+1 + qk .
If n = 0, then pk,n = pk and qk,n = qk . If n = ak+2 , then pk,n = pk+2 and qk,n = qk+2 . If 1 ≤ n ≤ ak+2 − 1, then the amount pk,n npk+1 + pk , = qk,n nqk+1 + qk is called an intermediary convergent of α = [a0 , a1 , . . . ]. From these, we have the following results. Proposition 3.4.3. If x, y > 0 are integers such that x α − < 1 , y y2 then x/y is either a convergent or an intermediary convergent of α. In fact, x/y = pk,n /qk,n with n = 0, 1 or ak+2 − 1. Proof. For the proof see [15]. Proposition 3.4.4. Let α be an irrational α −
number and let x, y ≥ 2 be integers such that x 2 < 2, y y
then x ∈ y
pk pk+1 ± pk 2pk+1 ± pk 3pk+1 ± pk pk+1 ± 2pk pk+1 − 3pk , , , , , qk qk+1 ± qk 2qk+1 ± qk 3qk+1 ± qk qk+1 ± 2qk qk+1 − 3qk
for some k ≥ 0. Proof. For proof see [16].
Section 3.4. Convergents of continued fractions
Page 13
The following result is a variation of a lemma of Baker and Davenport [9] and is due to Dujella and Peth˝o [3]. For a real number x, we put ||x|| = min{|x − n| : n ∈ Z} for the distance from x to the nearest integer. Lemma 3.4.5 (Dujella and Peth˝o, 1998). Let M be a positive integer, let p/q be a convergent of the continued fraction of the irrational τ such that q > 6M , and let A, B, µ be some real numbers with A > 0 and B > 1. Let := ||µq|| − M ||τ q||. If > 0, then there is no solution to the inequality 0 < mτ − n + µ < AB −k , in positive integers m, n and k with m≤M
and
k≥
log(Aq/) . log B
Proof. Suppose that 0 ≤ m ≤ M . Then we have m(τ q − p) + mp − nq + µq < qAB −k . Thus we get qAB −k > |µq − (nq − mp)| − mkτ qk ≥ kµqk − M kτ qk := ε, from which we get, after taking log on both sides of the above relation, k
1 be an integer which is not a perfect square and let (x1 , y1 ) be the minimal solution in positive integers of the Pell equation x2 − ny 2 = ±1.
(4.2.5)
We put ζ1 = x1 +
√
where ζ1 + ζ2 = 2x1 and ζ1 ζ2 = ±1.
ny1 and ζ2 = x1 −
√
ny1 ,
(4.2.6)
Section 4.2. Examples of binary recurrent sequences
Page 18
Then, there is a theorem from the theory of Pell equations which tells us that all positive integer solutions (x, y) of the equation (4.2.5) are of the form (x, y) = (xj , yj ) for some positive integer j, where √ √ (4.2.7) xj + nyj = (x1 + ny1 )j = ζ1j . On conjugating the above relation in (4.2.7), we get √ √ xj − nyj = (x1 − ny1 )j = ζ2j .
(4.2.8)
Thus, we can easily deduce from the relations (4.2.7) and (4.2.8) that xj =
ζ1j + ζ2j 2
and yj =
ζ1j − ζ2j √ 2 n
for all j ≥ 1.
(4.2.9)
It also turns out to be useful if we put (x0 , y0 ) = (1, 0) so that the relations in (4.2.9) hold with j = 0. It is now easy to see that the sequences {xj }j≥0 and {yj }j≥0 are binary recurrent sequences with the characteristic equation given by f (X) = X 2 − a1 X − a2 = (X − ζ1 )(X − ζ2 ) = X 2 − (ζ1 + ζ2 )X + ζ1 ζ2 ∴ f (X) = X 2 − 2x1 X ± 1.
