by mahadi ddamulira ([email protected])

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

[email protected]

Declaration

This work was carried out at AIMS-Ghana in partial fulfilment of the requirements for a Masterof Science Degree in Mathematical Sciences.

I hereby declare that except where due acknowledgement is made, this work has never beenpresented wholly or in part for the award of a degree at AIMS-Ghana or any other University.

Student: Mahadi Ddamulira

Supervisor: Prof. Florian Luca

i

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 immea-surable supports, inspirations, criticisms, comments and suggestions rendered through out theprocess 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. Tumwesigyein the department of Mathematics at Makerere University, for it is by their efforts that I cameto know about AIMS, in addition to a firm mathematical foundation they laid for me at theundergraduate 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, GraceOmollo Misereh, Frantisek Hajnovic, Amsalework Ejigu and all those who are not mentioned butwhose help are truly contributing in adding some significant information will not be forgotten, Ithank you all and may the Almighty reward you.

ii

Dedication

To my parents and relatives - the family of Sheikh Ali Muwoomya.

iii

Abstract

There are many Diophantine equations whose aim is to determine the intersection of two binaryrecurrent sequences (Fibonacci numbers, Pell numbers, etc.). In this research project, we findall Fibonacci numbers which are products of two Pell numbers and all Pell numbers which areproducts of two Fibonacci numbers.

Key words: Diophantine equations, Continuity fractions, Binary recurrent sequences, Fibonaccinumbers, Pell numbers, Lower bounds for linear forms in logarithms.

iv

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 21

5.1 Basic definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

v

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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

References 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 Lucassequences, respectively, given by{

Fn+2 = Fn+1 + Fn, F0 = 0, F1 = 1,

Ln+2 = Ln+1 + Ln, L0 = 2, L1 = 1.(1.1.2)

The Binet formulas for the general terms of the Fibonacci and Lucas sequences are obtainedusing standard techniques for solving linear recurrent sequences that we discuss in Chapter 4, aregiven by

Fn =αn − βn

α− βand Ln = αn + βn, (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-Lucassequences, respectively, given by{

Pn+2 = 2Pn+1 + Pn, P0 = 0, P1 = 1,

Qn+2 = 2Qn+1 +Qn, Q0 = 2, Q1 = 2.(1.1.4)

The Binet formulas for the general terms of these sequences are given by

Pn =γn − δn

γ − δand Qn = γn + δn, (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 Fibonacciand Pell sequences, given below:

Fk = PmPn (1.1.6)

and

Pk = FmFn. (1.1.7)

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 Pellsequences, respectively, and 12 and 13 are Pell and Fibonacci numbers respectively. The tworesults 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 oftwo numbers from the other sequence.

The detailed proof of Theorem 1.1.1 is given in Chapter 6. In the process of proving the theoremwe also use the properties of linear forms in logarithms and the Dujella-Petho [3] reductionprocedure which are discussed in Chapter 5 and Chapter 3, respectively, to solve the Diophantineequations (1.1.6) and (1.1.7).

1.2 Literature Review

Diophantine equations are polynomial equations which contain two or more unknowns, and onlythe integer values of these unknowns are to be found. Some forms of Diophantine equationsinclude a linear Diophantine equation, which is an equation between two sums of monomials ofdegree zero or one, and an exponential Diophantine equation, which is one in which exponents onterms can be unknowns. Diophantine equations were named after the Hellenistic mathematicianof the 3rd century, Diophantus of Alexandria who studied such equations and was also the firstto introduce the usage of symbols in algebra [4].

While Diophantine equations are mostly studied as a form of theoretical mathematics, they dohave a few applications which include cryptography and also in the field of chemistry, whereDiophantine equations are used to balance out chemical equations and determine the molecularformulas of compounds. Some famous examples of Diophantine equations include the Pell equa-tion which is in the form x2 + dy2 = ±1 as well as the Erdos-Straus conjecture which states thatfor 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 equationis 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 reasonableconditions (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}, whichthen are reduced in practice either by using the LLL algorithm or by using a procedure originallydiscovered by Baker and Davenport [9] and perfected by Dujella and Petho [3].

In this research project, after identifying this gap in literature, we concentrate on solving thediophantine equations of the forms

Uk = VmVn and Vk = UmUn 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. Theseforms represent the equations (1.1.6) and (1.1.7). We also use linear forms in logarithms and theDujella-Petho’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 thebasic properties of these, that will later on be used in the subsequent chapters of this researchproject.

In Chapter 3, we introduce all the aspects of the continued fractions that will be useful in thesubsequent chapters. We shall also insist on the property of convergents of an irrational numberα 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 alsodefine the Lucas sequences and observe that the sequence of Fibonacci numbers is an exampleof such a sequence. Various examples are given on expressing these sequences in terms of theirBinet forms.

In Chapter 5, we give various specific statements of which belong to what is known as the theoryof lower bounds for linear forms in logarithms of algebraic numbers. We give various exampleson 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 applyingthe theorem of Matveev [10] to determine the effective lower bounds on k,m, n and also applythe Dujella-Petho reduction procedure to reduce these bounds.

There are certain little computations along the way, in the subsequent sections of the researchproject, 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 thebasic properties of these, that will later on be used in this project. We assume familiarity withthe 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 multi-plication (×) 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 ofevery non-zero element to have an inverse) then the structure is called ring. If additionallythe multiplication is not commutative it is non-commutative ring for instance, matrices form anon-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 thesame addition (+) and multiplication (×). Thus, Q is a subfield of R which is in turn a subfieldof C an so on.

4

Section 2.2. Algebraic numbers and algebraic integers Page 5

2.2 Algebraic numbers and algebraic integers

Definition 2.2.1 (Algebraic number). An algebraic number is a root of a monic polynomial withrational coefficients. That is α is an algebraic number if and only if there exists n ≥ 1 anda1, a2, . . . , an ∈ Q such that

αn + a1αn−1 + a2α

n−2 + · · ·+ an−1α + an = 0. (2.2.1)

Definition 2.2.2 (Algebraic integer). An algebraic integer is a root of a monic polynomial withinteger coefficients. That is β is an algebraic integer if and only if there exists n ≥ 1 andb1, b2, . . . , bn ∈ Z such that

βn + b1βn−1 + b2β

n−2 + · · ·+ bn−1β + 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 andrational numbers, respectively. Thus, all integers are algebraic integers and all rational numbersare algebraic numbers [11].

