Linear Forms in Logarithms

Abstract

Hilbert’s problems form a list of twenty-three problems in mathematics published by David Hilbert, a German mathematician, in 1900. The problems were all unsolved at the time and several of them were very influential for 20th century mathematics. Hilbert believed it was essential for mathematicians to find new machineries and methods in order to solve the mentioned problems. The seventh problem deals with the transcendence of $\alpha ^\beta$ for algebraic $\alpha \ne 0,1$ and irrational algebraic $\beta$. This problem was solved by Gelfond and (independently) Schneider. In 1935, Gelfond found a lower bound for the absolute value of the linear form

$$\varLambda =\beta _1\log \alpha _1+\beta _2\log \alpha _2\ne 0.$$

He proved that

$$\log |\varLambda |\gg -h(\varLambda )^\kappa ,$$

where $h(\varLambda )$ is logarithmic height of the linear form $\varLambda$, $\kappa >5$ and $\gg$ denotes inequality that is valid up to an unspecified constant factor. He noticed that generalization of his results could prove a huge amount of unsolved problems in number theory.

In 1966 and 1967, in his papers “Linear forms in logarithms of algebraic numbers I, II, III”, A. Baker gave an effective lower bound on the absolute value of a nonzero linear form in logarithms of algebraic numbers, that is, for a nonzero expression of the form

$$\sum _{i=1}^{n}b_i\log \alpha _i,$$

where $\alpha _1, \dots , \alpha _n$ are algebraic numbers and $b_1, \dots , b_n$ are integers.

In these notes, we introduce definitions and theorems that are crucial for understanding and applications of linear forms in logarithms. Some Baker type inequalities that are easy to apply are introduced. In order to illustrate this very important machinery, we present some examples and show, among other things, that the largest Fibonacci number having only one repeated digit in its decimal expression is 55, that $d=120$ is the only positive integer such that the set $\{d+1, 3d+1, 8d+1\}$ consists of all perfect squares and that some parametric families of $D(-1)$-triples cannot be extended to $D(-1)$-quadruples.

Introduction

These lecture notes cover the course Linear Forms in Logarithms created for the Summer School Diophantine Analysis organized in Würzburg, Germany in 2014. The notes intend to be an introduction to Diophantine approximation and linear forms in logarithms.

We begin with the theory of Diophantine approximation which has an extremely important application in the study of Diophantine equations. One of the main topics is the question how well a given real number $\alpha$ can be approximated by rational numbers. By placing certain constraints on the rational numbers used in the approximation, we are able to classify the real number $\alpha$ as either a rational or an irrational number, or as an algebraic or a transcendental number. Diophantine approximation and transcendence theory are very close areas that share many theorems and methods which will be useful in the second part of these lecture notes.

There, we introduce linear forms in logarithms and provide lower bounds for linear forms in logarithms of algebraic numbers due to Alan Baker, one of the most famous mathematician in this field of mathematics. Baker was awarded the Fields Medal in 1970 because of his profound and significant contributions to number theory. To illustrate the importance of his machinery, many useful and interesting applications of the introduced concepts are presented.

1 Diophantine Approximation

Rational numbers are in every interval of the real line, no matter how small that interval is because the set of rational numbers is dense in the set of real numbers. As a consequence, for any given real number $\alpha$, we are able to find a rational number as close as we like to $\alpha$.

Approximating a real number by rational numbers helps us to better understand the set of real numbers and gives us a surprising insight in the properties of real numbers. By placing certain constraints on the rational numbers used in the approximation of the real number $\alpha$, properties of the real number $\alpha$ can be observed such that classify it as either a rational or an irrational number, or as an algebraic or a transcendental number.

As rational numbers approach a fixed real number, their denominators grow arbitrarily large. We study how closely real numbers can be approximated by rational numbers that have a fixed bound on the growth of their denominators.

1.1 Dirichlet’s Theorem

One of the main questions in Diophantine approximation is whether there exists any rational number $\frac{p}{q}$ satisfying the inequality

$$\left| \frac{a}{b}-\frac{p}{q}\right| <\frac{1}{q^2}.$$

The affirmative answer follows from the fundamental result due to Dirichlet¹ on rational approximation of real numbers.

Theorem 1.1.1

(Dirichlet 1842) Let $\alpha$ be a real number and n a positive integer. There exists a rational number $\frac{p}{q}$, $0<q\le n,$ satisfying the inequality

$$\begin{aligned} \left| \alpha -\frac{p}{q}\right| \le \frac{1}{(n+1)q}. \end{aligned}$$

(1.1)

Proof

For $n=1$, $\frac{p}{q}=\frac{\lfloor \alpha \rfloor }{1}$ or $\frac{p}{q}=\frac{\lfloor \alpha \rfloor +1}{1}$ satisfies

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{2},$$

where $\lfloor \ \rfloor$ stands for floor function. For $n\ge 2$, we consider $n+2$ elements

$$0, \ \alpha -\lfloor \alpha \rfloor , \ 2\alpha -\lfloor 2\alpha \rfloor , \ \dots , \ n\alpha -\lfloor n\alpha \rfloor , \ 1,$$

of the interval [0, 1].

Assume first that these elements are distinct as they are in the case $\alpha$ is an irrational number. The interval [0, 1] can be subdivided into $n+1$ subintervals of the length $\frac{1}{n+1}$ and the pigeonhole principle guarantees that two of these numbers differ in absolute value by at most $\frac{1}{n+1}$. If one of the numbers is 0 and the other is $i\alpha -\lfloor i\alpha \rfloor$, then $i\le n$, $|i\alpha -\lfloor i\alpha \rfloor |<\frac{1}{n+1}$ and

$$\left| \alpha -\frac{\lfloor i\alpha \rfloor }{i}\right| \le \frac{1}{(n+1)i}.$$

After $\frac{\lfloor i\alpha \rfloor }{i}$ is reduced to lowest terms $\frac{p}{q}$, the rational number $\frac{p}{q}$ satisfies (1.1). Similarly, if the two numbers in question are $j\alpha -\lfloor j\alpha \rfloor$ and 1, then $j\le n$ and reducing $\frac{\lfloor j\alpha \rfloor +1}{j}$ to lowest terms $\frac{p}{q}$, we have that $\frac{p}{q}$ satisfies (1.1). Finally, if the two numbers are $i\alpha -\lfloor i\alpha \rfloor$ and $j\alpha -\lfloor j\alpha \rfloor$ with $i<j$, then

$$|j\alpha -\lfloor j\alpha \rfloor -(i\alpha -\lfloor i\alpha \rfloor )|=|(j-i)\alpha -(\lfloor j\alpha \rfloor -\lfloor i\alpha \rfloor )|\le \frac{1}{n+1}.$$

Consequently, $j-i<n$ and

$$\left| \alpha -\frac{\lfloor j\alpha \rfloor -\lfloor i\alpha \rfloor }{j-i}\right| \le \frac{1}{(n+1)(j-i)}.$$

Thus, after $\frac{\lfloor j\alpha \rfloor -\lfloor i\alpha \rfloor }{j-i}$ is reduced to lowest terms $\frac{p}{q}$, the rational number $\frac{p}{q}$ satisfies (1.1).

If the $n+2$ numbers from the beginning are not distinct, then $\alpha$ itself is a rational number with denominator at most n. In this case, there exist $i<j$ so that $\alpha$ is equal to one of the following fractions

$$\frac{\lfloor i\alpha \rfloor }{i}, \frac{\lfloor j\alpha \rfloor -\lfloor i\alpha \rfloor }{j-i}$$

reduced to lowest terms. If the numbers are not distinct, the required inequality (1.1) is trivially satisfied by $\alpha$ itself.

Corollary 1.1.2

For $\alpha$ irrational, there exist infinitely many relatively prime numbers p, q such that

$$\begin{aligned} \left| \alpha -\frac{p}{q}\right| <\frac{1}{q^2}.\end{aligned}$$

(1.2)

Proof

Suppose there are only finitely many rationals

$$\frac{p_1}{q_1}, \frac{p_2}{q_2}, \dots , \frac{p_k}{q_k}$$

satisfying (1.2). In this case,

$$\left| \alpha -\frac{p_i}{q_i}\right| >0$$

for $1\le i\le k$. Consequently, since $\alpha$ is irrational, there exists a positive integer n such that the inequality

$$\left| \alpha -\frac{p_i}{q_i}\right| >\frac{1}{n+1}$$

holds for $1\le i\le k$. However, this contradicts Dirichlet’s Theorem 1.1.1 which asserts that, for this n, there exists a rational number $\frac{p}{q}$ with $q\le n$ such that

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{(n+1)q}<\frac{1}{q^2}.$$

Remark 1.1.3

Corollary 1.1.2 does not hold for $\alpha$ rational. Assume $\alpha =\frac{u}{v}.$ For $\frac{p}{q}\ne \alpha$, we have

$$\left| \alpha -\frac{p}{q}\right| =\left| \frac{u}{v}-\frac{p}{q}\right| \ge \frac{1}{vq},$$

so (1.1) implies $q<v,$ hence the inequality (1.1) can be satisfied only for finitely many relatively prime integers p, q.

Corollary 1.1.4

A real number $\alpha$ is irrational if and only if there are infinitely many rational numbers $\frac{p}{q}$ such that

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^2}.$$

A proof can be found in [44]. It appears that irrational numbers can be distinguished from rational numbers by the fact that they can be approximated by infinitely many rational numbers $\frac{p}{q}$ with an error less than $\frac{1}{q^2}.$

We may ask for the best possible value $C>0$ such that the statement of the Corollary 1.1.4 holds with $\frac{1}{Cq^2}$ in place of $\frac{1}{q^2}$? The answer will be given in Hurwitz’s² Theorem 1.3.5 which characterizes the best Dirichlet-type inequality.

1.2 Continued Fractions

Definition 1.2.1

• An infinite generalized continued fraction is an expression of the form

  $$\begin{aligned} a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{\ddots }}}, \end{aligned}$$

  (1.3)

  where $a_0, a_1, a_2, \dots$ and $b_1, b_2, \dots$ are either rational, real or complex numbers or functions of such variables.

• For $b_i=1, \ i\in \mathbb {N}$, (1.3) is called an infinite simple continued fraction. Its abbreviated notation is

  $$[a_0, a_1, a_2, \dots ].$$

• An expression

  $$[a_0, \ a_1, \dots , a_n]=a_0+\frac{1}{a_1+\frac{1}{a_2+\dots +\frac{1}{a_{n-1}+\frac{1}{a_n}}}}$$

  is called a finite simple continued fraction with $a_i\ge 1, \ i=1, \dots , n$ and $a_n\ge 2$ integers that are called partial quotients . The rational numbers

  $$\frac{p_0}{q_0}=[a_0], \ \frac{p_1}{q_1}=[a_0, a_1], \ \frac{p_2}{q_2}=[a_0, a_1, a_2], \ \dots , \ \frac{p_n}{q_n}=[a_0, a_1, a_2, \dots , a_n]$$

  are called the convergents of $\frac{p}{q}$ and n is its length.

Remark 1.2.2

We set $p_{-2}:=0, \ p_{-1}:=1, \ q_{-2}:=1, \ q_{-1}=0.$

Lemma 1.2.3

(Law of formation of the convergents) For $n\ge 0$,

$$\begin{aligned} p_n=a_np_{n-1}+p_{n-2}, \ \ q_n=a_nq_{n-1}+q_{n-2}. \end{aligned}$$

(1.4)

Proof

The equalities (1.4) are satisfied for $n=0$. We assume that (1.4) is satisfied for $n-1$, i.e.,

$$p_{n-1}=a_{n-1}p_{n-2}+p_{n-3}, \ \ q_{n-1}=a_{n-1}q_{n-2}+q_{n-3}.$$

Then we get

$$\begin{aligned}{p_n\over q_n}&=[a_0,a_1,\ldots ,a_{n-1}+1/a_n]\\ {}&={(a_{n-1}+{1\over a_n})p_{n-1}+p_{n-2}\over (a_{n-1}+{1\over a_n})q_{n-1}+q_{n-2}}\\ {}&={(a_na_{n-1}+1)p_{n-2}+a_np_{n-1}\over (a_na_{n-1}+1)q_{n-2}+a_nq_{n-1}}={a_np_{n-1}+p_{n-2}\over a_nq_{n-1}+q_{n-2}}. \end{aligned}$$

Lemma 1.2.4

For $n\ge -1$,

$$\begin{aligned} q_np_{n-1}-p_nq_{n-1}=(-1)^n. \end{aligned}$$

(1.5)

Proof

Let $n=-1$. We observe that $q_{-1}p_{-2}-p_{-1}q_{-2}=(-1)^{-1}.$ We assume that (1.5) is satisfied for $n-1$. Using Lemma 1.2.3, we find

$$\begin{aligned} q_np_{n-1}-p_nq_{n-1}&=(a_nq_{n-1}+q_{n-2})p_{n-1}-(a_np_{n-1}+p_{n-2})q_{n-1}=\\&=-(q_{n-1}p_{n-2}-p_{n-1}q_{n-2})=(-1)^n. \end{aligned}$$

Lemma 1.2.5

For $n\ge 0$,

$$q_np_{n-2}-p_nq_{n-2}=(-1)^{n-1}a_n.$$

Proof

It follows from Lemmas 1.2.3 and 1.2.4 that

$$\begin{aligned} q_np_{n-2}-p_nq_{n-2}&=(a_nq_{n-1}+q_{n-2})p_{n-2}-(a_np_{n-1}+p_{n-2})q_{n-2}=\\&=a_n(q_{n-1}p_{n-2}-p_{n-1}q_{n-2})=(-1)^{n-1}a_n. \end{aligned}$$

Theorem 1.2.6

The convergents $\frac{p_k}{q_k}$ satisfy the following inequalities:

• (i) $\frac{p_0}{q_0}<\frac{p_2}{q_2}<\frac{p_4}{q_4}<\dots ,$

• (ii) $\frac{p_1}{q_{1}}>\frac{p_3}{q_3}>\frac{p_5}{q_5}>\dots ,$

• (iii) for n even and m odd,

  $$\frac{p_n}{q_n}<\frac{p_m}{q_m}.$$

Proof

Using Lemma 1.2.5, we find

$$\frac{p_{n-2}}{q_{n-2}}-\frac{p_n}{q_n}=\frac{(-1)^{n-1}a_n}{q_{n-2}q_n}.$$

For $n\ge 2$ and n even, we obtain $\frac{p_{n-2}}{q_{n-2}}<\frac{p_n}{q_n}$, while $\frac{p_{n-2}}{q_{n-2}}>\frac{p_n}{q_n}$, for $n\ge 3$ and n odd.

It remains to prove the last inequality. Let $n<m$. Since $\frac{p_n}{q_n}\le \frac{p_{m-1}}{q_{m-1}},$ it is sufficient to prove $\frac{p_{m-1}}{q_{m-1}}<\frac{p_m}{q_m},$ which is satisfied by Lemma 1.2.4,

$$q_mp_{m-1}-p_mq_{m-1}=(-1)^m=-1<0.$$

The proof for $n>m$ follows analogously.

Lemma 1.2.7

For an integer $a_0$ and positive integers $a_1, a_2, \dots , a_n$, the continued fraction $[a_0, a_1, \dots , a_n]$ is rational. Conversely, for every rational number $\frac{u}{v}$ there exist $n\ge 0,$ an integer $a_0$ and positive integers $a_1, a_2, \dots , a_n$ such that

$$\frac{u}{v}=[a_0, a_1, \dots , a_n].$$

For $\frac{u}{v}\ge 1$, we have $a_0\ge 1.$

Proof

Using Lemma 1.2.3, we find

$$[a_0, a_1, \dots , a_n]=[a_0, a_1, \dots , a_{k-1}, [a_k, a_{k+1}, \dots , a_n]]=\frac{p_{k-1}r_k+p_{k-2}}{q_{k-1}r_k+q_{k-2}},$$

where $r_k:=[a_k, a_{k+1}, \dots , a_n].$

Conversely, let $v>0$ and $gcd (u, v)=1$. For $v=1$, clearly $\frac{u}{v}\in \mathbb {Z}$, hence, setting $a_0=\frac{u}{v},$ we obtain $\frac{u}{v}=[a_0]$.