(4.2.10)
4.2.7 Lucas sequences of the first kind. Definition 4.2.8. A Lucas sequence {un } is a linearly recurrent sequence of order 2 un+2 = run+1 + sun , for all n ≥ 0
(4.2.11)
such that gcd(r, s) = 1, u0 = 0, u1 = 1, (r, s) = 1, and the ratio α/β of the roots of the characteristic equation X 2 − rX − s = 0
(4.2.12)
is not a root of unity. From the characteristic equation in (4.2.12), we can easily notice that √ √ r + r2 + 4s r − r2 + 4s and β = . α= 2 2
(4.2.13)
We can also note that, s = α + β and s = −αβ. This initial conditions lead to the Binet formula un =
αn − β n , α−β
n = 0, 1, . . . .
(4.2.14)
Section 4.3. Some important inequalities involving Fibonacci and Pell sequences Example 4.2.9. The Fibonacci sequence {Fn }n≥0 is a Lucas sequence since √ √ ! αn − β n 1+ 5 1− 5 Fn = : (α, β) = , . α−β 2 2
Page 19
(4.2.15)
Its companion Lucas sequence {Ln }n≥0 has Ln = α n + β n =
F2n . Fn
(4.2.16)
Example 4.2.10. Let (xm , ym ) be all positive integer solutions to the Pell equation x2 −ny 2 = ±1 for some positive integer n > 1 which is not a perfect square. Let (x1 , y1 ) be the smallest of such solution. Then we have √ √ xm + nym = (x1 + ny1 )m . (4.2.17) √ √ We put α = x1 + ny1 and β = x1 − ny1 , from which we have α + β = 2x1 and αβ = ±1. Furthermore, we have xm = Thus, we have
n
ym y1
o
αm + β m 2
and ym =
αm − β m √ . 2 n
(4.2.18)
is the Lucas sequence with the characteristic polynomial X 2 − rX − s
m≥1
with roots α and β, where r = 2x and s = ±1.
4.3
Some important inequalities involving Fibonacci and Pell sequences
In this section, we state and prove some of the important inequalities associated with the Fibonacci and Pell sequences that will be used in solving the Diophantine equations (1.1.6) and (1.1.7). √ Proposition 4.3.1. Let α = (1 + 5)/2 and Fn the Fibonacci sequence, then we have αn−2 ≤ Fn ≤ αn−1
for all n ≥ 0.
(4.3.1)
Proof. We prove this inequality by induction on n. We split into two parts which we prove independently as follows: First we prove the the inequality: Fn ≤ αn−1 for all n ≥ 0. √ If n = 0, F0 = 0 and α−1 = 2/(1 + 5) > 0. Thus F0 ≤ α−1 . If n = 1, F1 = 1 and α0 = 1. Then, F1 ≤ α0 . Now by induction, we prove that the inequality still holds for all n ≥ 2. Assuming that it holds for n = k, then we prove that it holds for n = k + 1. Fk+1 = Fk + Fk−1 ≤ αk−1 + αk−2 = αk−2 (α + 1) = αk−2 · α2 = αk = α(k+1)−1 ,
Section 4.3. Some important inequalities involving Fibonacci and Pell sequences
Page 20
where we have used the fact that α2 = α + 1 since α is a root of the equation x2 − x − 1 = 0. Next we prove the the inequality: Fn ≥ αn−2 for all n ≥ 0. √ If n = 1, F1 = 1 and α−1 = 2/(1 + 5) < 1. Thus F1 ≥ α−1 . If n = 2, F2 = 2 and α0 = 1. Then, F2 ≥ α0 . Now by induction, we prove that the inequality still holds for all n ≥ 3. Assuming that it holds for n = k, then we prove that it holds for n = k + 1. Fk+1 = Fk + Fk−1 ≥ αk−2 + αk−3 = αk−3 (α + 1) = αk−3 · α2 = αk−1 = α(k+1)−2 . Combining these two inequalities, then this shows that we have proved the required result. This completes the proof. √ Proposition 4.3.2. Let γ = 1 + 2 and Pn the Pell sequence, then we have γ n−2 ≤ Pn ≤ γ n−1
for all n ≥ 0.