Some examples of algebraic integers include:

√2, i,

3√

5,1

2(1 +

√5), 2 cos

(2π

9

), . . . .

The examples of algebraic numbers include:√1

5,

1

2(1 + i),

1

2(1 +

√3i), 2 sin

(2π

7

), . . . .

Remark 2.2.3. The numbers e and π are neither algebraic numbers nor algebraic integers, theyare called transcendental numbers.

Definition 2.2.4 (Minimum polynomial). Let α be an algebraic number. Then among all poly-nomials with rational coefficients with α as a root, there exists a unique polynomial f(X) of thelowest degree, which is monic (with leading coefficient equal to one) and also irreducible (cannot be factorised further non-trivially). This polynomial is called the minimum polynomial of thealgebraic number α.

Definition 2.2.5. The degree of the algebraic number α is the degree of its minimum polynomialf(X).

Remark 2.2.6. The existence of minimum polynomials of any degree n implies the existence ofalgebraic numbers of degree n.

For example, all rational numbers are algebraic numbers of degree one. The complex number iis an algebraic number of degree two since it is a root of the irreducible polynomial x2 + 1 overthe 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 Page 6

2.3 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 callK 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 afinite dimensional vector space over F . The degree [K : F ] of a finite extension K/F is thedimension 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 nonzeropolynomial f ∈ F [x] (This is the minimum polynomial of α over F ). A field extension K/F isalgebraic 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 theextension 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 afinite 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 forK over E and B2 = {β1, β2, . . . , βm} is a basis for E over F . Then every element α ∈ K canbe written as a linear combination

α = a1α1 + a2α2 + · · ·+ anαn =n∑i=1

aiαi, ai ∈ E.

By writing each element ai ∈ E as a linear combination of the elements of the basis B2, that isto say,

ai = bi1β1 + bi2β2 + · · ·+ bimβm =m∑j=1

bijβj,

we get

α =n∑i=1

aiαi =n∑i=1

(m∑j=1

bijβj

)αi =

n∑i=1

m∑j=1

bijαiβj, bij ∈ F.

Then to prove that αiβj ∈ K,where 1 ≤ i ≤ n and 1 ≤ j ≤ m are linearly independent, we setα = 0. Then we have

n∑i=1

(m∑j=1

bijβj

)αi = 0 which implies

m∑j=1

bijβj = 0,

because B1 is a linearly independent set of K over E. We also have that, B2 is a linearlyindependent set of E over F , this also implies that bij = 0 for all i and j. Thus it immediatelyfollows 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 thesequence 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 theTower 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 isx2 − 2 because x2 − 2 is a monic polynomial in Q[x] for which

√2 is a root. This polynomial is

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[

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 theminimum 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[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[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 vectorspace over Q is simply called its degree.

Example 2.3.8. We consider the following examples of number fields which are considered inthe 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 arequadratic 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 5to the field of rational numbers. The degree of this number field is the degree of the fieldextension 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 inthe course of the project.

3.1 Finite continued fractions

A continued fraction is an expression obtained from an iterative process of representing a numberas a sum of its integer part and the reciprocal of another number, then writing this other numberas a sum of its integer part and another reciprocal, and so on. In a finite continued fraction, theiteration is terminated after finitely many steps by using an integer in lieu of another continuedfraction. 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

a0 +1

a1 +1

a2 + · · ·+ 1

an−1 +1

an

,

(3.1.1)

where a0 ∈ R and ai ∈ R+ for all 1 ≤ i ≤ n. For the purpose of simplicity, we shall usethe 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

47

17= 2 +

13

17= 2 +

117

13

= 2 +1

1 +4

13

= 2 +1

1 +113

4

= 2 +1

1 +1

3 +1

4

= [2, 1, 3, 4].

It is clear that every simple continued fraction is a rational number. Conversely, by the Euclideanalgorithm, every rational number can be represented as a simple continued fraction. In fact, if pand q > 0 are relatively prime and we write

p = r0a0 + r1, 1 ≤ r1 < r0 := q;

r0 = r1a1 + r2, 1 ≤ r2 < r1;

. . . = . . . ,

rn−1 = rnan,

(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 Page 9

3.2 Some properties of continued fractions

For each continued fraction [a0, . . . , an] we define p0, . . . , pn and q0, . . . , qn via

p0 = a0, q0 = 1;p1 = a0a1 + 1, q1 = a1;pk = akpk−1 + pk−2, qk = akqk−1 + qk−2

for all k = 2, . . . , n.

Proposition 3.2.1. With the previous notation, we have:

(i) Ck = pk/qk;

(ii) pkqk−1 − pk−1qk = (−1)k−1 for all k ≥ 1;

(iii) Ck − Ck−1 =(−1)k−1

qkqk−1

for 1 ≤ k ≤ n;

(iv) Ck − Ck−2 =(−1)kakqkqk−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 +1

a1

=a0a1 + 1

a1

=p1

q1

.

If k = 2, then

C2 = [a0, a1, a2] = a0 +1

a1 +1

a2

= a0 +a2

a1a2 + 1

=a2(a0a1 + 1) + a0

a2a1 + 1=a2p1 + p0

a2q1 + q0

=p2

q2

.

Now, we let k ≥ 2 and assume that Ci = pi/qi for all i = 0, . . . , k. We also observe thatpk−2, qk−2, pk−1 and qk−1 depend only on a0, . . . , ak−1. Thus we have,

Ck+1 =

[a0, a1, . . . , ak−1, ak +

1

ak+1

]=

(ak + 1/ak+1)pk−1 + pk−2

(ak + 1/ak+1)qk + qk−1

=ak+1(akpk−1 + pk−2) + pk−1

ak+1(akqk−1 + qk−2) + pk−1

=ak+1pk + pk−1

ak+1qk + qk−1

=pk+1

qk+1

.

(ii) If k = 1, then we have

p1q0 − p0q1 = (a0a1 + 1)1− a0a1 = 1.