If $v>1$, then there exist $q, r\in \mathbb {Z}$ such that $u=vq+r, \ 1\le r <v.$ We assume $\frac{v}{r}=[a_1, \dots , a_n].$ Since $\frac{v}{r}>1$, we conclude that $a_1, \dots , a_n\in \mathbb {N}.$ Hence,

$$\frac{u}{v}=q+\frac{1}{\frac{v}{r}}=q+\frac{1}{[a_1, \dots , a_n]}=[q, a_1, \dots , a_n],$$

where $a_0=q.$ Clearly, if $\frac{u}{v}\ge 1,$ then $a_0=q\ge 1.$

Remark 1.2.8

There is a one-to-one correspondence between rational numbers and finite simple continued fractions.

Remark 1.2.9

Let $\frac{u}{v}$ be rational, $\mathrm{{gcd}}(u, v)=1$ and $u>v>0.$ It follows from Euclid’s algorithm that

$$u=vq_1+r_1, \ v=r_1q_2+r_2, \ \dots , \ r_{j-1}=r_jq_{j+1},$$

hence

$$\frac{u}{v}=q_1+\frac{1}{\frac{c}{r_1}}=q_1+\frac{1}{q_2+\frac{1}{\frac{r_1}{r_2}}}=\dots =[q_1, q_2,\dots , q_{j+1}].$$

Example 1.2.10

$$\begin{aligned} \frac{7}{11}= & {} 0+\frac{1}{\frac{11}{7}}=0+\frac{1}{1+\frac{4}{7}}=0+\frac{1}{1+\frac{1}{\frac{7}{4}}}=\\= & {} 0+\frac{1}{1+\frac{1}{1+\frac{3}{4}}}=0+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{4}{3}}}}=0+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{3}}}}\ , \end{aligned}$$

which can be written as

$$\frac{7}{11}=[0, 1, 1, 1, 3].$$

Example 1.2.11

The number e is irrational.

The proof goes as follows: if we assume that e is rational, then it can be represented as

$$e=\sum _{i=0}^{\infty }\frac{1}{i!}=\frac{u}{v}.$$

Let n be an integer such that $n>v.$ We define $\frac{p}{q}=\sum _{i=0}^n\frac{1}{i!}$, with $q=n!$. Using Remark 1.2.9, we get

$$\begin{aligned} \frac{1}{v}\le q\left| e-\frac{p}{q}\right|= & {} n!\sum _{j=1}^{\infty }\frac{1}{(n+j)!}\\< & {} n!\sum _{k=0}^{\infty }\frac{1}{(n+1)!}\frac{1}{(n+1)^k}\le \frac{1}{n+1}\frac{1}{1-\frac{1}{n+1}}=\frac{1}{n}, \end{aligned}$$

a contradiction to $n>v.$

Lemma 1.2.12

For every integer r, there exist exactly two different simple continued fraction expansions representing r, namely $r=[r]$ and $r=[r-1, 1]$. For r rational, there exist exactly two different simple continued fraction expansions representing r, namely $[a_0, a_1, \dots , a_n]$ with $a_n\ge 2$ and $[a_0, a_1, \dots , a_{n-1}, a_n-1, 1].$

Proof

Every $r\in \mathbb {Q}$ can be represented as a finite simple continued fraction, namely

$$r=[a_0, a_1, \dots , a_n], \ \ a_0\in \mathbb {Z}, \ \ a_1, a_2, \dots , a_n\in \mathbb {N}.$$

If $r\in \mathbb {Z}$, then $n=0$ and $r=[a_0]=a_0$, resp. $r=[r].$ Otherwise, when $n>0$, then

$$r=a_0+\frac{1}{[a_1, \dots , a_n]},$$

with $[a_1, \dots , a_n]\ge 1.$ In view of $r-a_0\in \mathbb {Z}$, we conclude that $[a_1, \dots , a_n]=1.$ Since $a_1\ge 1,$ we get $a_1=1, n=1,$ so $a_0=r-1$ and $r=[r-1,1].$

Now consider $r=\frac{u}{v}, \ gcd (u, v)=1, \ v>0.$ We use induction on v in order to prove the second claim. The case when $v=1$ has already been considered. If $v>1,$ we get $\frac{u}{v}=a_0+\frac{u_1}{v},$ where $u_1<v$. Let $\alpha _1:=\frac{u_1}{v}$. Using the hypothesis of the induction, we conclude that $\alpha _1$ has two continued fraction expansions $[a_1, a_2, \dots , a_{n-1}, a_n], \ a_n\ge 2$ and $[a_1, a_2, \dots , a_n-1, 1].$

Lemma 1.2.13

Let $a_0$ be an integer and $a_1, a_2, \dots ,$ be positive integers. Then, the limit $\lim _{n\rightarrow \infty }\frac{p_n}{q_n}$ exists and its value is irrational. Conversely, for $\alpha$ irrational, there exist a unique integer $a_0$ and unique positive integers $a_1, a_2, \dots$ such that $\alpha =\lim _{n\rightarrow \infty }\frac{p_n}{q_n}.$

Proof

In view of $\frac{p_0}{q_0}<\frac{p_2}{q_2}<\dots <\frac{p_1}{q_1}$, it is clear that both limits

$$\lim _{\begin{array}{c} n\rightarrow \infty \\ n \ \text {even} \end{array}}\frac{p_n}{q_n} \qquad \text{ and }\qquad \lim _{\begin{array}{c} n\rightarrow \infty \\ n \ \text {odd} \end{array}}\frac{p_n}{q_n}$$

exist, and it is easily shown that both limits are equal. We put $\alpha =\lim _{n\rightarrow \infty }\frac{p_n}{q_n}$ and compute

$$\left| \alpha -\frac{p_n}{q_n}\right|<\left| \frac{p_{n+1}}{q_{n+1}}-\frac{p_n}{q_n}\right| =\frac{1}{q_nq_{n+1}}<\frac{1}{q_n^2}.$$

Since $p_n, q_n$ are relatively prime, there exist infinitely many rational numbers $\frac{p}{q}$ such that $|\alpha -\frac{p}{q}|<\frac{1}{q^2}$, so $\alpha$ is irrational.

Conversely, let $\alpha$ be irrational, $a_0=\lfloor \alpha \rfloor$, and let $\alpha _1:=a_0+\frac{1}{\alpha _1}.$ We notice that $\alpha _1>1$ is irrational. For $k\ge 1$ let $a_k=\lfloor \alpha _k \rfloor$ and $\alpha _k=a_k+\frac{1}{\alpha _{k+1}}.$ We observe that $a_k\ge 1, \ \alpha _{k+1}>1$, and $\alpha _{k+1}$ is irrational. Our goal is to show

$$\alpha =[a_0, a_1, a_2, \dots ].$$

Using Lemmas 1.2.3 and 1.2.4 with $\alpha =[a_0, a_1, \dots , \alpha _{n+1}]$, we find

$$\begin{aligned} q_n\alpha -p_n= & {} q_n\frac{\alpha _{n+1}p_n+p_{n-1}}{\alpha _{n+1}q_n+q_{n-1}}-p_n\\= & {} {q_n(\alpha _{n+1}p_n+p_{n-1})-p_n(\alpha _{n+1}q_n+q_{n-1})\over \alpha _{n+1}q_n+q_{n-1}}=\frac{(-1)^n}{\alpha _{n+1}q_n+q_{n-1}}. \end{aligned}$$

Hence,

$$\begin{aligned} \left| \alpha -\frac{p_n}{q_n}\right| <\frac{1}{q_n^2}, \end{aligned}$$

(1.6)

which implies $\lim _{n\rightarrow \infty }\frac{p_n}{q_n}=\alpha .$ Finally, it remains to prove that the integers $a_0, a_1\ge 1, a_2\ge 1, \dots$ are uniquely determined. In view of

$$\alpha =[a_0, a_1, a_2, \dots ]=a_0+\frac{1}{[a_1, a_2, \dots ]},$$

and $0\le \alpha - a_0<1$, we find $a_0=\lfloor \alpha \rfloor$ which implies that $a_0$ is unique and $\alpha _1=[a_1, a_2, \dots ]$ is uniquely determined by $\alpha .$ Because $a_1=\lfloor \alpha _1\rfloor$, $a_1$ is unique, etc. This proves the lemma.

1.3 Hurwitz’s Theorem

In the sequel we assume $\alpha$ to be irrational. According to (1.6), we conclude that each convergent of $\alpha$ satisfies the inequality

$$\left| \alpha -\frac{p}{q}\right| <\frac{1}{q^2}.$$

Vahlen³ and Borel⁴ have proved the following theorems that deal with approximation properties of two and three consecutive convergents, respectively.

Theorem 1.3.1

(Vahlen 1895, [48]) Let $\alpha$ be an irrational number and denote by $\frac{p_{n-1}}{q_{n-1}}, \frac{p_n}{q_n}$ two consecutive convergents of $\alpha$. Then, at least one of them satisfies

$$\left| \alpha -\frac{p}{q}\right| <\frac{1}{2q^2}.$$

Proof

We observe that

$$\left| \alpha -\frac{p_n}{q_n}\right| +\left| \alpha -\frac{p_{n-1}}{q_{n-1}}\right| =\left| \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\right| =\frac{1}{q_nq_{n-1}}<\frac{1}{2q_n^2}+\frac{1}{2q_{n-1}^2}.$$

Thus,

$$\left| \alpha -\frac{p_n}{q_n}\right|<\frac{1}{2q_n^2} \ \ \ or \ \ \ \left| \alpha -\frac{p_{n-1}}{q_{n-1}}\right| <\frac{1}{2q_{n-1}^2}.$$

Theorem 1.3.2

(Borel 1903, [9]) Let $\alpha$ be an irrational number and denote by $\frac{p_{n-1}}{q_{n-1}}, \frac{p_n}{q_n}, \frac{p_{n+1}}{q_{n+1}}$ three consecutive convergents of $\alpha .$ Then at least one of them satisfies the inequality

$$\left| \alpha -\frac{p}{q}\right| <\frac{1}{\sqrt{5}q^2}.$$

Proof

Let $\alpha =[a_0, a_1, \dots ], \ \alpha _i=[a_i, a_{i+1}, \dots ], \ \beta _i=\frac{q_{i-2}}{q_{i-1}}, \ q\ge 1.$ It is not difficult to deduce that

$$\begin{aligned} \left| \alpha -\frac{p_n}{q_n}\right| =\frac{1}{q_n^2(\alpha _{n+1}+\beta _{n+1})}. \end{aligned}$$

We show that there does not exist a positive integer n satisfying

$$\begin{aligned} \alpha _i+\beta _i<\sqrt{5} \end{aligned}$$

(1.7)

for $i=n-1, n, n+1$. Our reasoning is indirect. We assume that (1.7) is satisfied for $i=n-1, n.$ It follows from

$$\alpha _{n-1}=a_{n-1}+\frac{1}{\alpha _n}, \ \ \frac{1}{\beta _n}=\frac{q_{n-1}}{q_{n-2}}=a_{n-1}+\frac{q_{n-3}}{q_{n-2}}=a_{n-1}+\beta _{n-1}$$

that

$$\frac{1}{\alpha _n}+\frac{1}{\beta _n}=\alpha _{n-1}+\beta _{n-1}\le \sqrt{5}.$$

Hence, $1=\alpha _n\cdot \frac{1}{\alpha _n}\le (\sqrt{5}-\beta _n)(\sqrt{5}-\frac{1}{\beta _n})$ or, equivalently, $\beta _n^2-\sqrt{5}\beta _n+1\le 0$ which implies $\beta _n\ge \frac{\sqrt{5}-1}{2}.$ For $\beta _n$ rational, we conclude that $\beta _n>\frac{\sqrt{5}-1}{2}$. Now, if (1.7) is satisfied for $i=n, n+1$, then again $\beta _{n+1}>\frac{\sqrt{5}-1}{2}$, so we deduce

$$1\le a_n=\frac{q_n}{q_{n-1}}-\frac{q_{n-2}}{q_{n-1}}=\frac{1}{\beta _{n+1}}-\beta _n<\frac{2}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{2}<1,$$

the desired contradiction.

Legendre⁵ proved with the following important theorem a converse to the previous results.

Theorem 1.3.3

(Legendre, [32]) Let p, q be integers such that $q\ge 1$ and

$$\left| \alpha -\frac{p}{q}\right| <\frac{1}{2q^2}.$$

Then $\frac{p}{q}$ is a convergent of $\alpha$ .

Proof

Let $\alpha -\frac{p}{q}=\frac{\varepsilon \nu }{q^2}$ with $0<\nu <\frac{1}{2}$ and $\varepsilon =\pm 1.$ In view of Lemma 1.2.12 there exists a simple continued fraction

$$\frac{p}{q}=[b_0, b_1, \dots , b_{n-1}]$$

satisfying $(-1)^{n-1}=\varepsilon .$ We define $\omega$ by

$$\alpha =\frac{\omega p_{n-1}+p_{n-2}}{\omega q_{n-1}+q_{n-2}},$$

such that $\alpha =[b_0, b_1, \dots , b_{n-1}, \omega ]$. Hence,

$$\frac{\varepsilon \nu }{q^2}=\alpha -\frac{p}{q}=\frac{1}{q_{n-1}}(\alpha q_{n-1}-p_{n-1})=\frac{1}{q_{n-1}}\cdot \frac{(-1)^{n-1}}{\omega q_{n-1}+q_{n-2}},$$

giving $\nu =\frac{q_{n-1}}{\omega q_{n-1}+q_{n-2}}$ and $\omega = \frac{1}{\nu }-\frac{q_{n-2}}{q_{n-1}}>1.$ Next we consider the finite or infinite continued fraction of $\omega$,

$$\omega =[b_n, b_{n+1}, b_{n+2}, \dots ].$$

Since $\omega >1$, we conclude $b_j\in \mathbb {N}, \ j=n, n+1, n+2, \dots .$ Consequently,

$$\alpha =[b_0, b_1, \dots , b_{n-1}, b_n, b_{n+1}, \dots ]$$

which is the continued fraction expansion for $\alpha$ and

$$\frac{p}{q}=\frac{p_{n-1}}{q_{n-1}}=[b_0, b_1, \dots , b_{n-1}]$$

is indeed a convergent to $\alpha$.

Lemma 1.3.4

Assume the continued fraction expansion for $\alpha$ is given by

$$\begin{aligned} \alpha =[a_0, a_1, \dots , a_N, 1, 1, \dots ]. \end{aligned}$$

(1.8)

Then

$$\lim _{n\rightarrow \infty }q_n^2\left| \alpha -\frac{p_n}{q_n}\right| =\frac{1}{\sqrt{5}}.$$

A proof can be found in [45].

Theorem 1.3.5

(Hurwitz, [31]) Let $\alpha$ be an irrational number.

• (i) Then there are infinitely many rational numbers $\frac{p}{q}$ such that

  $$\left| \alpha -\frac{p}{q}\right| <\frac{1}{\sqrt{5}q^2}.$$

• (ii) If $\sqrt{5}$ is replaced by $C>\sqrt{5}$, then there are irrational numbers $\alpha$ for which statement (i) does not hold .

Proof

Claim (i) follows directly from Theorem 1.3.2, while claim (ii) follows from Theorem 1.3.3 and Lemma 1.3.4. Namely, if $\alpha$ is irrational and of the form (1.8), then according to Theorem 1.3.3, all solutions of $|\alpha -\frac{p}{q}|<\frac{1}{Cq^2}$ with $C>\sqrt{5}$ can be found among the convergents to $\alpha$, however, in view of Lemma 1.3.4 this inequality is satisfied by only finitely many convergents to $\alpha .$

Definition 1.3.6

An irrational number $\alpha$ is said to be a quadratic irrational number if it is the solution of some quadratic equation with rational coefficients.