(4.3.2)
Proof. The proof is similar to that of Proposition 4.3.1 which is given as follows. First we prove the the inequality: Pn ≤ γ n−1 for all n ≥ 0. √ If n = 0, P0 = 0 and γ −1 = 1/(1 + 2) > 0. Thus P0 ≤ γ −1 . If n = 1, P1 = 1 and γ 0 = 1. Then, P1 ≤ γ 0 . Now by induction, we prove that the inequality still holds for all n ≥ 2. Assuming that it holds for n = k, then we prove that it holds for n = k + 1. Pk+1 = 2Pk + Pk−1 ≤ 2γ k−1 + γ k−2 = γ k−2 (2γ + 1) = γ k−2 · γ 2 = γ k = γ (k+1)−1 , where we have used the fact that γ 2 = 2γ + 1 since γ is a root of the equation x2 − 2x − 1 = 0. Next we prove the the inequality: Pn ≥ γ n−2 for all n ≥ 0. √ If n = 1, P1 = 1 and γ −1 = 1/(1 + 2) < 1. Thus P1 ≥ γ −1 . If n = 2, P2 = 2 and γ 0 = 1. Then, P2 ≥ γ 0 . Now by induction, we prove that the inequality still holds for all n ≥ 3. Assuming that it holds for n = k, then we prove that it holds for n = k + 1. Pk+1 = 2Pk + Pk−1 ≥ 2γ k−2 + γ k−3 = γ k−3 (2γ + 1) = γ k−3 · γ 2 = γ k−1 = γ (k+1)−2 . Combining these two inequalities, then this shows that we have proved the required result as before. This completes the proof.
5. Lower Bounds for Linear Forms in Logarithms 5.1
Basic definitions and examples
Definition 5.1.1. A linear form in logarithms of algebraic numbers is an expression of the form λ = β0 + β1 log α1 + β2 log α2 + · · · + βn log αn ,
(5.1.1)
where the α0 s and the β 0 s denote complex algebraic numbers and log denotes any determination of the logarithm. Definition 5.1.2. The height of a rational number a/b ∈ Q is defined as H(a/b) = max{|a|, |b|}.
(5.1.2)
Definition 5.1.3. Let L be a number field of degree D, let α ∈ L be an algebraic number of degree d and let f (X) =
d X
ai X d−1 ∈ Z[X]
(5.1.3)
i=0
be the minimum polynomial of α over L with a0 > 0 and gcd(a0 , . . . , ad ) = 1. Now we write f (X) = a0
d Y (X − α(i) ),
(5.1.4)
i=1
where α = α(1) . The numbers α(i) are called the conjugates of α. Then we define the logarithmic height of α by ! d X 1 (5.1.5) log(|a0 |) + log(max{|α(i) |, 1}) . h(α) = d i=1 Example 5.1.4. The logarithmic height of a rational number x = a/b ∈ Q where b 6= 0 is given by h(a/b) = log (max{|a|, |b|}) (5.1.6) √ Example 5.1.5. The logarithmic height of x = 2 with minimum polynomial x2 − 2 over Q, is given by √ √ √ 1 h( 2) = log(1) + log(max{| 2|, 1}) + log(max{| − 2|, 1}) , 2 √ √ 1 (5.1.7) = log(1) + log(| 2|) + log(| − 2|) , 2 √ 1 ∴ h( 2) = log 2. 2 21
Section 5.2. Effective approximation the lower bound
Page 22
√ √ Below we also find the logarithmic heights of the numbers α = (1 + 5)/2 and γ = 1 + 2 which will be frequently used in the proof of Theorem 1.1.1 in Chapter 6. √ Example 5.1.6. The logarithmic height of α = (1 + 5)/2 with minimum polynomial f (X) = X 2 − X − 1 over Q, is given by 1 log(1) + log max{|α(1) |, 1} + log max{|α(2) |, 1} , 2 1 − √5 1 = (log(1) + log |α| + log(1)) , since |α(1) | = |α| > 1, |α(2) | = < 1, 2 2
h(α) =
∴ h(α) =
1 log α. 2
(5.1.8)
Example 5.1.7. The logarithmic height of γ = 1 + X 2 − 2X − 1 over Q, is given by
√ 2 with minimum polynomial g(X) =