By induction, if k ≥ 2, then

pkqk−1 − pk−1qk = (akpk−1 + pk−2)qk−1 − pk−1(akqk−1 + qk−2)

= −(pk−1qk−2 − pk−2qk−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 =pkqk− pk−1

qk−1

=pkqk−1 − pk−1qk

qkqk−1

=(−1)k−1

qkqk−1

.

(iv) Finally, we have

Ck − Ck−2 =pkqk− pk−2

qk−2

=pkqk−2 − pk−2qk

qkqk−2

=(akpk−1 + pk−2)qk−2 − (akqk−1 + qk−2)pk−2

qkqk−2

∴ Ck − Ck−2 =ak(pk−1qk−2 − pk−2qk−1)

qkqk−2

=(−1)k−2akqkqk−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

q2mq2m−1

< 0,

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 settingCk = [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 theinfinite 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. Wedefine the infinite continued fraction as the limit of the finite continued fraction

[a0, a1, . . .] = limn→∞

Cn.

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 = bαkc, αk+1 =1

αk − akfor all k ≥ 0.

Then α = [a0, a1, . . .].

Proof. Clearly,

α = α0 = a0 +1

α1

= [a0, α1] = a0 +1

a1 +1

α2

= [a0, a1, α2] = · · · = [a0, a1, . . . , ak, αk+1].

By (i) of Proposition 3.2.1,

α =αk+1pk + pk−1

αk+1qk + qk−1

,

therefore,

|α− Ck| =∣∣∣∣αk+1pk + pk−1

αk+1qk + qk−1

− pkqk

∣∣∣∣ =

∣∣∣∣−(pkqk−1 − qkpk−1)

(αk+1qk + qk−1)qk

∣∣∣∣ =1

(αk+1qk + qk−1)qk<

1

q2k

.

Since qk ≥ k for k ≥ 1, we have that 1/q2k → 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 bestapproximations 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 theconvergents of the continued fraction of α. If x, y ∈ Z with y > 0 and k is a positiveinteger such that

|yα− x| < |qkα− pk|,

then y ≥ qk+1.


(ii) If α is irrational and x/y is a rational number with y > 0 such that∣∣∣∣α− x

y

∣∣∣∣ < 1

2y2,

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/yto be a convergent of α. This result was generalised by several authors [14]. Here we give acouple of examples in what follows. Let Ck = pk/qk be a convergent of the continued fractionof 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. If1 ≤ n ≤ ak+2 − 1, then the amount

pk,nqk,n

=npk+1 + pknqk+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∣∣∣∣α− x

y

∣∣∣∣ < 1

y2,

then x/y is either a convergent or an intermediary convergent of α. In fact, x/y = pk,n/qk,nwith n = 0, 1 or ak+2 − 1.

Proof. For the proof see [15].

Proposition 3.4.4. Let α be an irrational number and let x, y ≥ 2 be integers such that∣∣∣∣α− x

y

∣∣∣∣ < 2

y2,

then

x

y∈{pkqk,pk+1 ± pkqk+1 ± qk

,2pk+1 ± pk2qk+1 ± qk

,3pk+1 ± pk3qk+1 ± qk

,pk+1 ± 2pkqk+1 ± 2qk

,pk+1 − 3pkqk+1 − 3qk

}for some k ≥ 0.

Proof. For proof see [16].


The following result is a variation of a lemma of Baker and Davenport [9] and is due to Dujellaand Petho [3]. For a real number x, we put ||x|| = min{|x− n| : n ∈ Z} for the distance fromx to the nearest integer.

Lemma 3.4.5 (Dujella and Petho, 1998). Let M be a positive integer, let p/q be a convergentof the continued fraction of the irrational τ such that q > 6M , and let A,B, µ be some realnumbers with A > 0 and B > 1. Let ε := ||µq|| −M ||τq||. If ε > 0, then there is no solutionto the inequality

0 < mτ − n+ µ < AB−k,

in positive integers m,n and k with

m ≤M and k ≥ log(Aq/ε)

logB.

Proof. Suppose that 0 ≤ m ≤M . Then we have

m(τq − p) +mp− nq + µq < qAB−k.

Thus we get

qAB−k > |µq − (nq −mp)| −m‖τq‖ ≥ ‖µq‖ −M‖τq‖ := ε,

from which we get, after taking log on both sides of the above relation,

k <log(Aq/ε)

logB.

A related version of the proof for this lemma is given in [3] and [14].

4. Binary Recurrent Sequences

4.1 Basic definitions and properties

Definition 4.1.1. Let k ≥ 1 be an integer. A sequence {Hn}n≥0 ⊆ C is called a linearlyrecurrent sequence of order k if the sequence

Hn+k = a1Hn+k−1 + a2Hn+k−2 + · · ·+ akHn (4.1.1)

holds for all n ≥ 0 with some fixed coefficients a1, . . . , ak ∈ C.

We suppose that ak 6= 0 (for if not, the sequence {Hn}n≥0 satisfies a linear recurrence of ordersmaller than k). If a1, . . . , ak ∈ Z and H0, . . . , Hk−1 ∈ Z, then it can easily be proved byinduction on n that Hn is an integer for all n ≥ 0. The polynomial

f(X) = Xk − a1Xk−1 − a2X

k−2 − · · · − ak ∈ C (4.1.2)

is called the characteristic polynomial of {Hn}n≥0. Suppose that

f(X) =m∏i=1

(X − αi)σi , (4.1.3)

where α1, . . . , αm are distinct roots of f(X) with multiplicities σ1, . . . , σm, respectively.

Theorem 4.1.2. Suppose that f(X) ∈ Z[X] has distinct roots. Then there exists constantsc1, . . . , ck ∈ K = Q(α1, . . . , αk) such that the formula

Hn =k∑i=1

ciαni holds for all n ≥ 0. (4.1.4)

This formula is called the Binet formula.

Proof. Let

H(z) =∑n≥0

Hnzn.