Definition 1.3.7

The continued fraction expansion $[a_0, a_1, \dots ]$ of a real number $\alpha$ is said to be eventually periodic if there exist integers $m\ge 0$ and $h>0$ such that

$$a_n=a_{n+h}, \ \ \text {for all} \ \ n\ge m.$$

In this case the continued fraction expansion is denoted by

$$[a_0, a_1, \dots , a_{m-1}, \overline{a_m, a_{m+1}, \dots , a_{m+h-1}}].$$

The continued fraction expansion is said to be periodic if it is eventually periodic with $m=0.$

Euler⁶ and Lagrange⁷ have proved an important characterization of quadratic irrationals in terms of their continued fraction expansion:

Theorem 1.3.8

(Euler 1737; Lagrange 1770) A real number $\alpha$ is quadratic irrational if and only if its continued fraction expansion is eventually periodic.

A proof can be found in [46].

Theorem 1.3.9

Let $d>1$ be an integer which is not a perfect square. Then the continued fraction expansion of $\sqrt{d}$ is of the form

$$[a_0, \overline{a_1, \dots , a_{n-1}, 2a_0}]$$

with $a_0=\lfloor \sqrt{d}\rfloor , a_1=a_{n-1}, a_2=a_{n-2}, \dots .$

A proof can be found in [41].

Remark 1.3.10

Let $E>1$ be an integer which is not a perfect square. Let

$$\alpha _0=\frac{s_0+\sqrt{E}}{t_0}$$

be a quadratic irrational with $s_0, t_0\in \mathbb {Z}$, $t_0\ne 0$ such that $t_0\mid (E-s_0^2)$. Then the partial quotients $a_i$ are given by the recursion

$$a_i=\lfloor \sqrt{\alpha _i}\rfloor , \ \ s_{i+1}=a_it_i-s_i, \ \ t_{i+1}=\frac{E-s_{i+1}^2}{t_i}, \ \ \alpha _{i+1}=\Bigl \lfloor \frac{s_{i+1}+\sqrt{E}}{t_{i+1}}\Bigr \rfloor .$$

Example 1.3.11

For $d\in \mathbb {N},$

$$\sqrt{2d(2d-1)}=[2d-1;\overline{2, 4d-2}],$$

$$\sqrt{2d(8d-1)}=[4d-1;\overline{1, 2, 1, 8d-2}].$$

Definition 1.3.12

An irrational number $\alpha$ is said to be badly approximable if there is a real number $c(\alpha )>0$, depending only on $\alpha$, such that

$$\left| \alpha -\frac{p}{q}\right| >\frac{c(\alpha )}{q^2}$$

for all rational numbers $\frac{p}{q}$.

Proposition 1.3.13

If an irrational number $\alpha$ is badly approximable, then for every $\varepsilon >0,$ there are only finitely many rational numbers $\frac{p}{q}$ satisfying

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{2+\varepsilon }}.$$

Proof

Let $c(\alpha )$ be a positive real number such that

$$\left| \alpha -\frac{p}{q}\right| >\frac{c(\alpha )}{q^2}$$

for all rational numbers $\frac{p}{q}$. Now suppose that there are infinitely many rational numbers $\frac{p}{q}$ satisfying

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{2+\varepsilon }}.$$

Consequently, there exists a rational number $\frac{p_1}{q_1}$ for which the inequality

$$q_1^{\varepsilon }>\frac{1}{c(\alpha )}$$

holds. It follows that

$$\left| \alpha -\frac{p_1}{q_1}\right| \le \frac{1}{q_1^2q_1^{\varepsilon }}<\frac{c(\alpha )}{q_1^2},$$

which is a contradiction.

Corollary 1.3.14

For quadratic irrational numbers $\alpha$ and any $\varepsilon >0,$ there are only finitely many rational numbers $\frac{p}{q}$ satisfying

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{2+\varepsilon }}.$$

A proof can be found in [44]. Hence, the exponent in the upper bound $\frac{1}{q^2}$ in Dirichlet’s Theorem 1.1.1 cannot be improved for real quadratic irrational numbers.

More on Diophantine approximation can be found in [13, 19, 20, 44, 45].

1.4 Algebraic and Transcendental Numbers

1.4.1 Basic Theorems and Definitions

Definition 1.4.1

Given fields F and E such that $F\subseteq E$, then E is called an extension of F , denoted by $F\le E$ or E / F. The dimension of the extension is called the degree of the extension and it is abbreviated by [E : F]. If $[E:F]=n<\infty ,$ E is said to be a finite extension of F.

Remark 1.4.2

It can be shown that, if f is a nonconstant polynomial over the field F, then there is an extension E of the field F and $\alpha \in E$ such that

$$f(\alpha )=0.$$

Hence, the field extension of the initial field F is often obtained from the field F by adjoining a root $\alpha$ of a nonconstant polynomial f over the field F; in this case the minimal field containing $\alpha$ and F is denoted by $F(\alpha ).$

Definition 1.4.3

A real number $\alpha$ is called algebraic over $\mathbb {Q}$ if it is a root of a polynomial equation with coefficients in $\mathbb {Q}$. A real number is said to be transcendental if it is not algebraic.

Definition 1.4.4

The minimal polynomial p of an algebraic number $\alpha$ over $\mathbb {Q}$ is the uniquely determined irreducible monic polynomial of minimal degree with rational coefficients satisfying

$$p(\alpha )=0.$$

Elements that are algebraic over $\mathbb {Q}$ and have the same minimal polynomial are called conjugates over $\mathbb {Q}$ .

Definition 1.4.5

Let $\alpha$ be an algebraic number. Then the degree of $\alpha$ is the degree of the minimal polynomial of $\alpha$ over $\mathbb {Q}$.

Remark 1.4.6

If

$$X^d+r_{d-1}X^{d-1}+\dots +r_1X+r_0$$

is the minimal polynomial of an algebraic number $\alpha$ over $\mathbb {Q},$ then multiplication by the least common multiple of the denominators of the coefficients $r_i, \ i=0, \dots , d-1,$ produces a unique polynomial P with $P(\alpha )=0$ having the form

$$P(X)=a_dX^d+a_{d-1}X^{d-1}+\dots +a_1X+a_0,$$

where the coefficients $a_i, \ i=0, \dots , d,$ are relatively prime integers and $a_d>0.$ We will call this polynomial the minimal polynomial of $\alpha$ over $\mathbb {Z}.$

Example 1.4.7

A real number is rational if and only if it is an algebraic number of degree 1.

Example 1.4.8

We shall show that

$$x=\frac{10^{\frac{2}{3}}-1}{\sqrt{-3}}$$

is an algebraic number.

To see that, we write $\sqrt{-3}x + 1 = 10^{2/3}$ and get $(\sqrt{-3}x+1)^3 = 100.$ Expanding the left-hand side, it follows that

$$3\sqrt{-3}x^3 + 9x^2 - 3\sqrt{-3}x + 99 = 0.$$

Dividing by $3\sqrt{-3}$, we get

$$x^3 - \sqrt{-3}x^2 - x -11\sqrt{-3}=0.$$

From this we deduce the minimal polynomial of x as

$$\begin{aligned}&P(x)=\Bigl ((x^3-x) - \sqrt{-3}(x^2+11)\Bigr )\Bigl ((x^3-x)+\sqrt{-3}(x^2+11)\Bigr )=\\&\qquad = x^6 + x^4 + 67x^2 + 363. \end{aligned}$$

1.4.2 Liouville’s Theorem

The main task in Diophantine approximation is to figure out how well a real number $\alpha$ can be approximated by rational numbers. In view of this problem, as we have mentioned earlier, a rational number $\frac{p}{q}$ is considered to be a “good” approximation of a real number $\alpha$ if the absolute value of the difference between $\frac{p}{q}$ and $\alpha$ may not decrease when $\frac{p}{q}$ is replaced by another rational number with a smaller denominator. This problem was solved during the 18th century by means of continued fractions.

Knowing the “best” approximation of a given number, the main problem is to find sharp upper and lower bounds of the mentioned difference, expressed as a function of the denominator.

It appears that these bounds depend on the nature of the real numbers to be approximated: the lower bound for the approximation of a rational number by another rational number is larger than the lower bound for algebraic numbers, which is itself larger than the lower bound for all real numbers. Thus, a real number that may be better approximated than an algebraic number is certainly a transcendental number. This statement had been proved by Liouville⁸ in 1844, and it produced the first explicit transcendental numbers. The later proofs on the transcendency of $\pi$ and e were obtained by a similar idea.

Theorem 1.4.9

(Liouville 1844) Let $\alpha$ be a real algebraic number of degree $d\ge 2$. Then there exists a constant $c(\alpha )>0$, depending only on $\alpha$, such that

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha )}{q^d}$$

for all rational numbers $\frac{p}{q}$ .

There are a few ways of proving Liouville’s Theorem 1.4.9: one can be found in [12]; a second, and maybe the most familiar one, uses the mean value theorem, an interesting variant concerning the constant $c(\alpha )$ is given in [45].

Corollary 1.4.10

Let $\alpha$ be a real algebraic number of degree $d\ge 2$. For every $\delta >0$, there are only finitely many rational numbers $\frac{p}{q}$ satisfying the inequality

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{d+\delta }}.$$

A proof can be found in [44].

We observe that the denominators of rational numbers which are getting closer and closer to a fixed real number $\alpha$ grow arbitrarily large. Liouville’s Theorem 1.4.9 proves that this is indeed the case for real algebraic numbers $\alpha$. For example, Liouville’s Theorem 1.4.9 implies that, if $\frac{p}{q}$ is within a distance $\frac{1}{10^{10}}$ of $\alpha$, then

$$q^d\ge 10^{10}c(\alpha ).$$

Remark 1.4.11

Consider the rational numbers $\frac{p}{q}$ and $\frac{p'}{q'}$ satisfying

$$q\left| \alpha -\frac{p}{q}\right| <q'\left| \alpha -\frac{p'}{q'}\right| .$$

The theory of continued fractions provides an efficient algorithm for constructing the best approximations to $\alpha$ in this case.

Liouville’s Theorem 1.4.9 can be used to find transcendental numbers explicitly. His construction of transcendental numbers (1844) predates Cantor’s⁹ proof (1874) of their existence.

Example 1.4.12

Let

$$\alpha =\sum _{n=1}^{\infty }\frac{1}{10^{n!}}=0.1100010000000000000000001000....$$

The digit 1 appears in the 1st, 2nd, 6th, 24th, ..., (n!)th,... decimal place. For $k\ge 1$, let $q(k)=10^{k!}$ and $p(k)=10^{k!}\sum _{n=1}^k\frac{1}{10^{n!}}.$ We notice that p(k) and q(k) are relatively prime integers, $\frac{p_k}{q_k}=\sum _{n=1}^{k}\frac{1}{10^{n!}}$ and

$$\left| \alpha -\frac{p(k)}{q(k)}\right| =\sum _{n=k+1}^{\infty }\frac{1}{10^{n!}}.$$

Comparing with a geometric series, we find

$$\sum _{n=k+1}^{\infty }\frac{1}{10^{n!}}<\frac{1}{10^{(k+1)!}}\sum _{n=0}^{\infty }\frac{1}{10^n}=\frac{10}{9}\frac{1}{q(k)^{k+1}},$$

and

$$\left| \alpha -\frac{p(k)}{q(k)}\right| <\frac{\frac{10}{9}}{q(k)^{k+1}}.$$

Finally, we observe that $\alpha$ does not satisfy Liouville’s Theorem 1.4.9. It follows from the calculations above that for any $c>0$ and any $d>0$, selecting k such that $\frac{10}{9}<c\cdot q(k)^{k+1-d},$ leads to

$$\left| \alpha -\frac{p(k)}{q(k)}\right| <\frac{c}{q(k)^d}$$

for large k. Thus, $\alpha$ must be a transcendental number.

In 1873, Charles Hermite¹⁰ proved that e is transcendental and nine years later Ferdinand von Lindemann¹¹ proved the transcendence of $\pi$. Hermite even showed that $e^a$ is transcendental when a is algebraic and nonzero. This approach was generalized by Weierstrass¹² to the Lindemann-Weierstrass theorem.

Theorem 1.4.13

(Hermite) e is transcendental .

A proof can be found in [4, 28].

Theorem 1.4.14

(Lindemann) $\pi$ is transcendental .

A proof can be found in [34].

Theorem 1.4.15

(Lindemann, Weierstrass) Let $\beta _1, \dots , \beta _n$ be nonzero algebraic numbers and $\alpha _1, \dots , \alpha _n$ distinct algebraic numbers. Then

$$\beta _1e^{\alpha _1}+\dots +\beta _ne^{\alpha _n}\ne 0.$$

A proof can be found in [4].

Corollary 1.4.16

If $\alpha$ is a nonzero algebraic integer, then

$$e^{\alpha }, \ \sin \alpha , \ \cos \alpha $$

are transcendental numbers.

Definition 1.4.17

A real number $\alpha$ is a Liouville number if for every positive integer n, there exist integers p, q with $q>1$ such that

$$0<\left| \alpha -\frac{p}{q}\right| <\frac{1}{q^n}.$$

Remark 1.4.18

It is known that $\pi$ and e are not Liouville numbers (see [37, 41], respectively). Mahler [38] found conditions on the expansion of a real number $\alpha$ in base g that imply that $\alpha$ is transcendental but not a Liouville number. One such example is the decimal number called Champernowne’s constant (Mahler’s number).

Theorem 1.4.19

All Liouville numbers are transcendental.

A proof can be found in [46].

Example 1.4.20

We show that

$$\alpha =\sum _{j=0}^{\infty }\frac{1}{2^{j!}}$$

is a Liouville number.

First, we observe that the binary expansion of $\alpha$ has arbitrarily long strings of 0’s, so it cannot be rational. Fix a positive integer n and consider $\frac{p}{q}=\sum _{j=0}^n\frac{1}{2^{j!}}$ with p and $q=2^{j!}>1$ integers. Then

$$0<\left| \alpha -\frac{p}{q}\right| =\sum _{j=n+1}^{\infty }\frac{1}{2^{j!}}<\sum _{j=(n+1)!}^{\infty }\frac{1}{2^{j!}}=\frac{1}{2^{(n+1)!-1}}\le \frac{1}{2^{n(n!)}}=\frac{1}{q^n},$$

which proves that $\alpha$ is indeed a Liouville number.

1.4.3 Roth’s Theorem

It is natural to ask for stronger versions of Liouville’s Theorem 1.4.9. First improvements were made by Thue,¹³ Siegel¹⁴ and Dyson.¹⁵ In 1955, K.F. Roth¹⁶ proved the most far-reaching extension, now known as the Thue-Siegel-Roth theorem, but also just as Roth’s theorem, for which he was awarded a Fields Medal in 1958. We quote from Roth’s paper [43].

Theorem 1.4.21

(Roth 1955) Let $\alpha$ be a real algebraic number of degree $d\ge 2$. Then, for every $\delta >0,$ the inequality

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{2+\delta }}$$

has only finitely many rational solutions $\frac{p}{q}$ .

A complete proof of Roth’s Theorem 1.4.21 can be found in [45], and for a generalized version we refer to [30].

Corollary 1.4.22

Let $\alpha$ be an algebraic number of degree $d\ge 2$. Then, for every $\delta >0,$ there is a constant $c(\alpha , \delta )>0$ such that

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha , \delta )}{q^{2+\delta }}$$

for all rational numbers $\frac{p}{q}$.

A proof can be found in [44]. Note that, if $\alpha$ is a real algebraic number of degree 2, then Liouville’s Theorem 1.4.9 is stronger than Roth’s Theorem 1.4.21. Whereas the latter one gives the estimate

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha , \delta )}{q^{2+\delta }}$$

for every $\delta >0$, Liouville’s Theorem 1.4.9 states that there is a constant $c(\alpha )>0$ such that

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha )}{q^2}$$

for all rational numbers $\frac{p}{q}$. The analogous result for real algebraic numbers $\alpha$ of degree $\ge$3, namely, that there is a constant $c(\alpha )>0$ such that an inequality of the form

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha )}{q^2}$$

holds for all rational numbers $\frac{p}{q},$ is conjectured to be false for all such $\alpha .$ However, this is not known at the present time to be false for a single real algebraic number.

Champernowne’s Constant

The transcendental numbers (except e and $\pi$) we have encountered so far are transcendental because their decimal (or dyadic) expansions have infinitely many runs of zeros whose lengths grow so quickly that the simple truncation of the decimal (or dyadic) expansion before each run of zeros leads to amazingly good approximations. However, for most numbers a collection of best rational approximations is not so easily detected from their decimal expansions. For most numbers even the best rational approximations are not close enough to allow us to conclude transcendency.