1 log(1) + log max{|γ (1) |, 1} + log max{|γ (2) |, 1} , 2 √ 1 = (log(1) + log |γ| + log(1)) , since |γ (1) | = |γ| > 1, |γ (2) | = |1 − 2| < 1, 2 1 ∴ h(γ) = log γ. (5.1.9) 2 h(γ) =
5.2
Effective approximation the lower bound
In 1966, A. Baker [17] gave an effective lower bound on the absolute value of a nonzero linear form in logarithms of algebraic numbers. His result marked the dawn of the era of effective resolution of the Diophantine equations of certain types, namely the ones that can be reduced to exponential ones, that is to say, where the unknown variables are in exponents. Many of the computer programs today which are used to solve Diophantine equations (PARI, MAGMA, KASH, ...) use some version of Baker’s inequality [14]. For our purpose, we shall give some of the Baker type inequalities available today which are easy to apply. Definition 5.2.1. Let L be an algebraic number field of degree dL ; that is, a finite extension of degree dL of Q. Let η1 , . . . , ηl be nonzero elements of L and let b1 , . . . , bl be integers. Put D = max{|b1 |, . . . , |bl |},
(5.2.1)
and Λ=
l Y
ηibi − 1.
(5.2.2)
i=1
Let A1 , A2 , . . . , Al be positive integers such that Aj ≥ h0 (ηj ) := max{dL h(ηj ), | log(ηj )|, 0.16} for all j = 1, . . . , l. We call h0 the modified height with respect to the field L.
(5.2.3)
Section 5.3. Reducing the bounds
Page 23
With the above notations, Matveev proved the following theorem which is a fundamental result we shall apply in the proof of Theorem 1.1.1. This result is discussed in details in his paper [10]. Theorem 5.2.2 (Matveev, 2000). Assume that Λ 6= 0, then log(|Λ|) > −3 · 30l+4 (l + 1)5.5 d2L (1 + log(dL ))(1 + log(nD))A1 A2 · · · Al .
(5.2.4)
If furthermore, L ⊂ R, then log(|Λ|) > −1.4 · 30l+3 l4.5 d2L (1 + log(dL ))(1 + log(D))A1 A2 . . . Al .
(5.2.5)
Proof. See [10].
5.3
Reducing the bounds
The upper bounds provided by the lower bounds for the linear forms in logarithms are in general too large to allow any meaningful computation, so they should be reduced. There is an entire theory concerning reducing such bounds, an algorithm called LLL, from the names of its inventors (Lenstra-Lenstra-Lovasz, etc) [14]. We will resume ourselves to the case of two logarithms, case in which the reduction lemma is based on continued fractions expansions of the involved numbers is surprisingly effective. This is done by using Lemma 3.4.5 that is frequently applied in the proof of Theorem 1.1.1 in the last section of Chapter 6.
6. Proof of Theorem 1.1.1 We ran a computation for k ≤ 400 in sage and got only the indicated solutions. We now assume that k > 400 and that n > m. We do not consider the case n = m since they lead to Fk = and Pk = whose largest solutions are k = 12 and k = 7, respectively, as we already pointed out in the introduction. In this context, stands for a square number.
6.1
Approximation of lower bounds on k, m and n
6.1.1 Case I. We deal with equation (1.1.6) first. We use the well known inequalities which were proved by induction in Chapter 4, that αn−2 ≤ Fn ≤ αn−1
and γ n−2 ≤ Pn ≤ γ n−1
for all n ≥ 0.