We observe that

H(z)(1− a1z − a2z2 − · · · − akzk) = H0 + (H1 − a1H0)z + (H2 − a1H1 − a2H0)z2

+ · · ·+∑m≥k

(Hm − a1Hm−1 − · · · − akHm−k)zm

:= Q(z),

14

Section 4.2. Examples of binary recurrent sequences Page 15

where Q(z) =k−1∑m=0

(Hm − a1Hm−1 − · · · − amH0)zm ∈ C[z]. Thus we have,

H(z) =Q(z)

1− a1z − · · · − akzk=

Q(z)

zkf(1/z)=

Q(z)

zk∏k

i=1(1/z − αi)

=Q(z)∏k

i=1(1− zαi)=

k∑i=1

ci1− zαi

for some coefficients ci ∈ K. For the last step we have used the theory of partial fractionstogether with the fact that the roots αi, . . . , αk are distinct and the degree of Q(z) is smallerthan k. So if

|z| < ρ := min{|αi|−1 : i = 1, . . . , k},

then we can write

1

1− zαi=∑n≥0

(zαi)n =

∑n≥0

αni zn for all n ≥ 0.

Thus, for |z| < ρ we get that

∑n≥0

Hnzn = H(z) =

k∑i=1

ci∑n≥0

αni zn =

∑n≥0

(k∑i=1

ciαni

)zn.

Thus, by identifying coefficients in the equation above, we get the relation (4.1.4).

Remark 4.1.3. If k = 2, the sequence {Hn}n≥0 is called binary recurrent. In this case, thecharacteristic polynomial is of the form

f(X) = X2 − a1X − a2 = (X − α1)(X − α2).

Suppose that α1 6= α2, then Theorem 4.1.2 tells us that

Hn = c1αn1 + c2α

n2 for all n ≥ 0. (4.1.5)

Definition 4.1.4. A binary recurrent sequence {Hn}n≥0 whose general term is given by formula(4.1.5) is called nondegenerate if c1c2α1α2 6= 0 and α1/α2 is not a root of unity.

4.2 Examples of binary recurrent sequences

4.2.1 Fibonacci sequences.The Fibonacci sequence {Fn}n≥0 is the sequence given by F0 = 0, F1 = 1 and

Fn+2 = Fn+1 + Fn for all n ≥ 2.


The first few terms of the Fibonacci sequence are

{Fn}n≥0 = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . . .

The characteristic polynomial of the Fibonacci sequence is given by

f(X) = X2 −X − 1 = (X − α)(X − β),

where α = (1 +√

5)/2 and β = (1 −√

5)/2. Clearly, α + β = 1, α − β =√

5 and αβ = −1.In order to find the constants c1 and c2 starting with formula (4.1.5), we give to n the values 0and 1 and obtain the following system of linear equations

c1 + c2 = F0 = 0, c1α + c2β = F1 = 1.

Solving this system, we get c1 = 1/√

5 and c2 = −1/√

5. Since√

5 = α− β, then we can writethe Binet formula for the general terms of the Fibonacci sequence as follows:

Fn =αn − βn

α− βfor all n ≥ 0. (4.2.1)

4.2.2 Pell sequences.The Pell sequence {Pm}m≥0 is the sequence given by P0 = 0, P1 = 1 and

Pm+2 = 2Pm+1 + Pm for all m ≥ 2.

The first few terms of the Pell sequence are

{Pm}m≥0 = 0, 1, 2, 5, 12, 29, 70, 169, 408, 985, . . . .

The characteristic polynomial of the Pell sequence is given by

g(X) = X2 − 2X − 1 = (X − γ)(X − δ),

where γ = 1 +√

2 and δ = 1−√

2. Clearly, γ + δ = 2, γ − δ = 2√

2 and γδ = −1. In order tofind the constants c1 and c2 starting with formula (4.1.5), we give to m the values 0 and 1 andobtain the following system of linear equations

c1 + c2 = P0 = 0, c1γ + c2δ = P1 = 1.

Solving this system, we get c1 = 1/2√

2 and c2 = −1/2√

2. Since 2√

2 = γ − δ, then we canwrite the Binet formula for the general terms of the Pell sequence as follows:

Pm =γm − δm

γ − δfor all m ≥ 0. (4.2.2)


4.2.3 Lucas sequences of the second kind.A sequence related to the Fibonacci sequence is the Lucas sequence {Ln}n≥0 given by L0 = 2,L1 = 1 and

Ln+2 = Ln+1 + Ln for all n ≥ 2.

The Lucas sequence has the same characteristic equation as the Fibonacci sequence, thereforethere exists two constants d1 and d2 such that

Ln = d1αn + d2β

n for all n ≥ 0.

Giving to n values 0 and 1, we obtain the system of linear equations

d1 + d2 = L0 = 2, d1α1 + d2α2 = L1 = 1.

Solving this system of linear equations, we get d1 = d2 = 1, and thus the Binet formula for thegeneral terms of the Lucas sequence is given by

Ln = αn + βn for all n ≥ 0. (4.2.3)

Remark 4.2.4. Using the formulas (4.2.1), (4.2.2) and (4.2.3) it is easy to prove various formulaswhich involve Fn, Ln or Pm.

Proposition 4.2.5. This example appears in [6]. The formula

L2n − 5F 2

n = 4(−1)n (4.2.4)

holds for all n ≥ 0.

Proof. We begin the proof by replacing the Ln and Fn by their corresponding Binet formulas asfollows:

L2n − 5F 2

n = (αn + βn)2 − 5

(αn − βn

α− β

)2

= (αn + βn)2 − (αn − βn)2

= 4(αβ)n

∴ L2n − 5F 2

n = 4(−1)n,

where we used the fact that (α− β) =√

5 and that αβ = −1.

4.2.6 Binary recurrences associated to Pell equations with N = ±1.Let n > 1 be an integer which is not a perfect square and let (x1, y1) be the minimal solution inpositive integers of the Pell equation

x2 − ny2 = ±1. (4.2.5)

We put

ζ1 = x1 +√ny1 and ζ2 = x1 −

√ny1, (4.2.6)

where ζ1 + ζ2 = 2x1 and ζ1ζ2 = ±1.