As an illustration of the difficulty of finding suitable rational approximations in general, we consider Champernowne’s constant (resp. Mahler’s¹⁷ number).

Definition 1.4.23

Champernowne’s constant (Mahler’s number)

$$\mathcal {M}=0.123456789101112131415161718192021\dots $$

is¹⁸ the number obtained by concatenating the positive integers in base 10 and interpreting them as decimal digits to the right of a the decimal point .

We mention a method worked out in detail in [11]. Firstly, rational approximations created by long runs of zeros are used, and $\mathcal {M}$ is truncated just after the 1 that appears whenever a power of 10 is reached. Following this truncation procedure, we see that the number of decimal digits before each run of zeros exceeds the length of that run by far. For example, to get a run of just one zero, $\mathcal {M}$ has to be truncated after 10 digits, i.e., 0.123456789. In general, if we want to come across a run of k zeros, we have to travel on the order of $k\cdot 10^k$ digits from the previous run of $k-1$ zeros. Thus, this truncation method cannot generate rational approximations having relatively small denominators that are sufficiently close to $\mathcal {M}$ in order to prove transcendence via Liouville’s Theorem 1.4.9.

Using a more clever construction to build rational approximations to $\mathcal {M}$, described in [11], one can find approximations all having relatively small denominators that allow to apply Liouville’s Theorem 1.4.9 to derive the following partial result.

Theorem 1.4.24

(Mahler 1937) The number

$$\mathcal {M}=0.123456789101112131415161718192021\dots $$

is either a transcendental number or an algebraic number of a degree at least 5.

A proof can be found in [11, 38]. Theorem 1.4.24 does not guarantee that $\mathcal {M}$ is transcendental, however, it does imply that $\mathcal {M}$ is neither a quadratic irrational, nor a cubic or even quartic algebraic number. A more advanced analysis building on Liouville’s Theorem 1.4.9 and Roth’s Theorem 1.4.21 would ensure that $\mathcal {M}$ is transcendental.

Theorem 1.4.25

(Mahler 1937) Champernowne’s constant $\mathcal {M}$ is a transcendental number.

A proof can be found in [11, 38].

1.4.4 Thue’s Theorem

Thue’s work was already a major breakthrough for those kind of questions:

Theorem 1.4.26

(Thue) Let $\alpha$ be a real algebraic number of degree d. Then, for every $\delta >0$, the inequality

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{\frac{1}{2}d+1+\delta }}$$

has only finitely many rational solutions $\frac{p}{q}$ .

A proof can be found in [44].

Corollary 1.4.27

Let $\alpha$ be a real algebraic number of degree d. Then, for every $\delta >0$, there is a constant $c(\alpha , \delta )>0$ such that

$$\left| \alpha -\frac{p}{q}\right| \ge \frac{c(\alpha , \delta )}{q^{\frac{1}{2}d+1+\delta }}$$

for all rational numbers $\frac{p}{q}.$

A proof can be found in [44].

The proof of Roth’s Theorem 1.4.21 is immensely more complex than those for the theorems of Liouville and Thue, even though, the framework is in essence the same. The proof of Roth’s Theorem 1.4.21 is not effective, that is, as noted in [30], for a given $\alpha$, the proof does not provide a method that guarantees to find the finitely many rational numbers $\frac{p}{q}$ satisfying

$$\left| \alpha -\frac{p}{q}\right| \le \frac{1}{q^{2+\delta }}.$$

In other words, the proof does not give a lower bound on $c(\alpha , \delta ).$ It is with respect to the work of Thue and Siegel that Roth’s Theorem 1.4.21 is often named the Thue-Siegel-Roth theorem.

We give an application of the Thue–Siegel–Roth theorem to Diophantine equations. It follows from the fact that the approximation exponent of an algebraic number $\alpha$ of degree $d\ge 3$ is strictly less than d. It appears that the full force of Roth’s Theorem 1.4.21 is not needed, Thue’s Theorem 1.4.26 is sufficient.

Theorem 1.4.28

Let

$$\begin{aligned} a_dX^d+a_{d-1}X^{d-1}+\dots +a_1X+a_0\in \mathbb {Z}[X] \end{aligned}$$

(1.9)

be an irreducible polynomial over $\mathbb {Q}$ of degree $d\ge 3.$ Then, for every nonzero integer m, the Diophantine equation

$$\begin{aligned} a_dX^d+a_{d-1}X^{d-1}Y+\dots +a_1XY^{d-1}+a_0Y^d=m \end{aligned}$$

(1.10)

has only finitely many integer solutions (p, q).

A proof can be found in [47]. An equation of the form (1.10) is called a Thue equation. It is interesting to note that equations of the form $X^2-dY^2=1,$ where d is a positive and a square free integer, the so called Pellian equations, have infinitely many integer solutions, but equations of the form $X^3-dY^3=1$ with an arbitrary integer d have at most finitely many integer solutions. The books [11, 46] provide nice introductions to various aspects of Diophantine approximation, transcendence theory and Diophantine equations.

2 Linear Forms in Logarithms

2.1 Introduction

Hilbert’s problems form a list of twenty-three major problems in mathematics collected, proposed and published by D. Hilbert¹⁹ in 1900. The problems were all unsolved at the time and several of them turned out to be very influential for 20th century mathematics. Hilbert believed that new machinery and methods were needed for solving these problems. He presented ten of them at the International Congress of Mathematicians in Paris in 1900. The complete list of his 23 problems was published later, most notably an English translation appeared 1902 in the Bulletin of the American Mathematical Society.

Hilbert’s seventh problem, entitled “irrationality and transcendence of certain numbers”, is dealing with the transcendence of the number

$$\alpha ^\beta $$

for algebraic $\alpha \ne 0,1$ and irrational algebraic $\beta$. He believed that the proof of this problem would only be given in more distant future than proofs of Riemann’s hypothesis or Fermat’s last theorem. Even though Hilbert was mistaken, he was correct when he expressed his belief that the proof of that problem would be extremely intriguing and influential for 20th century mathematics. The seventh problem was solved independently by Gelfond²⁰ and Schneider²¹ in 1935. They proved that, if $\alpha _1, \alpha _2\ne 0$ are algebraic numbers such that $\log \alpha _1, \log \alpha _2$ are linearly independent over $\mathbb {Q}$, then

$$\beta _1\log \alpha _1+\beta _2\log \alpha _2\ne 0$$

for all algebraic numbers $\beta _1, \beta _2$.

In 1935, Gelfond found a lower bound for the absolute value of the linear form

$$\varLambda =\beta _1\log \alpha _1+\beta _2\log \alpha _2\ne 0.$$

He proved that

$$\log |\varLambda |\gg -h(\varLambda )^\kappa ,$$

where $h(\varLambda )$ is the logarithmic height of the linear form $\varLambda$, $\kappa >5$ and $\gg$ is Vinogradov’s notation for an inequality that is valid up to an unspecified constant factor. Gelfond also noticed that generalization of his results would lead to a powerful new analytic method by which mathematicians could prove a huge amount of unsolved problems in number theory.

2.2 Basic Theorems and Definitions

In 1966 in 1967, A. Baker²² gave in his papers “Linear forms in logarithms of algebraic numbers I, II, III”, [1,2,3] an effective lower bound on the absolute value of a nonzero linear form in logarithms of algebraic numbers, that is, for a nonzero expression of the form

$$\sum _{i=1}^{n}b_i\log \alpha _i,$$

where $\alpha _1, \dots , \alpha _n$ are algebraic numbers and $b_1, \dots , b_n$ are integers. This result initiated the era of effective resolution of Diophantine equations that can be reduced to exponential ones (where the unknown variables are in the exponents). The generalization of the Gelfond-Schneider theorem was only the beginning of a new and very interesting branch in number theory called Baker’s theory.

Definition 2.2.1

Let $\alpha _1, \alpha _2, \dots , \alpha _n$ be n (real or complex) numbers. We call $\alpha _1, \alpha _2, \dots , \alpha _n$ linearly dependent over the rationals (equivalently integers) if there are rational numbers (integer numbers) $r_1, r_2, \dots , r_n$, not all zero, such that

$$r_1\alpha _1+r_2\alpha _2+\dots +r_n\alpha _n=0.$$

If $\alpha _1, \alpha _2, \dots , \alpha _n$ are not linearly dependent over the rationals (integers), they are linearly independent over the rationals (integers).

Definition 2.2.2

A linear form in logarithms of algebraic numbers is an expression of the form

$$\varLambda =\beta _0+\beta _1\log \alpha _1+\beta _2\log \alpha _2+\dots +\beta _n\log \alpha _n,$$

where $\alpha _i, \ i=1, \dots , n$ and $\beta _i, \ i=0, \dots , n$ are complex algebraic numbers and $\log$ denotes any determination of the logarithm.

Remark 2.2.3

We are interested in the degenerate case when $\beta _0=0$ and $\beta _i\in \mathbb {Z}, i=1, \dots , n.$ In the sequel we write $\beta _i=b_i, \ i=1,\dots , n$ and $\log$ always represents the principal value of the complex logarithm.

A generalization of the Gelfond-Schneider theorem to arbitrarily many logarithms was obtained by Baker in 1966 [1]. In 1970, he was awarded the Fields Medal for his work in number theory, especially in the areas of transcendence and Diophantine geometry. One of his major contributions is the following

Theorem 2.2.4

(Baker 1966) If $\alpha _1, \alpha _2, \dots , \alpha _n\ne 0,1$ are algebraic numbers such that $\log \alpha _1, \log \alpha _2, \dots , \log \alpha _n, 2\pi i$ are linearly independent over the rationals, then

$$\beta _0+\beta _1\log \alpha _1+\dots +\beta _n\log \alpha _n\ne 0$$

for any algebraic numbers $\beta _0, \beta _1, \dots , \beta _n$ that are not all zero.

A proof can be found in [1].

Remark 2.2.5

Linear independence over the rationals implies linear independence over the algebraic numbers (see [4]).

Theorem 2.2.6

(Baker 1967) The number $e^{\beta _0}\alpha _1^{\beta _1}\dots \alpha _n^{\beta _n}$ is transcendental for all nonzero algebraic numbers $\alpha _i, \ i=1, \dots , n$ and $\beta _i, \ i=0, \dots , n$. Furthermore, the number $\alpha _1^{\beta _1}\dots \alpha _n^{\beta _n}$ is transcendental if $1, \beta _1, \dots , \beta _n$ are linearly independent over the rationals.

A proof can be found in [3].

Definition 2.2.7

The height H of a rational number $\frac{p}{q}$ is defined by

$$H\left( \frac{p}{q}\right) =\max \{|p|, |q|\}.$$

Definition 2.2.8

Let $\mathbb {L}$ be a number field of the degree D, $\alpha \in \mathbb {L}$ an algebraic number of degree $d\mid D$ and let $\sum _{0\le k\le d}a_kX^k$ be its minimal polynomial in $\mathbb {Z}[X]$ with $a_d\ne 0$. We define the absolute logarithmic height $h(\alpha )$ by

$$\begin{aligned} h(\alpha )=\frac{1}{d}\left( \log (|a_d|)+\sum _{1\le i\le d}\max \{\log (|\alpha _i|,0)\}\right) , \end{aligned}$$

(2.1)

where $\alpha _i$ are the conjugates of $\alpha .$

Example 2.2.9

The absolute logarithmic height h of a rational number $\frac{p}{q}$ is

$$h\left( \frac{p}{q}\right) =\log \max \{|p|, |q|\}.$$

Example 2.2.10

Let $\alpha =\sqrt{2}$. The absolute logarithmic height of $\alpha$ is

$$h(\alpha )=h(\sqrt{2})=\frac{1}{2}\left( \log |\sqrt{2}|+\log |-\sqrt{2}|\right) =\frac{1}{2}\log 2.$$

Example 2.2.11

Let

$$\alpha _1=\frac{1}{\sqrt{3}+\sqrt{5}}.$$

Before calculating the absolute logarithmic height of $\alpha _1$, we compute the degree of the field extension $\mathbb {Q}\left( \frac{1}{\sqrt{3}+\sqrt{5}}\right)$ over $\mathbb {Q}$ as

$$\left[ \mathbb {Q}\left( \frac{1}{\sqrt{3}+\sqrt{5}}\right) :\mathbb {Q}\right] =4,$$

so the degree of the algebraic number $\alpha _1=\frac{1}{\sqrt{3}+\sqrt{5}}$ over $\mathbb {Q}$ is 4.

The minimal polynomial of $\alpha _1$ is given by

$$\begin{aligned} P_{\alpha _1}(x)= & {} \left( x-\frac{1}{\sqrt{3}+\sqrt{5}}\right) \left( x-\frac{1}{-\sqrt{3}-\sqrt{5}}\right) \times \\&\times \left( x-\frac{1}{-\sqrt{3}+\sqrt{5}}\right) \left( x-\frac{1}{\sqrt{3}-\sqrt{5}}\right) \\= & {} x^4-4x^2+\frac{1}{4}, \end{aligned}$$

over the rationals, and

$$P_{\alpha _1}(x)=4x^4-16x^2+1$$

over the integers. The conjugates of $\alpha _1$ are

$$\alpha _1=\frac{1}{\sqrt{3}+\sqrt{5}}, \ \alpha _2=\frac{1}{-\sqrt{3}+\sqrt{5}}, \ \alpha _3=\frac{1}{\sqrt{3}-\sqrt{5}}, \ \alpha _4=\frac{1}{-\sqrt{3}-\sqrt{5}}.$$

Hence, using (2.1), the absolute logarithmic height of $\alpha _1$ is equal to

$$\begin{aligned}&h(\alpha _1)=h\left( \frac{1}{\sqrt{3}+\sqrt{5}}\right) \\&\,=\frac{1}{4}\bigg (\log |4|+\max \bigg \{\log \left| \frac{1}{\sqrt{3}+\sqrt{5}}\right| ,0\bigg \}+\max \bigg \{\log \left| \frac{1}{-\sqrt{3}+\sqrt{5}}\right| ,0\bigg \}+\\&\qquad +\max \bigg \{\log \left| \frac{1}{\sqrt{3}-\sqrt{5}}\right| ,0\bigg \}+\max \bigg \{\log \left| \frac{1}{-\sqrt{3}-\sqrt{5}}\right| ,0\bigg \}\bigg )\\&\,=0.689146. \end{aligned}$$

Let $\mathbb {L}$ be a number field of degree D, $\alpha _1, \alpha _2, \dots , \alpha _n$ nonzero elements of $\mathbb {L}$ and let $b_1, b_2, \dots , b_n$ be integers. We define

$$B=\max \{|b_1|, |b_2|, \dots , |b_n|\}$$

and

$$\varLambda ^*=\alpha _1^{b_1}\alpha _2^{b_2}\dots \alpha _n^{b_n}-1.$$

We wish to bound $|\varLambda ^*|$ from below, assuming that it is nonzero.

Since $\log (1+x)$ is asymptotically equal to x as |x| tends to 0, our problem consists of finding a lower bound for the linear form in logarithms

$$\varLambda =b_1\log \alpha _1+\dots +b_n\log \alpha _n+b_{n+1}\log (-1),$$

where $b_{n+1}=0$ if $\mathbb {L}$ is real and $|b_{n+1}|\le nB$, otherwise.

Definition 2.2.12

Let $A_1, A_2, \dots , A_n$ be real numbers such that

$$A_j\ge h'(\alpha _j):=\max \{Dh(\alpha _j),\ |\log \alpha _j|, \ 0.16\}, 1\le j\le n.$$

Then $h'$ is called the modified height with respect to the field $\mathbb {L}$.