(6.1.1)
Thus we have, αk−2 ≤ Fk = Pm Pn ≤ γ m+n−2
and αk−1 ≥ Fk = Pm Pn ≥ γ m+n−4 .
(6.1.2)
and αk−1 ≥ γ m+n−4 ,
(6.1.3)
This implies that, αk−2 ≤ γ m+n−2 from which we get 1 + c1 (m + n − 4) ≤ k ≤ 2 + c1 (m + n − 2), where c1 =
(6.1.4)
log γ = 1.83157092 . . . . log α
In particular, we take k < 4n. Here we replace the Fibonacci and Pell sequences in equation (1.1.6) by their corresponding Binet formulas as follows. m n αk − β k γ − δm γ − δn (6.1.5) = , α−β γ−δ γ−δ which simplifies to 1 1 √ αk − β k = (γ m − δ m ) (γ n − δ n ) . 8 5
(6.1.6)
Now, we rewrite the above equation by separating on one side the large terms and on the other side the smaller terms. That is, equation (6.1.6) can be re-written in the form k k m+n n m m n m+n α β γ γ δ + γ δ − δ √ − = √ − . (6.1.7) 5 8 5 8 24
Section 6.1. Approximation of lower bounds on k, m and n By using the fact that β = −α−1 and δ = −γ −1 , and the fact that
Page 25 3 8
−1.4 × 306 × 34.5 × 42 (1 + log 4)(1 + log(4n))(4 log 8)(2 log α)(2 log γ), which simplifies to after computation, log |Λ1 | > −7.8 × 1013 (1 + log(4n)).
(6.1.10)
Then also from (6.1.9) we have log |Λ1 | < log
16 √ 5
− 2m log γ.
Thus, by comparing the inequalities in (6.1.10) and (6.1.11) we get 16 2m log γ − log √ < 7.8 × 1013 (1 + log(4n)). 5
(6.1.11)
(6.1.12)
Section 6.1. Approximation of lower bounds on k, m and n
Page 26
Hence, m log γ < 4 × 1013 (1 + log(4n)).
(6.1.13)
We now return back to equation (1.1.6) and write it as n αk − β k γ − δn = Pm , α−β γ−δ
(6.1.14)
which is equivalent to 1 P √ αk − β k = √m (γ n − δ n ) , 5 2 2
(6.1.15)
which can be rewritten as follows by separating bigger terms from the smaller terms as before: n αk βk γ δ √ (6.1.16) 5P − 2√2 = √5P − 2√2 . m m Similarly, by using the fact that β = −α−1 and δ = −γ −1 , and the fact that
1 √ 2 2
4h(η1 ). Finally, we can take D = 4n. We also note that √ 2 2 k −n Λ2 = √ α γ − 1. 5Pm
(6.1.20)
By a similar argument we used to prove that Λ1 6= 0, one can easily justify that Λ2 6= 0. Thus Theorem 5.2.2 gives that, log |Λ2 | > −1.4 × 306 × 34.5 × 42 (1 + log 4)(1 + log(4n))2 × 16 × 1013 × (2 log α)(2 log γ), which simplifies to after some computation, log |Λ2 | > −1.5 × 1027 (1 + log(4n))2 .
(6.1.21)
From (6.1.19), we also have log |Λ2 | < log
√ ! 36 2 √ − 2n log γ. 5
By comparing (6.1.21) and (6.1.22), we get √ ! 36 2 √ 2n log γ − log < 1.5 × 1027 (1 + log(4n))2 , 5
(6.1.22)
(6.1.23)
which gives, after some computation, n < 5 × 1030 .
(6.1.24)
Section 6.1. Approximation of lower bounds on k, m and n
Page 28
6.1.2 Case II. We now deal with equation (1.1.7). The same arguments applied to equation (1.1.6) are applied to equation (1.1.7). This is done by just swapping the roles of the pairs (α, β) and (γ, δ), and that of √15 and 2√1 2 . Below we give the details. We assume m ≥ 3, otherwise m ∈ {1, 2}, Fm = 1 and the solutions of equation (1.1.7) are among the solutions to equation (1.1.6) with m = 1. Then we have, γ k−2 ≤ Pk = Fm Fn ≤ αm+n−2
and γ k−1 ≥ Pk = Fm Fn ≥ αm+n−4 .