Then, there is a theorem from the theory of Pell equations which tells us that all positive integersolutions (x, y) of the equation (4.2.5) are of the form (x, y) = (xj, yj) for some positive integerj, where

xj +√nyj = (x1 +

√ny1)j = ζj1 . (4.2.7)

On conjugating the above relation in (4.2.7), we get

xj −√nyj = (x1 −

√ny1)j = ζj2 . (4.2.8)

Thus, we can easily deduce from the relations (4.2.7) and (4.2.8) that

xj =ζj1 + ζj2

2and yj =

ζj1 − ζj2

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 withj = 0.

It is now easy to see that the sequences {xj}j≥0 and {yj}j≥0 are binary recurrent sequences withthe characteristic equation given by

f(X) = X2 − a1X − a2

= (X − ζ1)(X − ζ2)

= X2 − (ζ1 + ζ2)X + ζ1ζ2

∴ f(X) = X2 − 2x1X ± 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 thecharacteristic equation

X2 − 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

α =r +√r2 + 4s

2and β =

r −√r2 + 4s

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 − βn

α− β, n = 0, 1, . . . . (4.2.14)

Section 4.3. Some important inequalities involving Fibonacci and Pell sequences Page 19

Example 4.2.9. The Fibonacci sequence {Fn}n≥0 is a Lucas sequence since

Fn =αn − βn

α− β: (α, β) =

(1 +√

5

2,1−√

5

2

). (4.2.15)

Its companion Lucas sequence {Ln}n≥0 has

Ln = αn + βn =F2n

Fn. (4.2.16)

Example 4.2.10. Let (xm, ym) be all positive integer solutions to the Pell equation x2−ny2 = ±1for some positive integer n > 1 which is not a perfect square. Let (x1, y1) be the smallest ofsuch 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 =αm + βm

2and ym =

αm − βm

2√n

. (4.2.18)

Thus, we have{ymy1

}m≥1

is the Lucas sequence with the characteristic polynomial X2 − rX − swith roots α and β, where r = 2x and s = ±1.

4.3 Some important inequalities involving Fibonacci andPell sequences

In this section, we state and prove some of the important inequalities associated with the Fibonacciand 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 proveindependently 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 holdsfor 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.


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. Thiscompletes the proof.

Proposition 4.3.2. Let γ = 1 +√

2 and Pn the Pell sequence, then we have

γn−2 ≤ Pn ≤ γn−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 ≤ γn−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 γ0 = 1. Then, P1 ≤ γ0.


Pk+1 = 2Pk + Pk−1 ≤ 2γk−1 + γk−2 = γk−2(2γ + 1) = γk−2 · γ2 = γk = γ(k+1)−1,

where we have used the fact that γ2 = 2γ + 1 since γ is a root of the equation x2− 2x− 1 = 0.

Next we prove the the inequality: Pn ≥ γn−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 γ0 = 1. Then, P2 ≥ γ0.


Pk+1 = 2Pk + Pk−1 ≥ 2γk−2 + γk−3 = γk−3(2γ + 1) = γk−3 · γ2 = γk−1 = γ(k+1)−2.

Combining these two inequalities, then this shows that we have proved the required result asbefore. This completes the proof.

5. Lower Bounds for Linear Forms inLogarithms

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 logα1 + β2 logα2 + · · ·+ βn logαn, (5.1.1)

where the α′s and the β′s denote complex algebraic numbers and log denotes any determinationof 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 ofdegree d and let

f(X) =d∑i=0

aiXd−1 ∈ Z[X] (5.1.3)

be the minimum polynomial of α over L with a0 > 0 and gcd(a0, . . . , ad) = 1. Now we write

f(X) = a0

d∏i=1

(X − α(i)), (5.1.4)

where α = α(1). The numbers α(i) are called the conjugates of α. Then we define the logarithmicheight of α by

h(α) =1

d

(log(|a0|) +

d∑i=1

log(max{|α(i)|, 1})

). (5.1.5)

Example 5.1.4. The logarithmic height of a rational number x = a/b ∈ Q where b 6= 0 is givenby

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, isgiven by

h(√

2) =1

2

(log(1) + log(max{|

√2|, 1}) + log(max{| −

√2|, 1})

),

=1

2

(log(1) + log(|

√2|) + log(| −

√2|)),

∴ h(√

2) =1

2log 2.

(5.1.7)

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 +√

2which 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) =X2 −X − 1 over Q, is given by

h(α) =1

2

(log(1) + log

(max{|α(1)|, 1}

)+ log

(max{|α(2)|, 1}

)),

=1

2(log(1) + log |α|+ log(1)) , since |α(1)| = |α| > 1, |α(2)| =

∣∣∣∣∣1−√

5

2

∣∣∣∣∣ < 1,

∴ h(α) =1

2logα. (5.1.8)

Example 5.1.7. The logarithmic height of γ = 1 +√

2 with minimum polynomial g(X) =X2 − 2X − 1 over Q, is given by

h(γ) =1

2

(log(1) + log

(max{|γ(1)|, 1}

)+ log

(max{|γ(2)|, 1}

)),

=1

2(log(1) + log |γ|+ log(1)) , since |γ(1)| = |γ| > 1, |γ(2)| = |1−

√2| < 1,

∴ h(γ) =1

2log γ. (5.1.9)

5.2 Effective approximation the lower bound

In 1966, A. Baker [17] gave an effective lower bound on the absolute value of a nonzero linearform in logarithms of algebraic numbers. His result marked the dawn of the era of effectiveresolution of the Diophantine equations of certain types, namely the ones that can be reducedto exponential ones, that is to say, where the unknown variables are in exponents. Many ofthe 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 ofthe 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 ofdegree 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∏

i=1

ηbii − 1. (5.2.2)

Let A1, A2, . . . , Al be positive integers such that

Aj ≥ h′(ηj) := max{dLh(ηj), | log(ηj)|, 0.16} for all j = 1, . . . , l. (5.2.3)

We call h′ the modified height with respect to the field L.