A. Baker, E.M. Matveev²³ and G. Wüstholz²⁴ proved the following theorems.

Theorem 2.2.13

(Baker–Wüstholz 1993) Assume that

$$\varLambda =b_1\log \alpha _1+\dots + b_n\log \alpha _n\ne 0$$

for algebraic $\alpha _i$ and integers $b_i$, $i=1, \dots , n$. Then

$$\log |\varLambda |\ge -18(n+1)!n^{n+1}(32D)^{n+2}\log (2nD)h''(\alpha _1)\dots h''(\alpha _n)\log B,$$

where D is the degree of the extension $\mathbb {Q}(\alpha _1, \dots , \alpha _n)$, $B=\max \{|b_i|, \ i=1, \dots , n\}$ and $h''(\alpha )=\max \{h(\alpha ),\ \frac{1}{D}|\log (\alpha )|,\ \frac{1}{D}\}.$

A proof can be found in [6].

Theorem 2.2.14

(Matveev 2001) Assume that $\varLambda ^*$ is nonzero. Then

$$|\varLambda ^*|>-3\cdot 30^{n+4}(n+1)^{5.5}D^2A_1\dots A_n(1+\log D)(1+\log nB).$$

If $\mathbb {L}$ is real, then

$$\log |\varLambda ^*|>-1.4\cdot 30^{n+3}(n+1)^{4.5}D^2A_1\dots A_n(1+\log D)(1+\log B).$$

Theorem 2.2.15

(Matveev 2001) Assume that

$$\varLambda = b_1\log \alpha _1+\dots +b_n\log \alpha _n\ne 0$$

for algebraic $\alpha _i$ and integers $b_i$, $i=1, \dots , n$. Then

$$\log |\varLambda |>-2\cdot 30^{n+4}(n+1)^6D^2A_1\dots A_n(1+\log D)(1+\log B),$$

where $B=\max \{|b_i|, \ 1\le i\le n\}.$

Proofs for Matveev’s theorems can be found in [39].

2.3 A Variation of Baker–Davenport Lemma

For a real number x we introduce the notation

$$||x||=\min \{|x-n| : n\in \mathbb {Z}\}$$

for the distance from x to the nearest integer.

The following result is a variation of a lemma of Baker and Davenport²⁵ [5], it is due to Dujella²⁶ and Pethő,²⁷ [22].

Lemma 2.3.1

Let N be a positive integer, $\frac{p}{q}$ a convergent of the continued fraction expansion of an irrational number $\kappa$ such that $q>6N$ and let $\mu$ be some real number. Let $\varepsilon =||\mu q||-N||\kappa q||.$ If $\varepsilon >0$, then there is no solution to the inequality

$$0<m\kappa -n+\mu <AB^{-m}$$

in positive integers m and n with

$$\frac{\log (Aq/\varepsilon )}{\log B}\le m\le N.$$

Proof

Suppose that $0\le m\le N.$ Then

$$m(\kappa q-p)+mp-nq+\mu q<qAB^{-m}.$$

Thus,

$$qAB^{-m}>|\mu q-(nq-mp)|-m||\kappa q||\ge ||\mu q||-N||\kappa q||:=\varepsilon ,$$

from where we deduce that

$$m<\frac{\log (Aq/\varepsilon )}{\log B}.$$

Remark 2.3.2

The method from Lemma 2.3.1 is called Baker–Davenport reduction.

Example 2.3.3

Find all nonnegative integers that satisfy

$$\begin{aligned} 0<\left| x_1\log 2-x_2\log 3+\log 5\right| <40e^{-X}, \end{aligned}$$

(2.2)

where $X=\max \{x_1, x_2\}\le 10^{30}$.

Let

$$x_1\log 2-x_2\log 3+\log 5>0.$$

First, we divide (2.2) by $\log 3$, in order to get inequalities of the form as in Lemma 2.3.1. We get

$$0<x_1\frac{\log 2}{\log 3}-x_2+\frac{\log 5}{\log 3}<\frac{40}{\log 3}e^{-X}.$$

Now, we define

$$\kappa =\frac{\log 2}{\log 3}, \mu =\frac{\log 5}{\log 3}, A=\frac{40}{\log 3}, B=e.$$

We observe that the inequalities $A>0, B>1$ are satisfied.

We shall try to find a convergent $\frac{p}{q}$ of the continued fraction expansion of $\kappa$ that satisfies the condition $q>6N$. Since $\kappa$ does not have a finite or periodic continued fraction expansion, we give only the first 25 terms of its continued fraction expansion:

$$\kappa =[0, 1, 1, 1, 2, 2, 3, 1, 5, 2, 23, 2, 2, 1, 1, 55, 1, 4, 3, 1, 1, 15, 1, 9, 2,\dots ].$$

The first convergent $\frac{p}{q}$ that satisfies the inequality $q>6N$ is

$$\frac{p}{q}=\frac{35270892459770675836042178475339}{55903041915705101922536695520222}.$$

We therefore obtain

$$||\kappa q||N\approx 0.007651391,\qquad ||\mu q||\approx 0.466714899.$$

Hence,

$$\varepsilon =||\mu q||-||\kappa q||N=0.4590635078>0.$$

The given inequality does not have any solution in integers m such that

$$\frac{\ln \left( \frac{40}{\log 3}\cdot q\cdot \frac{1}{0.4590635}\right) }{\ln e}\le m\le N.$$

We observe that $m\le 77.4746$, so $N=77.$ Repeating Baker–Davenport reduction one more time, but now with $N=77$, we find convergents $\frac{p}{q}$ of the continued fraction expansion of $\kappa$ that satisfies the condition $>6N, N=77$. The first such convergent is

$$\frac{p}{q}=\frac{306}{485}.$$

We get

$$||\kappa q||N\approx 0.071647126,\qquad ||\mu q||\approx 0.487842451.$$

Finally, we obtain

$$\varepsilon \approx 0.4161953257.$$

After applying Baker–Davenport reduction, we get $m\le 10.6556$, resp. $N=10$. Therefore, we can find pairs $(x_1, x_2)$ satisfying the introduced inequalities. Using a simple computer algorithm, it turns out that the following pairs

$$(0,0), (0,1), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2), (3,0), (3,1), (3,2), (3,3), (4, 3)$$

satisfy (2.2).

For

$$x_1\log 2-x_2\log 3+\log 5<0,$$

we have

$$0<x_2\frac{\log 3}{\log 2}-x_1-\frac{\log 5}{\log 2}<\frac{40}{\log 2}e^{-X}.$$

We define

$$\kappa =\frac{\log 3}{\log 2}, \ \ \mu =-\frac{\log 5}{\log 2}, \ \ A=\frac{40}{\log 2}, \ \ B=e.$$

Then the inequalities $A>0,\ B>1$ are satisfied. The next step is to find a convergent $\frac{p}{q}$ of the continued fraction expansion of $\kappa$ satisfying $q>6N.$ For

$$\frac{p}{q}=\frac{1522076097743333607781100045449522888}{ 960323097266207036440783078900790949}$$

we get

$$\varepsilon =0.124406274>0.$$

Analogously, we find $N=89$ after the first reduction and $N=13$ after the second one. Finally, we get that the following pairs

$$(0,2), (0,3), (1,3), (2,3), (2,4), (3,4), (4,4), (5,5).$$

that satisfy (2.2), as well.

2.4 Applications

For the applications of linear forms in logarithms to Diophantine equations, the strategy is as follows: first, we use various algebraic manipulations to associate “relatively big” solutions of the equations to a “very small” value of the specific linear form in logarithms which implies that we are able to find upper bound for values of the linear form in the logarithms that corresponds to a solution of the equation. If we compare that upper bound with the lower bound (using the Baker–Wüstholz Theorem 2.2.13 or Matveev’s Theorems 2.2.14 and 2.2.15), we get an absolute upper bound M for the absolute values of the unknowns of the equations.

It often happens that the upper bound M is not too large and using various methods, including reductions and sieves, we can find the complete set of solutions below M. In order to realize this, it is crucial to get a reasonably small value for M. Its size is directly related to the size of the “numerical constant” that appears in Matveev’s Theorem 2.2.14 which is $1.4\cdot 30^{n+3}n^{4.5}.$ Many celebrated Diophantine equations lead to estimates of linear forms in two or three logarithms and in these cases Matveev’s Theorem 2.2.14 gives numerical constants around $10^{12}$ and $10^{14},$ respectively.

2.4.1 A Lower Bound for $|2^m-3^n|$

One of the simplest applications of linear forms in logarithms is to prove that $|2^m-3^n|$ tends to infinity with $m+n$; in addition one can even get an explicit lower bound for this quantity. The following material is presented in detail in [14].

Let $n\ge 2$ be an integer and m and $m'$ are defined by the conditions

$$2^{m'}<3^n<2^{m'+1}, |3^n-2^m|=\min \{3^n-2^{m'}, 2^{m'+1}-3^n\}.$$

Then

$$|2^m-3^n|<2^m, (m-1)\log 2<n\log 3<(m+1)\log 2,$$

and the problem of finding a lower bound for $|2^m-3^n|$ clearly reduces to this special case.

Consider the linear form

$$\varLambda =3^n2^{-m}-1.$$

Applying Matveev’s Theorem 2.2.14, we get

$$\log |\varLambda |>-c_0(1+\log m).$$

It is easy to verify that we can take $c_0=5.87\cdot 10^8$. Hence, the following theorem is proved.

Theorem 2.4.1

Let m, n be positive integers. Then

$$|2^m-3^n|>2^m(em)^{-5.87\cdot 10^8}$$

This theorem enables us to find the list of all powers of 3 that increased by 5 give a power of 2.

Corollary 2.4.2

The only integer solutions to the Diophantine equation

$$2^m-3^n=5$$

are $(m, n)=(3,1), (5, 3).$

Applying Theorem 2.4.1, we get

$$5>2^m(em)^{-5.87\cdot 10^8},$$

which implies

$$\log 5>m\log 2-5.87\cdot 10^8(1+\log m),$$

so that $m<2.1\cdot 10^{10}$ and $n<m\frac{\log 2}{\log 3}<1.4\cdot 10^{10}$. Moreover, the equality $2^m-3^n=5$ implies

$$\left| m-n\frac{\log 3}{\log 2}\right| <\frac{5}{\log 2}3^{-n}.$$

Since

$$\frac{5}{\log 2}\cdot 3^{-n}<\frac{1}{2n}$$

for $n\ge 4,$ we observe that, if (m, n) is a solution to our problem with $n\ge 4,$ then $\frac{m}{n}$ is a convergent of the continued fraction expansion of $\xi =\frac{\log 3}{\log 2}$. Also, for $n<N=1.4\cdot 10^{10}$ the smallest value of $|m-n\xi |$ is obtained for the largest convergent of the continued fraction expansion of $\xi$ with the denominator less then N. We thus get

$$\frac{5}{\log 2}\cdot 3^{-n}>\left| m-n\frac{\log 3}{\log 2}\right| >10^{-11}, 0<n<1.4\cdot 10^{10}.$$

Hence, $n\le 24$. Now, it is very easy to prove the initial statement.

More generally, the following result can be obtained.

Theorem 2.4.3

(Bennett, [7]) For given nonzero integers a, b, c the equation

$$a^m-b^n=c$$

has at most two integer solutions.

2.4.2 Rep-Digit of Fibonacci Numbers

The Fibonacci sequence $(F_n)_{n\ge 0}$ is given by

$$F_0=0, \ F_1=1, \ \dots , \ F_{n+2}=F_{n+1}+F_n, \ n\ge 0.$$

Its characteristic equation is

$$f(X)=X^2-X-1=(X-\alpha )(X-\beta ),$$

where $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}.$ We can also write

$$F_n=\frac{\alpha ^n-\beta ^n}{\alpha -\beta }, \ n\ge 0.$$

In this subsection we are concerned with those Fibonacci numbers $F_n$ that have equal digits in base 10. Putting d for the repeated digit and assuming that $F_n$ has m digits, the problem reduces to finding all solutions of the Diophantine equation

$$\begin{aligned} F_n=\overline{dd\dots d}_{(10)}=d10^{m-1}+d10^{m-1}+\dots +d=d\frac{10^m-1}{10-1}, d\in \{1,2,\dots 9\}.\end{aligned}$$

(2.3)

Theorem 2.4.4

The largest solution of Eq. (2.3) is $F_{10}=55$.

Proof

Suppose that $n>1000$. We start by proving something weaker. Our goal is to obtain some bound on n. We rewrite Eq. (2.3) as

$$\frac{\alpha ^n-\beta ^n}{\sqrt{5}}=d\frac{10^m-1}{9}.$$

Next we separate large and small terms on both sides of the equation. It is easy to obtain $\alpha =\frac{-1}{\beta }$ or $\beta =\frac{-1}{\alpha }$ which implies

$$\begin{aligned} {} \left| \alpha ^n-\frac{d\sqrt{5}}{9}10^m\right| =\left| \beta ^n-\frac{d\sqrt{5}}{9}\right| \le |\beta ^n|+\left| \frac{d\sqrt{5}}{9}\right| \le \alpha ^{-1000}+\sqrt{5}<2.5.\quad \end{aligned}$$

(2.4)

Our goal is to get some estimates for m in terms of n. By induction on n, it is easy to prove that

$$\alpha ^{n-2}<F_n<\alpha ^{n-1}, n\ge 3.$$

Thus,

$$\alpha ^{n-2}<F_n<10^m \,\text {or}\,n<m\frac{\log 10}{\log \alpha }+2,$$

and

$$10^{m-1}<F_n<\alpha ^{n-1}.$$

On the other hand,

$$n>\frac{\log 10}{\log \alpha }(m-1)+1=\frac{\log 10}{\log \alpha }m-\left( \frac{\log 10}{\log \alpha }-1\right) >\frac{\log 10}{\log \alpha }m-4.$$

We deduce that

$$n\in [c_1m-4, \ c_1m+2],\qquad c_1=\frac{\log 10}{\log \alpha }=4.78497..$$

Since $c_1>4$, for $n>1000,$ we have $n\ge m$. Hence,

$$|\varLambda |=\left| \frac{d\sqrt{5}}{9}\alpha ^{-n}10^m-1\right|<\frac{2.5}{\alpha ^n}<\frac{1}{\alpha ^{n-2}},$$

which leads to

$$\log |\varLambda |=\log \frac{d\sqrt{5}}{9}-n\log \alpha +m\log 10<-(n-2)\log \alpha .$$

Let

$$\alpha _1=\frac{d\sqrt{5}}{9}, \alpha _2=\alpha , \alpha _3=10, b_1=1, b_2=-n, b_3=m,$$

as well as $\mathbb {L}=\mathbb {Q}(\alpha _1, \alpha _2, \alpha _3)=\mathbb {Q}(\sqrt{5}),$ so $D=2$ and $B=n$. The minimal polynomial of $\alpha _1$ over $\mathbb {Z}$ is

$$P_{\alpha _1}(X)=81X^2-5d^2.$$

Hence,

$$h(\alpha _1)<\frac{1}{2}\left( \log 81+2\log \sqrt{5}\right) =\frac{1}{2}\log 405<3.01,$$

and

$$h(\alpha _2)=\frac{1}{2}(\log \alpha +1)<0.75, h(\alpha _3)=\log 10<2.31.$$

We may take

$$A_1=6.02,A_2=1.5, A_3=4.62.$$

Then Matveev’s Theorem 2.2.14 gives us a lower bound for $\varLambda$, namely

$$\log \varLambda >-1.4\cdot 30^6\cdot 3^{4.5}\cdot 4(1+\log 4)6.02\cdot 1.5\cdot 4.62(1+\log n).$$

Comparing the above inequality with

$$\log \varLambda <-(n-2)\log \alpha ,$$

we get

$$(n-2)\log \alpha <1.41\cdot 30^6\cdot 3^{4.5}\cdot 4(1+\log 4)6.02\cdot 1.5\cdot 4.62(1+\log n),$$

and

$$n<4.5\cdot 10^{15}.$$

Reducing the bound. Observe that the right-hand side of the inequality

$$1-\frac{d\sqrt{5}}{9}\alpha ^{-n}10^m\le \frac{1}{\alpha ^n}\left( \beta ^n-\frac{d\sqrt{5}}{9}\right) $$

is negative. Writing

$$z=\log \alpha _1-n\log \alpha _2+m\log \alpha _3,$$

we get that

$$-\frac{2.5}{\alpha ^n}<1-e^z<0.$$

In particular, $z>0$. Furthermore, we have $e^z<1.5$ for $n>1000$. Thus,

$$0<e^z-1<\frac{2.5e^z}{\alpha ^n}<\frac{4}{\alpha ^n}.$$

Since $e^z-1>z,$ we get

$$0<m\left( \frac{\log \alpha _3}{\log \alpha _2}\right) -n+\left( \frac{\log \alpha _1}{\log \alpha _2}\right)<\frac{4}{\alpha ^n\log \alpha _2}<\frac{9}{\alpha ^n}.$$

We notice that

$$\left( \frac{10^md\sqrt{5}}{\alpha ^n}\right) <2$$

and therefore

$$\alpha ^n>\frac{10^md\sqrt{5}}{2}>10^m.$$

Hence,

$$0<m\left( \frac{\log \alpha _3}{\log \alpha _2}\right) -n+\left( \frac{\log \alpha _1}{\log \alpha _2}\right) <\frac{9}{10^m}.$$

Since $n<4.5\cdot 10^{15},$ the previous inequality implies $m<9.5\cdot 10^{14}$. With

$$\kappa =\frac{\log \alpha _3}{\log \alpha _2}, \mu =\frac{\log \alpha _1}{\log \alpha _2}, A=9, B=10$$

we get

$$0<\kappa m-n+\mu <\frac{A}{B^m},$$

where $m<N:=10^{15}$. Observe that

$$\frac{p_{35}}{q_{35}}=C_{35}=\frac{970939497358931987}{202914354378543655}$$

and $q_{35}>202914354378543655>2\cdot 10^{17}>6N.$

For each one of the values of $d\in \{1,\dots ,9\},$ we compute $||q_{35}\mu ||$. The minimal value of this expression is obtained when $d=5$ and is

$$0.029...>0.02.$$

Thus, we can take $\varepsilon =0.01<0.02-0.01<||q_{35}\mu ||-N||q_{35}\kappa ||$. Since

$$\frac{\log (Aq_{35}/\varepsilon )}{\log B}=21.2313...,$$

we observe that there is no solution in the range $m\in \left[ 22, 10^{15} \right]$. Thus, $m\le 21,$ and $n\le 102.$ However, we have assumed that $n>1000$. To finish, we compute the values of all Fibonacci numbers modulo 10000 (their last four digits) and convince ourselves that there are no Fibonacci numbers with the desired pattern in the range $11\le n\le 1000.$ This example stems from [25, 36].