(6.1.25)
and γ k−1 ≥ αm+n−4 .
(6.1.26)
This implies that, γ k−2 ≤ αm+n−2
With these inequalities, inequality (6.1.4) becomes 1 + c2 (m + n − 4) ≤ k ≤ 2 + c2 (m + n − 2), where c2 =
(6.1.27)
log α = 0.545979 . . . . log γ
This implies in particular that k < 3n. The analogue of inequality (6.1.8) in this case is given by k k m+n n m m n m+n γ √ −α = δ√ − α β + α β − β 2 2 5 2 2 5 (6.1.28) 6 6αn−m k n−m ≤ max{|δ| , α }= . 5 5 By dividing through by
αm+n 5
leads to 5 k −n−m 6 √ γ α − 1 < 2m , 2 2 α
(6.1.29)
which is the analogue of inequality (6.1.9). We check that the amount Λ3 on the left hand side of the above equation is also non-zero through similar arguments as before. We then apply Theorem 5.2.2 with data 5 5 Λ3 = √ γ k α−n−m − 1, l = 3, η1 = √ , η2 = γ, 2 2 2 2 η3 = α, d1 = 1, d2 = k, d3 = −m − n. With this data, Theorem 5.2.2 gives log |Λ3 | > −1.4 × 306 × 34.5 × 42 (1 + log 4)(1 + log(3n))(4 log 5)(2 log α)(2 log γ), which simplifies to log |Λ3 | > −5.98 × 1013 (1 + log(3n)).
(6.1.30)
Also from inequality (6.1.29) we have log |Λ3 | < log 6 − 2m log α.
(6.1.31)
Section 6.1. Approximation of lower bounds on k, m and n
Page 29
By comparing these two inequalities (6.1.30) and (6.1.31), we get the analogue of (6.1.12) as 2m log α − log 6 < 5.98 × 1013 (1 + log(3n)),
(6.1.32)
m log α + 1 < 3 × 1013 (1 + log(3n)),
(6.1.33)
which gives
which is the analogue of inequality (6.1.13). Returning to equation (1.1.7), we get γk αn δ k β n 2 1 1 √ 2 2F − √5 = 2√2F − √5 ≤ √5 max γ k , αn . m
(6.1.34)
m
By (6.1.27), we get γ k ≥ γαm+n−4 ≥ γα−3 αn , so
1 α3 /γ 2 ≤ < n. k n γ α α Hence, by (6.1.34) and (6.1.35), we get √ 5 4 k −n √ γ α − 1 < 2n . 2 2Fm α
(6.1.35)
(6.1.36)
This is the analogue of (6.1.19). Writing Λ4 for the amount under the absolute value in the left– hand side above, we get that it is non-zero by arguments similar to the ones used to √ prove√that Λi 6= 0 for i = 1, 2, 3. We apply Matveev’s theorem as we did for Λ2 . Here, η1 = 2 2Fm / 5 is a root of 5X 2 − 8Fm2 . Its height therefore satisfies √ h(η1 ) ≤ log Fm + log 2 2, √ ≤ (m − 1) log α + log 2 2, < m log α + 1, h(η1 ) < 3 × 1013 (1 + log(3n)), by (6.1.33). We get that log |Λ4 | > −1.4 × 306 × 34.5 × 42 (1 + log 4)(1 + log(3n))2 12 × 1013 × (2 log α)(2 log γ), which simplifies to after computation, log |Λ4 | > −1.2 × 1027 (1 + log(3n))2 ,
(6.1.37)
which together with (6.1.36) leads to 2n log α − log 4 < 1.2 × 1027 (1 + log(3n))2 . This gives after some computation, n < 7 × 1030 .