Section 5.3. Reducing the bounds Page 23

With the above notations, Matveev proved the following theorem which is a fundamental resultwe 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)5.5d2L(1 + log(dL))(1 + log(nD))A1A2 · · ·Al. (5.2.4)

If furthermore, L ⊂ R, then

log(|Λ|) > −1.4 · 30l+3l4.5d2L(1 + log(dL))(1 + log(D))A1A2 . . . 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 generaltoo large to allow any meaningful computation, so they should be reduced. There is an entiretheory 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, casein which the reduction lemma is based on continued fractions expansions of the involved numbersis surprisingly effective. This is done by using Lemma 3.4.5 that is frequently applied in the proofof 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 assumethat 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 pointedout 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 whichwere proved by induction in Chapter 4, that

αn−2 ≤ Fn ≤ αn−1 and γn−2 ≤ Pn ≤ γn−1 for all n ≥ 0. (6.1.1)

Thus we have,

αk−2 ≤ Fk = PmPn ≤ γm+n−2 and αk−1 ≥ Fk = PmPn ≥ γm+n−4. (6.1.2)

This implies that,

αk−2 ≤ γm+n−2 and αk−1 ≥ γm+n−4, (6.1.3)

from which we get

1 + c1(m+ n− 4) ≤ k ≤ 2 + c1(m+ n− 2), (6.1.4)

where c1 =log γ

logα= 1.83157092 . . . .

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.

αk − βk

α− β=

(γm − δm

γ − δ

)(γn − δn

γ − δ

), (6.1.5)

which simplifies to

1√5

(αk − βk

)=

1

8(γm − δm) (γn − δn) . (6.1.6)

Now, we rewrite the above equation by separating on one side the large terms and on the otherside the smaller terms. That is, equation (6.1.6) can be re-written in the form∣∣∣∣ αk√5

− γm+n

8

∣∣∣∣ =

∣∣∣∣ βk√5− γnδm + γmδn − δm+n

8

∣∣∣∣ . (6.1.7)

24

Section 6.1. Approximation of lower bounds on k, m and n Page 25

By using the fact that β = −α−1 and δ = −γ−1, and the fact that 38< 1√

5, we get that∣∣∣∣ αk√5

− γm+n

8

∣∣∣∣ < 2√5

max{|β|k, γn−m} =2γn−m√

5. (6.1.8)

By dividing across by γm+n

8, we get∣∣∣∣ 8√

5αkγ−n−m − 1

∣∣∣∣ < 16√5γ2m

. (6.1.9)

Then we apply Theorem 5.2.2 due to Matveev [10], on the left hand side of (6.1.9) with the data:

Λ1 =8√5αkγ−n−m − 1, l = 3, η1 =

8√5,

η2 = α, η3 = γ, d1 = 1, d2 = k, d3 = −m− n.

First, we show that Λ1 6= 0. The fact that Λ1 6= 0 follows from the fact that if it were zero, thenwe would get that α−kγm+n = 8√

5. However, Λ1 is a unit in L, whereas 8√

5is not as its norm

over L is 212/52. Thus, Λ1 6= 0.

Then, we take L = Q(√

2,√

5), for which dL = 4. Since

h(η1) =1

2

(log 5 + log

∣∣∣∣ 8√5

∣∣∣∣+ log

∣∣∣∣ 8

−√

5

∣∣∣∣) = log 8,

h(η2) =1

2logα,

h(η3) =1

2log γ,

we take

A1 = 4 log 8, A2 = 2 logα, and A3 = 2 log γ.

Finally, we take D = 4n. Thus Theorem 5.2.2 gives

log |Λ1| > −1.4× 306 × 34.5 × 42(1 + log 4)(1 + log(4n))(4 log 8)(2 logα)(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. (6.1.11)

Thus, by comparing the inequalities in (6.1.10) and (6.1.11) we get

2m log γ − log

(16√

5

)< 7.8× 1013(1 + log(4n)). (6.1.12)


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

αk − βk

α− β= Pm

(γn − δn

γ − δ

), (6.1.14)

which is equivalent to

1√5

(αk − βk

)=

Pm

2√

2(γn − δn) , (6.1.15)

which can be rewritten as follows by separating bigger terms from the smaller terms as before:∣∣∣∣ αk√5Pm

− γ

2√

2

∣∣∣∣ =

∣∣∣∣ βk√5Pm

− δn

2√

2

∣∣∣∣ . (6.1.16)

Similarly, by using the fact that β = −α−1 and δ = −γ−1, and the fact that 12√

2< 1√

5, we get∣∣∣∣ αk√

5Pm− γ

2√

2

∣∣∣∣ < 2√5

max

{1

αk,

1

γn

}. (6.1.17)

Then, we divide both sides above by γn

2√

2to get∣∣∣∣∣ 2

√2√

5Pmαkγ−n − 1

∣∣∣∣∣ < 4√

2√5

max

{1

αkγn,

1

γ2n

}. (6.1.18)

From (6.1.2), we have that

1

αkγn=

1/α

αk−1γn≤ 1/α

γ2n+m−4=

γ3/α

γ2n+m−1<

9

γ2n,

because γ3/α < 9 and that m ≥ 1. Therefore, we have∣∣∣∣∣ 2√

2√5Pm

αkγ−n − 1

∣∣∣∣∣ < 2√

2× 9√5γ2n

=36√

2√5γ2n

. (6.1.19)

Then we apply Theorem 5.2.2 on the left hand side of (6.1.19) with the data below

Λ2 =2√

2√5Pm

αkγ−n − 1, l = 3, η1 =

√5Pm

2√

2,

η2 = α, η3 = γ, d1 = −1, d2 = k, d3 = −n.

So we take L = Q(√

2,√

5), for which dL = 4. As before, we have

h(η2) =1

2logα, h(η3) =

1

2log γ,


so that we can take A2 = 2 logα and A3 = 2 log γ. As for h(η1), the polynomial

8X2 − 5P 2m

has η1 as a root. Thus we get

h(η1) =1

2

(log 8 + log

∣∣∣∣∣√

5Pm

2√

2

∣∣∣∣∣+ log

∣∣∣∣∣−√

5Pm

2√

2

∣∣∣∣∣),

= logPm +1

2log 5,

≤ (m− 1) log γ +1

2log 5,

∴ h(η1) < m log γ.

Then using (6.1.13), we can take

A1 = 16× 1013(1 + log(4n)) > 4h(η1).