2.4.3 Simultaneous Pellian Equations

The following result is due to Baker and Davenport and was historically the first example of a successful use of lower bounds for linear forms in logarithms of algebraic numbers; it actually allowed the effective computation of all common members of two binary recurrent sequences with real roots; for more details see [19, 25, 36].

Theorem 2.4.5

The only positive integer d such that $d+1, \ 3d+1, \ 8d+1$ are all perfect squares is $d=120$.

Proof

If $d+1, \ 3d+1, \ 8d+1$ are all perfect squares, then we write

$$d+1=x^2, 3d+1=y^2, 8d+1=z^2.$$

Eliminating d from the above equations, we get

$$3x^2-y^2=2, 8x^2-z^2=7,$$

which is a system of simultaneous Pellian equations since it consists of two Pellian equations with a component in common. If we want x to be positive, the solutions of the above system are

$$\begin{aligned} y+x\sqrt{3}= & {} (1+\sqrt{3})(2+\sqrt{3})^m,\\ z+x\sqrt{8}= & {} (\pm 1+\sqrt{8})(3+\sqrt{8})^n, \end{aligned}$$

where m, n are nonnegative integers. Let the sequence $(v_m)$ be given by the recursion formula

$$v_0=1, v_1=3, v_{m+2}=4v_{m+1}-v_m,$$

and put $x=w_n^{+,-}$, for some $n\ge 0,$ where the sequences $(w_n^+), (w_n^-)$ are defined by

$$w_0^+=1, w_1^+=4, w_{n+2}^+=6w_{n+1}^+-w_n^+, \ n\in \mathbb {N},$$

$$w_0^-=1, w_1^-=2, w_{n+2}^-=6w_{n+1}^--w_n^-, \ n\in \mathbb {N}.$$

We want to solve the equation

$$v_m=w_n^{+,-}.$$

For this aim we shall use the following lemmas.

Lemma 2.4.6

If $v_m=w_n^{+,-}$, $m,n>2$, then

$$0<|\varLambda |<7.3(2+\sqrt{3})^{-2m},$$

where $\varLambda$ is

$$\varLambda =m\log (2+\sqrt{3})-n\log (3+2\sqrt{2})+\log \frac{2\sqrt{2}(1+\sqrt{3})}{\sqrt{3}(2\sqrt{2}\pm 1)}.$$

Proof

The expression $v_m=w_n^{+,-}$ implies

$$\begin{aligned}&\frac{(1+\sqrt{3})(2+\sqrt{3})^m-(1-\sqrt{3})(2-\sqrt{3})^m}{2\sqrt{3}}\nonumber \\&=\frac{(2\sqrt{2}\pm 1)(3+2\sqrt{2})^n+(2\sqrt{2}\mp 1)(3-2\sqrt{2})^n}{4\sqrt{2}}. \end{aligned}$$

(2.5)

Obviously,

$$v_m>\frac{(1+\sqrt{3})(2+\sqrt{3})^m}{2\sqrt{3}},$$

$$w_{n}^{+,-}<\frac{(2\sqrt{2}+1)(3+2\sqrt{2})^n}{2\sqrt{2}},$$

hence

$$\frac{(1+\sqrt{3})(2+\sqrt{3})^m}{2\sqrt{3}}<\frac{(2\sqrt{2}+1)(3+2\sqrt{2})^n}{2\sqrt{2}},$$

$$(3-2\sqrt{2})^n<\frac{\sqrt{3}(2\sqrt{2}+1)}{\sqrt{2}(\sqrt{3}+1)}(2-\sqrt{3})^m<1.7163(2-\sqrt{3})^m.$$

Dividing (2.5) by $\frac{2\sqrt{2}\pm 1}{4\sqrt{2}}(3+2\sqrt{2})^n$, we obtain

which proves Lemma 2.4.6.

Lemma 2.4.7

Let $a\in \mathbb {R}\backslash \{0\}, a>1$. If $|x|<a$, then

$$\begin{aligned} |\log (1+x)|<\frac{-\log (1-a)}{a}|x|. \end{aligned}$$

(2.6)

Proof

We observe that the function

$$\frac{\log (1+x)}{x}$$

is positive and strictly decreasing for $|x|<1$. Consequently, for $|x|<a,$ Inequality (2.6) holds for $x=-a.$

We shall investigate the following linear form in three logarithms:

$$\varLambda =m\log (2+\sqrt{3})-n\log (2\sqrt{2}+3)+\log \frac{2\sqrt{2}(\sqrt{3}+1)}{\sqrt{3}(2\sqrt{2}\pm 1)}.$$

Let

$$\alpha _1=2+\sqrt{3}, \alpha _2=3+2\sqrt{2}, \alpha _3=\frac{2\sqrt{2}(1+\sqrt{3})}{\sqrt{3}(2\sqrt{2}\pm 1)},$$

$$b_1=m, b_2=-n, b_3=1, D=[\mathbb {Q}(\alpha _1, \alpha _2, \alpha _3):\mathbb {Q}]=4.$$

The minimal polynomials over $\mathbb {Z}$ are

$$\begin{aligned} P_{\alpha _1}(x)= & {} x^2-4x+1,\\ P_{\alpha _2}(x)= & {} x^2-6x+1,\\ P_{\alpha _3}(x)= & {} 441x^4-2016x^3+2880x^2-1536x+256, \end{aligned}$$

hence

$$\begin{aligned} h''(\alpha _1)= & {} \frac{1}{2}\log (2+\sqrt{3})<0.6585,\\ h''(\alpha _2)= & {} \frac{1}{2}\log (3+2\sqrt{2})<0.8814, \end{aligned}$$

and

$$h''(\alpha _3)=\frac{1}{4}\log \left( 441\frac{2(4+\sqrt{2})(3+\sqrt{3})}{21}\frac{2(4-\sqrt{2})(3+\sqrt{3})}{21}\right) <1.7836.$$

Applying the Baker–Wüstholz Theorem 2.2.13, we get a lower bound for $\varLambda$, namely

$$\log |\varLambda |\ge -3.96\cdot 10^{15}\log m.$$

According to Lemma 2.4.6, we may conclude that

$$m<6\cdot 10^{16}.$$

This upper bound is rather big so we reduce it using Baker–Davenport reduction. Applying Lemma 2.4.6, we get the upper bound

$$\varLambda =m\log (2+\sqrt{3})-n\log (2\sqrt{2}+3)+\log \frac{2\sqrt{2}(\sqrt{3}+1)}{\sqrt{3}(2\sqrt{2}\pm 1)}<7.29(2+\sqrt{3})^{-2m}.$$

Let

$$N=6\cdot 10^{16}, \kappa =\frac{\log \alpha _1}{\log \alpha _2}, \mu =\frac{\log \alpha _3}{\log \alpha _2}, A=\frac{7.3}{\log \alpha _2}, B=(2+\sqrt{3})^2.$$

The next step is to find a convergent $\frac{p}{q}$ of the continued fraction expansion of $\kappa =\frac{\log \alpha _1}{\log \alpha _2}$ such that $q>6N$. The first such convergent is

$$\frac{p}{q}=\frac{742265900639684111}{993522360732597120}.$$

Before calculating $\varepsilon$, we observe that

$$||\kappa q||N\approx 0.0187822,\qquad ||\mu q||\approx 0.00762577.$$

Unfortunately,

$$\varepsilon =||\mu q||-||\kappa q||N<0.$$

If we want to use Baker–Davenport reduction, we have to find another convergent $\frac{p}{q}$ of the continued fraction expansion for which the condition $\varepsilon >0$ is satisfied.

The next convergent $\frac{p}{q}$ of $\kappa$ that satisfies condition $q>6N$ is

$$\frac{p}{q}=\frac{2297570640187354392}{3075296607888933649}.$$

It follows that

$$\varepsilon =||\mu q||-||\kappa q||N\approx 0.296651>0.$$

The given inequality does not have any solutions in integers m such that

$$\frac{\log \left( \frac{Aq}{\varepsilon }\right) }{\log B}\le m <N.$$

Thus, the new upper bound for m is 17. Repeating the procedure once again, we get $m\le 4$, and there are only two solutions, namely

$$v_0=w_0^{+,-}=1,$$

which is the trivial solutions of our Diophantine equation with $d=0$, and

$$v_2=w_2^{-}=11,$$

which suits the case when $d=120$.

2.4.4 Fibonacci Numbers and the Property of Diophantus

The next result is due to Dujella [17]. He proved that, if k and d are positive integers such that the set

$$\{F_{2k}, F_{2k+2}, F_{2k+4}, d\}$$

is a D(1)-quadruple, then $d=4F_{2k+1}F_{2k+2}F_{2k+3},$ where $F_k$ is k-th Fibonacci number. This is a generalization of the Theorem 2.4.5 of Baker and Davenport for $k=1$. Here a set $\{a_1,a_2,a_3,a_4\}$ of distinct positive integers is called a D(1)-quadruple , if $a_ia_j+n$ is a perfect square for every i, j with $1 \le i<j \le 4$.

Proof

(Sketch) Let $k\ge 2$ be a positive integer and

$$a=F_{2k}, \ b=F_{2k+2}, \ c=F_{2k+4}.$$

Then $c=3b-a$. Furthermore,

$$ab+1=(b-a)^2, \ ac+1=b^2, \ bc+1=(a+b)^2.$$

If we assume that d is a positive number such that $\{a, b, c, d\}$ has the property D(1) of Diophantus, it implies that there exist positive integers x, y, z such that

$$ad+1=x^2, \ bd+1=y^2, \ (3b-a)d+1=z^2.$$

Eliminating d, we get a system of Pellian equations

$$\begin{aligned} ay^2-bx^2=a-b,\quad az^2-(3b-a)x^2=2a-3b. \end{aligned}$$

(2.7)

Dujella proved the following lemmas.

Lemma 2.4.8

Let x, y, z be positive integer solutions of the system of Pellian equations (2.7). Then there exist integers m and n such that

$$x=v_m=w_n,$$

where $(v_m)$ is given by

$$v_0=1, \ v_1=b, \ v_{m+2}=2(b-a)v_{m+1}-v_m, \ m\in \mathbb {Z},$$

and the two-sided sequence $(w_n)$ is defined by

$$w_0=1, \ w_1=a+b, \ w_{n+2}=2bw_{n+1}-w_n, \ n\in \mathbb {Z}.$$

In order to apply Baker’s method, it is convenient to consider the two-sided sequence as two ordinary sequences. Therefore, instead of the sequence $(v_m)_{m\in \mathbb {Z}}$, Dujella considered two sequences $(v_m)_{m\ge 0}$ and $(v_m)_{m\le 0}$ and applied the same method for the sequence $(w_n)_{n\in \mathbb {Z}}$. Thus, four equations of the form

$$v_m=w_n$$

have to be considered.

Lemma 2.4.9

If $v_m=w_n,$ and $m\ne 0,$ then

$$\begin{aligned} 0< & {} m\log (b-a+\sqrt{ab})-n\log (b+\sqrt{ac})+\\&+\log \frac{\sqrt{c}(\pm \sqrt{a}+\sqrt{b})}{\sqrt{b}(\pm \sqrt{a}+\sqrt{c})}<4(b-a+\sqrt{ab})^{-2m}. \end{aligned}$$

In the present situation, $l=3, d=4, B=m$ and

$$\alpha _1=b-a+\sqrt{ab}, \ \alpha _2=b+\sqrt{ac}, \ \alpha _3=\frac{\sqrt{c}(\pm \sqrt{a}+\sqrt{b})}{\sqrt{b}(\pm \sqrt{a}+\sqrt{c})},$$

$$h'(\alpha _1)=\frac{1}{2}\log \alpha _1<1.05\log , \ h'(\alpha _2)=\frac{1}{2}\log \alpha _2<1.27\log a,$$

$$h'(\alpha _3)=\frac{1}{4}\log (bc(c-a)(\sqrt{a}+\sqrt{b})^2)<2.52\log a,$$

$$\log 4(b-a+\sqrt{ab})^{-2m}<\log a^{-2m}=-2m\log a.$$

Hence,

$$2m\log a<3.822\cdot 10^{15}\cdot 3.361\log ^3a\log m,$$

and

$$\begin{aligned} \frac{m}{\log m}<6.423\cdot 10^{15}\log ^2 a. \end{aligned}$$

(2.8)

Applying Lemma 2.4.8, we get

$$\begin{aligned} |m|\ge 2b-2>4a. \end{aligned}$$

(2.9)

Comparing (2.8) and (2.9), we obtain

$$\frac{m}{\log ^3m}<6.423\cdot 10^{15},$$

which implies $m<8\cdot 10^{20},$ $a=F_{2k}<2\cdot 10^{20}$. The author has proved the theorem for $k\ge 49.$

It remains to prove the theorem for $2\le k\le 48$. Dujella used Lemma 2.3.1 with

$$\kappa =\frac{\log \alpha _1}{\log \alpha _2}, \ \mu =\frac{\log \alpha _3}{\log \alpha _2}, \ A=\frac{4}{\log \alpha _3}, \ B=(b-a+\sqrt{ab})^2$$

as well as $N=8\cdot 10^{20}$ and gets a new bound $m\le N_0$, where $N_0\le 12.$ Repeating the method one more time, he obtained a new upper bound $m\le 2$ which completes the proof.

2.4.5 The Non-extendibility of Some Parametric Families of $D(-1)$-Triples

Definition 2.1

Let n be a nonzero integer. A set $\{a_1,\dots ,a_m\}$ of m distinct positive integers is called a Diophantine m-tuple with the property D(n), or simply a D(n)-m-tuple, if $a_ia_j+n$ is a perfect square for any i, j with $1 \le i<j \le m$.