(6.1.38)
So, comparing the above bound in (6.1.38) with that (6.1.24), we conclude that both in equation (1.1.6) and (1.1.7), we get the bound on n as n < 7 × 1030 . We record what we proved as a lemma. Lemma 6.1.3. If (k, m, n) are positive integers satisfying one of the equations (1.1.6) or (1.1.7) with m ≤ n, then k < 4n and n < 7 × 1030 .
Section 6.2. Reducing the bounds
6.2
Page 30
Reducing the bounds
Now we need to reduce the bounds. To do so, we make use several times of the result in Lemma 3.4.5, which is a slight variation of a result due to Dujella and Peth˝o [3] which itself is a generalization of a result of Baker and Davenport [9]. 6.2.1 Case 1. We look at (6.1.9). Assume that m ≥ 20. We put √ Γ1 := k log α − (n + m) log γ + log(8/ 5). Then we have from (6.1.9) that |eΓ1 − 1| = |Λ1 | < √
16 1 < , 4 5γ 2m
(6.2.1)
which implies that 1 |Γ1 | < . 2
(6.2.2)
Since |x| < 2|ex − 1| whenever x ∈ (−1/2, 1/2), we get from Λ1 = eΓ1 − 1 and (6.1.9) that |Γ1 | < √
32 . 5γ 2m
(6.2.3)
If Γ1 > 0 (we observe that Γ1 6= 0 since Λ1 6= 0), then √ log α 32 17 log(8/ 5) 0 4n > k), √ log α log(8/ 5) τ= , µ= , A = 17, B = γ 2 . log γ log γ Writing τ = [a0 , a1 , . . .] as a continued fraction, we get [a0 , . . . , a74 ] =
p74 2037068391552562960855777461929676271 = , q74 3731035235978315437343082205475618926
and we get q74 > 3 × 1036 > 6M . We compute ε = kµq74 k − M kτ q74 k > 0.4.
(6.2.4)
Thus, by Lemma 3.4.5, we get m ≤ 49. A similar conclusion is reached if we assume that Γ1 < 0. This is in the case of inequality (6.1.9).
Section 6.2. Reducing the bounds
Page 31
6.2.2 Case 2. In the case of inequality (6.1.29), assuming again that m ≥ 20, we get that √ 12 (n + m) log α − k log γ − log(5/2 2) < 2m . 5α Let Γ3 be the expression under the absolute value of the left–hand side above. If Γ3 > 0, we get √ log α log(2 2/5) 12 3 0 < (n + m) −k+ < < 2m . 2m log γ log γ (5 log γ)α α We keep the same values for M, τ, q and only change µ to √ log(2 2/5) 0 , A = 3, B = α2 . µ = log γ We get ε > 0.2, and by Lemma 3.4.5, m ≤ 90. A similar conclusion is reached if Γ3 < 0. Thus, m ≤ 90 in all cases. 6.2.3 Case 3. Now we move on to (6.1.19). Assume n > 100. We then get √ √ √ 72 2 k log α − n log γ + log(2 2/ 5Pm ) < √ 2n . 5δ Let Γ2 be the expression under the absolute value in the left–hand side above. If Γ2 > 0, we then get √ √ √ log α 52 72 2 log(2 2/( 5Pm )) < 2n . 0 0.019, so n ≤ 53. A similar conclusion is reached if Γ2 < 0. 6.2.4 Case 4. Finally, if instead of (6.1.19), we have (6.1.36), then a similar argument leads to √ √ 4 n log α − k log γ + log(2 2Fm / 5) < 2n . α Putting Γ4 for the amount under the absolute value in the left–hand side above, we get in case Γ4 > 0 that √ √ log(2 2Fm / 5) 4 5 log α −k+ 0 0.005, so n ≤ 94. So, in all cases n ≤ 94, so k < 400. We used sage to generate {Fk }1≤k≤400 and {Pm Pn }1≤m