Finally, we can take D = 4n. We also note that

Λ2 =2√

2√5Pm

αkγ−n − 1. (6.1.20)

By a similar argument we used to prove that Λ1 6= 0, one can easily justify that Λ2 6= 0. ThusTheorem 5.2.2 gives that,

log |Λ2| > −1.4× 306 × 34.5 × 42(1 + log 4)(1 + log(4n))2 × 16× 1013 × (2 logα)(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√5

)− 2n log γ. (6.1.22)

By comparing (6.1.21) and (6.1.22), we get

2n log γ − log

(36√

2√5

)< 1.5× 1027(1 + log(4n))2, (6.1.23)

which gives, after some computation,

n < 5× 1030. (6.1.24)


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 1√

5and 1

2√

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,

γk−2 ≤ Pk = FmFn ≤ αm+n−2 and γk−1 ≥ Pk = FmFn ≥ αm+n−4. (6.1.25)

This implies that,

γk−2 ≤ αm+n−2 and γk−1 ≥ αm+n−4. (6.1.26)

With these inequalities, inequality (6.1.4) becomes

1 + c2(m+ n− 4) ≤ k ≤ 2 + c2(m+ n− 2), (6.1.27)

where c2 =logα

log γ= 0.545979 . . . .

This implies in particular that k < 3n. The analogue of inequality (6.1.8) in this case is given by∣∣∣∣ γk2√

2− αm+n

5

∣∣∣∣ =

∣∣∣∣ δk2√

2− αnβm + αmβn − βm+n

5

∣∣∣∣≤ 6

5max{|δ|k, αn−m} =

6αn−m

5.

(6.1.28)

By dividing through by αm+n

5leads to∣∣∣∣ 5

2√

2γkα−n−m − 1

∣∣∣∣ < 6

α2m, (6.1.29)

which is the analogue of inequality (6.1.9). We check that the amount Λ3 on the left hand side ofthe above equation is also non-zero through similar arguments as before. We then apply Theorem5.2.2 with data

Λ3 =5

2√

2γkα−n−m − 1, l = 3, η1 =

5

2√

2, η2 = γ,

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 logα)(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 logα. (6.1.31)


By comparing these two inequalities (6.1.30) and (6.1.31), we get the analogue of (6.1.12) as

2m logα− log 6 < 5.98× 1013(1 + log(3n)), (6.1.32)

which gives

m logα + 1 < 3× 1013(1 + log(3n)), (6.1.33)

which is the analogue of inequality (6.1.13). Returning to equation (1.1.7), we get∣∣∣∣ γk

2√

2Fm− αn√

5

∣∣∣∣ =

∣∣∣∣ δk

2√

2Fm− βn√

5

∣∣∣∣ ≤ 2√5

max

{1

γk,

1

αn

}. (6.1.34)

By (6.1.27), we getγk ≥ γαm+n−4 ≥ γα−3αn,

so1

γk≤ α3/γ

αn<

2

αn. (6.1.35)

Hence, by (6.1.34) and (6.1.35), we get∣∣∣∣∣√

5

2√

2Fmγkα−n − 1

∣∣∣∣∣ < 4

α2n. (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 thatΛi 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 5X2 − 8F 2m. Its height therefore satisfies

h(η1) ≤ logFm + log 2√

2,

≤ (m− 1) logα + log 2√

2,

< m logα + 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))212× 1013 × (2 logα)(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 logα− 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 alemma.

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.


6.2 Reducing the bounds

Now we need to reduce the bounds. To do so, we make use several times of the result inLemma 3.4.5, which is a slight variation of a result due to Dujella and Petho [3] which itself is ageneralization 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 logα− (n+m) log γ + log(8/√

5).

Then we have from (6.1.9) that

|eΓ1 − 1| = |Λ1| <16√5γ2m

<1

4, (6.2.1)

which implies that

|Γ1| <1

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√5γ2m

. (6.2.3)

If Γ1 > 0 (we observe that Γ1 6= 0 since Λ1 6= 0), then

0 < k

(logα

log γ

)− (n+m) +

log(8/√

5)

log γ<

32

(√

5 log γ)γ2m<

17

γ2m.

We apply Lemma 3.4.5 with M = 3× 1031 (note that M > 4n > k),

τ =logα

log γ, µ =

log(8/√

5)

log γ, A = 17, B = γ2.

Writing τ = [a0, a1, . . .] as a continued fraction, we get

[a0, . . . , a74] =p74

q74

=2037068391552562960855777461929676271

3731035235978315437343082205475618926,

and we get q74 > 3× 1036 > 6M . We compute

ε = ‖µq74‖ −M‖τq74‖ > 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).


6.2.2 Case 2. In the case of inequality (6.1.29), assuming again that m ≥ 20, we get that∣∣∣(n+m) logα− k log γ − log(5/2√

2)∣∣∣ < 12

5α2m.

Let Γ3 be the expression under the absolute value of the left–hand side above. If Γ3 > 0, we get

0 < (n+m)

(logα

log γ

)− k +

log(2√

2/5)

log γ<

12

(5 log γ)α2m<

3

α2m.

We keep the same values for M, τ, q and only change µ to

µ′ =log(2

√2/5)

log γ, A = 3, B = α2.

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∣∣∣k logα− n log γ + log(2√

2/√

5Pm)∣∣∣ < 72

√2√

5δ2n.

Let Γ2 be the expression under the absolute value in the left–hand side above. If Γ2 > 0, wethen get

0 < k

(logα

log γ

)− n+

log(2√

2/(√

5Pm))

log γ<

72√

2

(√

5 log γ)γ2n<

52

γ2n.


µm =log(2

√2/(√

5Pm))

log γ, A = 52, B = γ2 for all m = 1, . . . , 90.

We get ε > 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∣∣∣n logα− k log γ + log(2√

2Fm/√

5)∣∣∣ < 4

α2n.

Putting Γ4 for the amount under the absolute value in the left–hand side above, we get in caseΓ4 > 0 that

0 < n

(logα

log γ

)− k +

log(2√

2Fm/√

5)

log γ<

4

log γα2n<

5

α2n.


µm =log(2

√2Fm/

√5)

log γ, A = 5, B = α2, for all m = 1, . . . , 90.

We get ε > 0.005, so n ≤ 94. So, in all cases n ≤ 94, so k < 400.