The set $\{1, 3, 8, 120\}$ considered in the previous Sect. 2.4.3 is known as the first example of a D(1)-quadruple found by Fermat.²⁸ In 1969, Baker and Davenport proved that $\{1,3,8\}$ cannot be extended to a D(1)-quintuple (see [5]). This result was generalized by Dujella [15], who showed that the D(1)-triple $\{k-1, k+1, 4k\}$ for an integer k cannot be extended to a D(1)-quintuple, and by Dujella and Pethő [22], who proved that the D(1)-pair $\{1,3\}$ cannot be extended to a D(1)-quintuple. It is a conjectured that there does not exist a D(1)-quintuple. The most general results on this conjecture are due to Dujella [18] who proved that there does not exist a D(1)-sextuple and that there exist at most finitely many D(1)-quintuples. There have been some improvements on those results recently, but the conjecture still remains open.

In contrast to the case $n=1$, it is conjectured that there does not exist a $D(-1)$-quadruple (see [16]). The first important step in this direction was done by Dujella and Fuchs [21] who showed that, if $\{a,b,c,d\}$ is a $D(-1)$-quadruple with $a<b<c<d$, then $a=1$. Later Dujella, Filipin and Fuchs [23] showed that there exist at most finitely many $D(-1)$-quadruples. The number of $D(-1)$-quadruples is now known to be bounded by $5\cdot 10^{60}$ (see [24]). However, this bound is too large for verifying the conjecture by present day computers. Recently, He and Togbé [29] proved that the $D(-1)$-triple $\{1,k^2+1,k^2+2k+2\}$ cannot be extended to a $D(-1)$-quadruple. Their result and the proof appears to be very important because of their use of a linear form in two logarithms (instead of three) for the first time; this leads to a much better upper bound for the solutions which shortens the reduction time significantly. In this subsection, we extend their method and apply it to several other families of $D(-1)$-triples. Let us also mention that it is not always possible to use linear forms in two logarithms. In the sequel we only explain the idea and give a sketch of the proofs; more details can be found in [26].

Introduction

Let $\{1,b,c\}$ be a $D(-1)$-triple with $b<c$. We define positive integers r, s and t by

$$b-1=r^2, \ c-1=s^2, \ bc-1=t^2.$$

Then, s and t satisfy

$$\begin{aligned} t^2-bs^2=r^2. \end{aligned}$$

(2.10)

It can be proven that Diophantine equation (2.10) has at least three classes of solutions belonging to

$$(t_0,s_0)=(r,0), \ (b-r,\pm (r-1))$$

(see [27, p. 111]). We call a positive solution (t, s) of (2.10) regular if (t, s) belongs to one of these three classes. However, it is possible for a solution (t, s) not to be regular. In general, we do not know in advance how many classes of solutions we have, except for some special type of b.

Remark 2.4.10

An example having non-regular solutions can be found in the case of $r=2q^2$, where q is a positive integer. Then (2.10) has two more classes of solutions belonging to

$$(t_0',s_0')=(2q^3+q,\pm q).$$

Our goal is to prove the following theorem.

Theorem 2.2

Let (t, s) be a regular solution of (2.10) and let $c=s^2+1$. Then, the system of Diophantine equations

$$\begin{aligned} y^2-bx^2&=r^2,\\ z^2-cx^2&=s^2 \end{aligned}$$

has only trivial solutions $(x,y,z)=(0,\pm r,\pm s)$. Furthermore, if $r=2q^2$ for some positive integer q, then the same is true for any positive solution (t, s) of (2.10) belonging to the same class as one of $(2q^3+q,\pm q)$.

By [23, Theorem 1] we have that $c<11b^6$ (using the hyper-geometric method), hence the above-mentioned result of He and Togbé shows that it is enough to prove Theorem 2.2 for $c=c_i$ with $2 \le i \le 7$, where

$$\begin{aligned} c_2&=4r^4+1,\\ c_3&=(4r^3-4r^2+3r-1)^2+1,\\ c_4&=(4r^3+4r^2+3r+1)^2+1,\\ c_5&=(8r^4+4r^2)^2+1,\\ c_6&=(16r^5-16r^4+20r^3-12r^2+5r-1)^2+1,\\ c_7&=(16r^5+16r^4+20r^3+12r^2+5r+1)^2+1 \end{aligned}$$

(see [27, p. 111]), and in the case of $r=2q^2$, additionally for $c=c_i'$ with $1 \le i \le 5$, where

$$\begin{aligned} c_1'&=(4q^3-q)^2+1,\\ c_2'&=(16q^5+4q^3+q)^2+1,\\ c_3'&=(64q^7-16q^5+8q^3-q)^2+1,\\ c_4'&=(256q^9+64q^7+48q^5+8q^3+q)^2+1,\\ c_5'&=(1024q^{11}-256q^9+256q^7-48q^5+12q^3-q)^2+1. \end{aligned}$$

It is easy to see that Theorem 2.2 immediately implies

Corollary 2.4.11

Let (t, s) be either a regular solution of (2.10) or, in the case of $r=2q^2$ for some positive integer q, a regular solution or a positive solution of (2.10) belonging to the same class as one of $(2q^3+q,\pm q)$. Let $c=s^2+1$. Then, the $D(-1)$-triple $\{1,b,c\}$ cannot be extended to a $D(-1)$-quadruple.

It was proven in [27, p. 111], that if r is prime, then (2.10) has only regular solutions. We can generalize this to find that, if $r=p^k$ or $2p^k$ for an odd prime p and a positive integer k, then (2.10) has only regular solutions, except in the case of $r=2p^k$ with k even. In the latter case, there are exactly five classes of solutions. Furthermore, if $b=p$ or $2p^k$, then (2.10) has only regular solutions ($b=p^k$ can occur only if $k=1$, since $b=r^2+1$; [35]). Hence, we get another corollary of the Theorem 2.2.

Corollary 2.4.12

Let r be a positive integer and let $b=r^2+1$. Suppose that one of the following assumptions holds for an odd prime p and a positive integer k:

(i) $b=p;$   (ii) $b=2p^k;$   (iii) $r=p^k;$   (iv) $r=2p^k$.

Then, the system of Diophantine equations

$$\begin{aligned} y^2-bx^2&=r^2,\\ z^2-cx^2&=s^2 \end{aligned}$$

has only the trivial solutions $(x,y,z)=(0,\pm r,\pm s)$, where (t, s) is a positive solution of (2.10) and $c=s^2+1$. Moreover, the $D(-1)$-pair $\{1,b\}$ cannot be extended to a $D(-1)$-quadruple.

The System of Pellian Equations

Let $\{1,b,c\}$ be a $D(-1)$-triple with $b<c$, and let r, s, t be positive integers defined by $b-1=r^2$, $c-1=s^2$, $bc-1=t^2$. Suppose that we can extend the triple $\{1,b,c\}$ to a $D(-1)$-quadruple with element d. Then, there exist integers x, y, z such that

$$ d-1=x^2,\quad bd-1=y^2,\quad cd-1=z^2. $$

Eliminating d, we obtain the system of simultaneous Diophantine equations

$$\begin{aligned} z^2-cx^2&=s^2, \end{aligned}$$

(2.11)

$$\begin{aligned} bz^2-cy^2&=c-b, \end{aligned}$$

(2.12)

$$\begin{aligned} y^2-bx^2&=r^2. \end{aligned}$$

(2.13)

We may assume that $c<11b^6$ (according to Theorem 1, [23]). The positive solutions (z, x) of Eq. (2.11) and (z, y) of Eq. (2.12) are respectively given by

$$\begin{aligned} z+x \sqrt{c}&=s(s+\sqrt{c})^{2m}\qquad (m \ge 0),\\ z \sqrt{b}+y \sqrt{c}&=(s \sqrt{b} \pm r \sqrt{c})(t+ \sqrt{bc})^{2n}\qquad (n \ge 0) \end{aligned}$$

(see Lemmas 1 and 5 in [23]). Using that y is a common solution of Eqs. (2.12) and (2.13), He and Togbé proved that the positive solutions (y, x) of Eq. (2.13) are given by

$$ y+x \sqrt{b} =r(r+\sqrt{b})^{2l},\qquad l \ge 0, $$

and, moreover, they proved the following proposition.

Proposition 2.4.13

([29, Proposition 2.1]) The $D(-1)$-triple $\{1,b,c\}$ can be extended to a $D(-1)$-quadruple if and only if the system of simultaneous Pellian equations

$$\begin{aligned} (z/s)^2-c(x/s)^2&=1, \\ (y/r)^2-b(x/r)^2&=1 \end{aligned}$$

has a positive integer solution (x, y, z).

Proposition 2.4.13 implies that we can write $x=sv_m=ru_l$, where

$$\begin{aligned} v_m=\frac{\alpha ^{2m}-\alpha ^{-2m}}{2\sqrt{c}}\quad \text {and}\quad u_l=\frac{\beta ^{2l}-\beta ^{-2l}}{2\sqrt{b}} \end{aligned}$$

are positive solutions of the Pellian equations $Z^2-cX^2=1$ and $Y^2-bW^2=1$, respectively, where $\alpha =s+\sqrt{c}$ and $\beta =r+\sqrt{b}$. We have mentioned before that we cannot always use linear forms in two logarithms. More precisely, for our approach $\alpha$ and $\beta$ should be near to each other or one of them has to be near to a power of the other one.

Gap Principles

We next consider the extension of $D(-1)$-triples $\{1,b,c\}$ with

$$c=c_2,c_3,c_4,c_5,c_6,c_7,c_1',c_2',c_3',c_4',c_5'$$

from above. We shall establish gap principles for these special cases.

Let us define the linear form $\varLambda$ in three logarithms

$$\varLambda =2m\log \alpha -2l\log \beta +\log \frac{s\sqrt{b}}{r\sqrt{c}}.$$

The proofs of the following lemmas can be found in [29]; the results rely on Baker’s theory of linear forms in logarithms.

Lemma 2.4.14

([29, Lemma 3.1]) If $sv_m=ru_l$ has a solution with $m\ne 0,$ then

$$0<\varLambda <\frac{b}{b-1}\cdot \beta ^{-4l}.$$

Lemma 2.4.15

([29, Lemma 3.3]) If $sv_m=ru_l$ has a solution with $m\ne 0,$ then $m\log \alpha <l\log \beta .$

The proofs of the following lemmas can be found in [26]; these results are based on the property that an algebraic number $\alpha$ is close to some power of $\beta$.

Lemma 2.4.16

Let $c=c_2=4r^4+1.$ If the equation $sv_m=ru_l$ has a solution with $m\ne 0$, then

$$m>\frac{\varDelta }{2}\cdot \alpha \log \beta ,$$

where $\varDelta$ is a positive integer.

Lemma 2.4.17

Let $c=c_3=(4r^3-4r^2+3r-1)^2+1$ or $c=c_4=(4r^3+4r^2+3r+1)^2+1.$ If the equation $sv_m=ru_l$ has a solution with $m\ne 0,$ then

$$m>\frac{3\varDelta -1}{3}\cdot \frac{8r}{9}\log \beta ,$$

where $\varDelta$ is a positive integer.

Lemma 2.4.18

Let $c=c_1'=(4q^3-q)^2+1.$ If the equation $sv_m=ru_l$ has a solution with $m\ne 0,$ then

$$m>\frac{\varDelta }{6}\cdot \alpha \log \beta ,$$

where $\varDelta$ is a positive integer.

Similar lemmas for other choices of c can be obtained. The following table contains information how to choose $\varDelta$ and a lower bound for m.

c	$\varDelta$	a lower bound for m
$c_5$	$4m-l$	$m>\varDelta \cdot \frac{8r^4}{4r^2+1}\log \beta$
$c_6$	$5m-l$	$m>\frac{5\varDelta -1}{5}\cdot \frac{32r}{33}\log \beta$
$c_7$	$l-5m$	$m>\frac{5\varDelta -1}{5}\cdot \frac{32r}{33}\log \beta$
$c_2'$	$2l-5m$	$m>\frac{10\varDelta -4}{5}\cdot q^2\log \beta$
$c_3'$	$7m-2l$	$m>\frac{3\varDelta }{2}\cdot q^2\log \beta$
$c_4'$	$2l-9m$	$m>\frac{18\varDelta -4}{9}\cdot q^2\log \beta$
$c_5'$	$11m-2l$	$m>\frac{3\varDelta }{2}\cdot q^2\log \beta$

Linear Forms in Two Logarithms

Next we shall apply the following result due to Laurent,²⁹ M. Mignotte³⁰ and Y. Nesterenko³¹ to our linear form $\varLambda .$

Lemma 2.4.19

([33, Corollary 2]) Let $\gamma _1$ and $\gamma _2$ be multiplicatively independent, positive algebraic numbers, $b_1$, $b_2\in \mathbb {Z}$ and

$$\varLambda =b_1\log \gamma _1+b_2\log \gamma _2.$$

Let $D:=[\mathbb {Q}(\gamma _1,\gamma _2):\mathbb {Q}],$ for $i=1,2$ let

$$h_i\ge \max \left\{ h(\gamma _i),\frac{|\log \gamma _i|}{D},\frac{1}{D}\right\} ,$$

where $h(\gamma )$ is the absolute logarithmic height of $\gamma ,$ and

$$b'\ge \frac{|b_1|}{Dh_2}+\frac{|b_2|}{Dh_1}.$$

If $\varLambda \ne 0,$ then

$$\log |\varLambda |\ge -24.34\cdot D^4\left( \max \left\{ \log b'+0.14,\frac{21}{D},\frac{1}{2}\right\} \right) ^2h_1h_2.$$

This lemma has also been used by He and Togbé in [29]. We are dealing with the same linear form, but only with a different c. Using the same method to transform our form to a linear form in two logarithms, then applying Lemma 2.4.19 for $c=c_2,c_3,\ldots ,c_7,c_1',\ldots ,c_5',$ and combining the lower bound for $|\varLambda |$ together with the gap principles, we can prove Theorem 2.2 for large values of r. That $\gamma _1$ and $\gamma _2$ are multiplicatively independent follows from the fact that $\alpha$ and $\beta$ are multiplicatively independent algebraic units and $\frac{r\sqrt{c}}{s\sqrt{b}}$ is not an algebraic unit.

As an example we consider the case $c=c_2=4r^4+1.$ We can write

$$\varLambda =2m\log \left( \frac{\alpha }{\beta ^2}\right) -\log \left( \beta ^{-2\varDelta }\cdot \frac{r\sqrt{c}}{s\sqrt{r^2+1}}\right) ,$$

where $\varDelta =2m-l$ is defined as in Lemma 2.4.16. In the notation of Lemma 2.4.19 we have

$$D=4, \ b_1=2m, \ b_2=-1, \ \gamma _1=\frac{\alpha }{\beta ^2}, \ \gamma _2=\beta ^{-2\varDelta }\cdot \frac{r\sqrt{c}}{s\sqrt{b}}.$$

Furthermore,

$$h(\gamma _1)\le h\left( \frac{\alpha }{\beta }\right) +h(\beta )=\frac{1}{2}\log \alpha +\frac{1}{2}\log \beta <\log \alpha ,$$

hence, for $h_1$, we can take $h_1=\log \alpha .$ Moreover,

$$h\left( \frac{r\sqrt{c}}{s\sqrt{b}}\right) =\frac{1}{2}\log ((c-1)b)<\frac{1}{2}\log \beta ^6=3\log \beta ,$$

which yields

$$ h(\gamma _2)<(\varDelta +3)\log \beta =:h_2.$$

For $r\ge 10$, we find

$$\frac{|b_2|}{Dh_1}=\frac{1}{4\log \alpha }<0.042,$$

and then

$$b'=\frac{m}{2(\varDelta +3)\log \beta }+0.042.$$

Now Lemma 2.4.16 implies

$$\frac{m}{2(\varDelta +3)\log \beta }>\frac{\varDelta }{4(\varDelta +3)}\cdot \alpha \ge \frac{\alpha }{16}>169$$

for $r\ge 26.$ Thus, for $r\ge 26$, we get $\log b'+0.14>\frac{21}{D}$ and applying Lemma 2.4.19 we conclude

$$\log |\varLambda |\ge -24.34\cdot 4^4\cdot (\log b'+0.14)^2\cdot \log \alpha \cdot (\varDelta +3)\log \beta .$$

On the other hand, Lemma 2.4.14 yields

$$\log |\varLambda |<0.002-4l\log \beta .$$

Combining these lower and upper bounds for $\varLambda$, we obtain

$$\frac{l}{\log \alpha }<\frac{0.002}{4\log \alpha \log \beta }+24.34\cdot 64(\log b'+0.14)^2(\varDelta +3).$$

Furthermore, $m\log \alpha <l \log \beta$ gives us

$$\frac{m}{2(\varDelta +3)\log \beta }<0.0001+24.34\cdot 32(\log b'+0.14)^2$$

and finally

$$b'<0.042+778.88(\log b'+0.14)^2,$$

which implies $b'<106996.$ It furthermore gives $m<213992(\varDelta +3)\log \beta$ and

$$\alpha<\frac{2m}{\varDelta \log \beta }<427983\cdot \frac{\varDelta +3}{\varDelta }<1.72\cdot 10^6,$$

from which we deduce $r<656.$ Thus, we have proved Theorem 2.2 for $c=c_2$ and $r\ge 656.$

The cases $c=c_3$, $c_4=(4r^3\mp 4r^2+3r\mp 1)^2+1$ are described in details in [26]. The upper bounds for r and q in the remaining cases are given in the following table.

c	an upper bound for r or q
$c_3$	$r<1.81\cdot 10^6$
$c_4$	$r<1.81\cdot 10^6$
$c_5$	$r<802$
$c_6$	$r<1.94\cdot 10^6$
$c_7$	$r<1.94\cdot 10^6$
$c_2'$	$q<846$
$c_3'$	$q<846$
$c_4'$	$q<949$
$c_5'$	$q<1000$

The Reduction Method and the Proof of Theorem 2.2

We have just proven Theorem 2.2 for large parameters r and q. We are left to consider the cases of small r and q. Using Baker–Davenport reduction, it turns out that in all remaining cases there is no extension of the triple $\{1,b,c\}$ to a quadruple $\{1,b,c,d\}.$

Some useful results from [23] can be used. We know that, if we have the extension of our triple with the element d, then $cd-1=z^2$, where $z=V_m=W_n$ such that

$$V_0=s,\,V_1=(2c-1)s,\,V_{m+2}=(4c-2)V_{m+1}-V_m$$

and

$$W_0=s,\,W_1=(2bc-1)s\pm 2rtc,\,W_{n+2}=(4bc-2)W_{n+1}-W_n.$$

We use the following lemmas.

Lemma 2.4.20

([23, Lemma 11]) If $V_m=W_n,$ $n\ne 0,$ then

$$0<2n\log (t+\sqrt{bc})-2m\log (s+\sqrt{c})+\log \frac{s\sqrt{b}\pm r\sqrt{c}}{2\sqrt{b}}<(3.96bc)^{-n+1}.$$

From the proofs of Propositions 2, 3, 4 in [23] we know that $n<10^{20}$ in all cases. Applying Baker–Davenport reduction with

$$\kappa =\frac{\log (t+\sqrt{bc})}{\log (s+\sqrt{c})},\quad \mu =\frac{\log \frac{s\sqrt{b}\pm r\sqrt{c}}{2\sqrt{b}}}{2\log (s+\sqrt{c})},\quad A=\frac{3.96bc}{2\log (s+\sqrt{c})},\quad B=3.96bc$$

and $N =10^{21}$ with any choice of r and c left, we get after two steps that $n<2.$ Here, one may also use that $D(-1)$-triples $\{1,b,c\}$ cannot be extended to a quadruple for $r\le 143000$, which was verified by computer. Hence, in some cases one can avoid to use reduction at all.

We are still left to deal with the cases of small indices m and n in the equation $z=V_m=W_n.$ From [21] we know that $n\ge 3;$ otherwise we have only the trivial solution (corresponding with an extension with $d=1,$ which is no real extension, because we ask for elements in D(n)-m-tuple to be distinct).

The following lemma, which was proved in [26], and which examines the fundamental solutions of (2.10) in the cases of $b=p,\,2p^k$ and $r=p^k,\,2p^k$, together with Theorem 2.2 implies Corollary 2.4.12.

Lemma 2.4.21

• (1) If $b=p$ or $2p^k$ for an odd prime p and a positive integer k, then Diophantine Equation (2.10) has only regular solutions.

• (2) If $r=p^k$ or $2p^k$ for an odd prime p and a positive integer k, then Diophantine Equation (2.10) has only regular solutions, except in the case of $r=2p^{2i}$ with i a positive integer, where it in addition has exactly two classes of solutions belonging to $(2p^{3i}+p^i,\pm p^i)$.

2.4.6 Pure Powers in Binary Recurrent Sequences

The Lucas numbers $(L_n)_{n\ge 0}$ are given by

$$L_0=2, \ L_1=1, \ \dots , \ L_{n+2}=L_{n+1}+L_n, \ n\ge 0.$$

Recall that Fibonacci numbers $(F_n)$ as well as Lucas numbers $(L_n)$ are defined by

$$F_n=\frac{\alpha ^n-\beta ^n}{\sqrt{5}}, L_n=\alpha ^n+\beta ^n, \alpha =\frac{1+\sqrt{5}}{2}, \beta =\frac{1-\sqrt{5}}{2},$$

respectively. Now suppose

$$F_n=y^p$$

is a pure power. Since

$$\alpha ^n-\sqrt{5}y^p=O(\alpha ^{-n}),$$

we find

$$\varLambda =n\log \alpha -p\log y-\log \sqrt{5}=O(\alpha ^{-2n})=O(y^{-2p}).$$

There exist integers k, r such that $n=kp+r$ with $|r|\le \frac{p}{2},$ hence we have

$$\varLambda =p\log \left( \frac{\alpha ^k}{y}\right) +r\log \alpha -\log \sqrt{5}$$

which is a linear form in three logarithms. If we apply Matveev’s Theorem 2.2.14, we get

$$\log |\varLambda |\ge -c^*\log y\log p.$$

Comparing both estimates of $|\varLambda |,$ we see that the exponent p is bounded. Matveev’s Theorem 2.2.14 implies $p<3\cdot 10^{13},$ but a special estimate for linear forms in three logarithms implies the sharper upper bound is $p<2\cdot 10^8$ which is suitable for computer calculations.

For Lucas numbers a similar study leads to a linear form in two logarithms and $p<300$ provided that $L_n=y^p.$ By this reasoning, it can be proved [10] that all perfect powers in the Fibonacci and Lucas sequences are

$$F_0 = 0, \ F_1 = F_2 = 1, \ F_6 = 8 = 2^3, \ F_{12} = 144 = 12^2; $$

$$L_1 = 1, \ L_3 = 4 = 2^2.$$

Using this method we can solve many similar problems, for example, all Pell numbers for which $P_n+4$ is a perfect square are given by $P_0=0$, $P_3=5$ and $P_4=12$. Recall that the Pell numbers are given by the recursion

$$P_0=0, \ P_1=1, \ \dots , \ P_{n+2}=2P_{n+1}+P_n, \ n\ge 0.$$

2.4.7 Lucas Numbers and the Biggest Prime Factor

Next we are interested in finding all Lucas numbers for which the biggest prime factor is less than or equal to 5.

We may express Lucas numbers as

$$L_n=\left( \frac{1+\sqrt{5}}{2}\right) ^n+\left( \frac{1-\sqrt{5}}{2}\right) ^n,n\in \mathbb {N}.$$

For Lucas numbers the length of the period of the sequence $(L_n\mod 5)$ is equal to 4 with the cycle $\{1, 3, 4, 2\}$; therefore 5 can never be a divisor of any Lucas number. We want to find all Lucas numbers such that

$$2^k3^l=\left( \frac{1+\sqrt{5}}{2}\right) ^n+\left( \frac{1-\sqrt{5}}{2}\right) ^n,n, k, l, m\in \mathbb {N}.$$

Let $\alpha =\frac{1+\sqrt{5}}{2},$ so the previous expression can be rewritten as

$$2^k3^l=\alpha ^n-\alpha ^{-n}, k,l,n\in \mathbb {N}.$$

Thus the corresponding linear form in logarithms is

$$\varLambda =2n\log \alpha -k\log 2-l\log 3.$$

Now, applying Matveev’s Theorem 2.2.14, we have $n=3$, $D=2$, and

$$A_1\ge h'(2)=\max \{2\log 2, |\log 2|, 0.16\}=1.38<2,$$

$$A_2\ge h'(3)=\max \{2\log 3, |\log 3|, 0.16\}=2.19<3,$$

$$A_3\ge h'\left( \frac{1+\sqrt{5}}{2}\right) =\max \{2\log \frac{1+\sqrt{5}}{2}, \left| \log \frac{1+\sqrt{5}}{2}\right| , 0.16\}=0.48<1.$$

We get

$$\log |\varLambda |\ge -7.28022\cdot 10^{15}(1+\log 2n).$$

After finding the upper bound from the expression

$$\frac{L_n}{\alpha ^n}-1=-\alpha ^{-2n},$$

and

$$\log |\varLambda |<-2n\log \frac{1+\sqrt{5}}{2},$$

we find that $n<3.17654\cdot 10^{17}$. After applying Baker–Davenport reduction, we get $n<14$, so we may conclude that the Lucas numbers for which the biggest prime factors are less or equal to 5 are

$$L_0=2, L_2=3, L_4=4=2^2, L_6=18=2\cdot 3^2.$$

2.4.8 Pillai’s Equation

Given positive integers $a>b>1$, Pillai³² [42] proved that there are only finitely many integers $c\ne 0$ admitting more than one representation of the form

$$c=a^x-b^y$$

in nonnegative integers x, y. In particular, the equation

$$\begin{aligned} a^x-b^y=a^{x_1}-b^{y_1}, \ \ with \ (x, y)\ne (x_1, y_1) \end{aligned}$$

(2.14)

has only finitely many integer solutions. We shall apply the technique of lower bounds for linear forms in logarithms of algebraic numbers to find all the solutions for

$$(a, b)=(3, 2).$$

Proposition 2.4.22

The only nontrivial solutions of Eq. (2.14) with $(a, b)=(3, 2)$ are

$$3^1-2^2=3^0-2^1, \ \ 3^2-2^4=3^0-2^3, \ \ 3^2-2^3=3^1-2^1,$$

$$3^3-2^5=3^1-2^3, \ \ 3^5-2^8=3^1-2^4.$$

Proof

The initial equation can be rewritten as

$$3^x-3^{x_1}=2^y-2^{y_1}.$$

After relabeling the variables, we may assume that $x>x_1.$ Consequently, $y>y_1$. Since

$$2\cdot 3^{x-1}=3^x-3^{x-1}\le 3^x-3^{x_1}=2^y-2^{y_1}<2^y,$$

we get $x<y$. Let $B=y.$ Now,

$$3^{x_1}\mid (2^y-2^{y_1})=2^{y_1}(2^{y-y_1}-1).$$

We observe that $3^m\mid (2^n-1)$ if and only if $2\cdot 3^{m-1}\mid n.$ In particular,

$$\begin{aligned} x_1\le 1+\frac{\log ((y-y_1)/2)}{\log 3}\le \frac{\log (3B/2)}{\log 3},\end{aligned}$$

(2.15)

therefore,

$$3^{x_1}<\frac{3B}{2}<2B.$$

Similarly,

$$2^{y_1}\mid (3^x-3^{x_1})=3^{x_1}(3^{x-x_1}-1).$$

Analogously, if $m\ge 3,$ then $2^m\mid (3^n-1)$ if and only if $2^{m-2}\mid n.$ Thus,

$$\begin{aligned} y_1\le 2+\frac{\log (x-x_1)}{\log 2}<\frac{\log (4B)}{\log 2},\end{aligned}$$

(2.16)

and therefore

$$2^{y_1}\le 4B.$$

The original equation may be rewritten in such a way that the large parts are on one side and the small parts are on the other, namely

$$|3^x-2^y|=|3^{x_1}-2^{y_1}|<2B,$$

which in turn gives an inequality of the form

$$|1-3^x2^{-y}|<\frac{2B}{2^B}.$$

Thus the linear form $\varLambda$ to study is given by

$$\varLambda =x\log 3-y\log 2.$$

If $\varLambda >0,$ then

$$e^{\varLambda }-1<\frac{2B}{2^B}.$$

If $\varLambda <0,$ assuming that $B>10$, we find $\frac{2B}{2^B}<\frac{1}{2}$ and therefore, $|1-e^{\varLambda }|<\frac{1}{2}$, which implies $e^{|\varLambda |}<2.$ In particular,

$$\begin{aligned} |\varLambda |<\frac{4B}{2^B}. \end{aligned}$$

(2.17)

The last inequality holds independent of the sign of $\varLambda$. We observe that $\varLambda \ne 0$, since in the opposite case, we would get $3^x=2^y$ which, by unique factorization, implies $x=y=0$, a contradiction. Put

$$\alpha _1=2, \ \alpha _2=3, \ b_1=y, \ b_2=x, \ B=y, \ A_1=1, \ A_2=\log 3$$

and

$$b'=\frac{y}{\log 3}+x<B\left( \frac{1}{\log 3}+1\right) <2B.$$

Since $\log 2$ and $\log 3$ are linearly independent, real and positive, Lemma 2.4.19 yields the estimate

$$\log |\varLambda |>-23.34(\max \{\log (2B)+0.14, 21\})^2\cdot \log 3.$$

Comparing the last inequality with (2.17), we get

$$B\log 2-\log (4B)<23.34\cdot 3\cdot (\max \{\log (2B)+0.14, 21\})^2.$$

If the above maximum is 21, we get

$$B\log 2-\log (4B)<25.7\cdot 21^2,$$

hence $B<17000.$ Otherwise, we have

$$B\log 2-\log (4B)<25.7(\log (2B)+0.14)^2,$$

yielding $B<2900.$ Thus, we consider the inequality $B<17000$. From (2.15) and (2.16), we get $x_1\le 9$ and $y_1\le 16$. Hence,

$$x-1\le (y-1)\frac{\log 2}{\log 3}<B\frac{\log 2}{\log 3}<11000.$$

Now, we reduce this bound. Suppose that $B\ge 30,$ then we get

$$3^x>3^x-3^{x_1}=2^y-2^{y_1}\ge 2^{B-1}\ge 2^{29},$$

which implies $x\ge 19.$ We check that the congruence

$$3^x-3^{x_1}-2^{y_1}\equiv 0\pmod {2^{30}}$$

does not hold for any triple $(x, x_1, y_1)$ with $11\le x\le 1100, 0\le x_1\le 9,$ and $0\le y_1\le 16$. This gives $B\le 29.$ Since $3^{x-1}<2^{y-1}\le 2^{28}$, we get $x\le 18.$ Now, it is easy to show that there are no solutions beyond those in the statement of the proposition. For details see [36].

2.4.9 The Diophantine Equation $ax^n-by^n=c$

We consider

$$ax^n-by^n=c,$$

where a, b are strictly positive and x, y, n are unknowns. If for some exponent n there exists a solution (x, y) with $|y|>1$, then

$$\varLambda =\log \left| \frac{a}{b}\right| -n\log \left| \frac{x}{y}\right| =O(|y|^{-n}).$$

In the other direction, Matveev’s Theorem 2.2.14 implies

$$\log |\varLambda |\ge -c_{*}\log |y|\log n.$$

Comparing both estimates, we get $n<c_{**}$, where $c_{**}$ depends only on a, b, c.

The following theorems give us explicit results.

Theorem 2.4.23

(Mignotte, [40]) Assume that the exponential Diophantine inequality

$$|ax^n-by^n|\le c, a,b,c\in \mathbb {Z}_+, a\ne b$$

has a solution in positive integers x, y with $\max \{x, y\}>1.$ Then

$$n\le \max \left\{ 3\log (1.5|c/b|), \ 7400\frac{\log A}{\log (1+(\log A)/\log |a/b|)}\right\} , A=\max \{a, b, 3\}.$$

Bennett obtained the following definitive result for $c=\pm 1$.

Theorem 2.4.24

(Bennett, [8]) For $n\ge 3,$ the equation

$$|ax^n-by^n|=1, a, b\in \mathbb {Z}_+$$

has at most one solution in positive integers x, y.

For more examples of applications of linear forms in logarithms we refer to [25, 36] which are highly recommended for this purpose.