We used sage to generate {Fk}1≤k≤400 and {PmPn}1≤m<n≤100 and intersected them, and alsogenerated {Pk}1≤k≤400 and {FmFn}1≤m<n≤100 and intersected them and got no other solutions.Hence, Theorem 1.1.1 is proved.

7. Comments and Conclusion

It is apparent from our proof that the method is more general and shows that every Diophantineequation of the form

Uk = VmVn

has only finitely effectively computable many positive integer solutions (k,m, n) provided that{Un}n≥0 and {Vn}n≥0 satisfy a few technical conditions such as:

(i) they are both non degenerate binary recurrent and have characteristic equations of realroots α, β and γ, δ with αβ = ±1 and γδ = ±1.

(ii) Q[α] and Q[γ] are distinct quadratic fields.

In fact, more is true, namely that for fixed k and s, the Diophantine equation

k∏i=1

Fni=

s∏j=1

Pmj

has only finitely many effectively computable positive integer solutions

(n1, . . . , nk,m1, . . . ,mk).

Such a statement is not very difficult to prove. A deeper conjecture made in [18] to the effectthat the intersection of the multiplicative group generated by {Fn}n≥1 with the multiplicativegroup generated by Pell numbers {Pn}n≥1 is finitely generated cannot unfortunately be attackedby these methods.

32

AppendixA.

A.1 Sage source code

###################### List of first 400 Pell numbers ########################

a = [0, 1]

for k in range(2, 400):

u = a[k - 1]

v = a[k - 2]

a.append(2*u + v)

print(a)

############# The product P_{m}P_{n} for first 100 Pell numbers ###############

b = set()

for m in range(1, 100):

for n in range(1, 100):

b.add(a[m]*a[n])

print(b)

#################### List of first 400 Fibonacci numbers #####################

c = [0, 1]

for k in range(2, 400):

u = c[k - 1]

v = c[k - 2]

c.append(u + v)

print(c)

########## The product F_{m}F_{n} for first 100 Fibonacci numbers ############

d = set()

for m in range(1, 100):

for n in range(1, 100):

d.add(c[m]*c[n])

print(d)

########### The intersection of F_{k} and P_{m}P_{n} #########################

for f in c:

if f in b:

print("Intersect1: " + str(f))

########### The intersection of P_{k} and F_{m}F_{n} #########################

for f in a:

if f in d:

print("Intersect2: " + str(f))

33

References[1] J. H. E. Cohn. On Square Fibonacci Numbers. Journal of the London Mathematical Society,

1(1):537–540, 1964.

[2] J. H. E. Cohn. Perfect Pell Powers. Glasgow Mathematical Journal, 38(1):19–20, 1996.

[3] A. Dujella and A. Petho. A generalization of a Theorem of Baker and Davenport. TheQuarterly Journal of Mathematics, 49(195):291–306, 1998.

[4] Wikipedia. Diophantine equation — Wikipedia, The Free Encyclopedia, 2015. http://en.wikipedia.org/w/index.php?title=Diophantine equation&oldid=653279286 [Online; ac-cessed 19-April-2015].

[5] D. G. Terzi. On a Conjecture by Erdos-Straus. BIT Numerical Mathematics, 11(2):212–216,1971.

[6] B. Demirturk and R. Keskin. Integer Solutions of some Diophantine Equations via Fibonacciand Lucas Numbers. Journal of Integer Sequences, 12(2):3, 2009.

[7] M. Mignotte. Intersection des images de certaines suites recurrentes lineaires. TheoreticalComputer Science, 7(1):117–121, 1978.

[8] M. Mignotte. Une extension du theoreme de skolem - mahler. C. R. Academic Science Paris,288(1):233–235, 1979.

[9] A. Baker and H. Davenport. The Equations 3x2− 2 = y2 and 8x2− 7 = z2. The QuarterlyJournal of Mathematics, 20(1):129–137, 1969.

[10] E. M. Matveev. An Explicit Lower Bound for a Homogeneous Rational Linear Form inLogarithms of Algebraic Numbers (II). Izvestiya: Mathematics, 64(6):1217, 2000.

[11] W. B. Jones and W. J. Thron Continued Fractions. Analytic Theory and Applications.Encyclopedia of Mathematics and its Applications, 1980.

[12] B. Everitt. Symmetries of Equations: An Introduction to Galois Theory, 2007. www-users.york.ac.uk/∼bje1/galnotes.pdf [Online; accessed 10-April-2015].

[13] Wikipedia. Continued fraction — Wikipedia, The Free Encyclopedia. http://en.wikipedia.org/w/index.php?title=Continued fraction&oldid=663737364, 2015. [Online; accessed 19-April-2015].

[14] F. Luca. Effective Methods for Diophantine Equations, 2009. https://math.dartmouth.edu/∼m105f12/lucaHungary1.pdf [Online; accessed 10-April-2015].

[15] S. Lang. Introduction to Diophantine Approximation, 1966. Addison-Wesley, Reading, MA.

[16] A. Dujella and B. Jadrijevic. A Family of Quartic Thue Inequalities. ACTA ARITHMETICA-WARSZAWA-, 111(1):61–76, 2004.

34

http://en.wikipedia.org/w/index.php?title=Diophantine_equation&oldid=653279286

http://en.wikipedia.org/w/index.php?title=Diophantine_equation&oldid=653279286

www-users.york.ac.uk/~bje1/galnotes.pdf

www-users.york.ac.uk/~bje1/galnotes.pdf

http://en.wikipedia.org/w/index.php?title=Continued_fraction&oldid=663737364

http://en.wikipedia.org/w/index.php?title=Continued_fraction&oldid=663737364

https://math.dartmouth.edu/~m105f12/lucaHungary1.pdf

https://math.dartmouth.edu/~m105f12/lucaHungary1.pdf

REFERENCES Page 35

[17] A. Baker. Linear Forms in the Logarithms of Algebraic Numbers (IV). Mathematika,15(02):204–216, 1968.

[18] F. Luca, C. Pomerance, and S. Wagner. On Fibonacci Integers. Journal of Number Theory,131(3):440–457, 2011.

by mahadi ddamulira ([email protected])

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