On Pillai's problem with X-coordinates of Pell equations and powers of 2

Abstract

In this paper we show that if $(X_n,Y_n)$ is the nth solution of the Pell equation $X^2-d{Y}^2=\pm 1$ for some non–square d, then given any integer c, the equation $c=X_n-2^m$ has at most 3 integer solutions $(n,m)$ with $n\ge 1$ and $m\ge 0$.

Introduction

In 1936, Pillai conjectured that for any given integer $c\ge 1$, the number of positive integer solutions $(a,b,x,y)$, with $x\ge 2$ and $y\ge 2$, to the Diophantine equation

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

is finite. This conjecture which is still open for all $c\ne 1$, amounts to saying that the distance between two consecutive terms in the sequence of all perfect powers tends to infinity. The work of Pillai was pursued by Herschfeld [3], [4] who showed that if c is an integer with sufficiently large absolute value, then the equation (1), in the special case $(a,b)=(3,2)$, has at most one solution $(x,y)$. For small $|c|$ this is not the case. Pillai [8], [9] extended Herschfeld's result to a more general exponential Diophantine equation (1) with fixed integers $a,b,c$ with $\text{gcd}(a,b)=1$ and $a>b\ge 1$. Specifically, Pillai showed that there exists a positive integer $c_0(a,b)$ such that, for $|c|>c_0(a,b)$, equation (1) has at most one integer solution $(x,y)$. That is to say that there are only finitely many integers c such that the equation (1) has more than one positive integer solution $(x,y)$. His method was ineffective. This was made effective by Stroeker and Tijdeman in [11] by using Baker's theory on linear forms in logarithms. In particular, they found all such c when $(a,b)=(3,2)$ together with their multiple representations of the form $c=3^x-2^y$.

The equation (1) was revisited recently by replacing the powers of a and b by members of sequences ${\{U_n\}}_{n\ge 0}$ and ${\{V_m\}}_{m\ge 0}$ satisfying certain properties. For instance, a consequence of the main result of [1] says that if ${\{U_n\}}_{n\ge 0}$ and ${\{V_m\}}_{m\ge 0}$ are linearly recurrent sequences of integers with dominant roots α and ${\alpha }_1$ which are multiplicatively independent (that is, the only solution of the equation ${\alpha }^x{\alpha }_1^y=1$ in integers $(x,y)$ is $(0,0)$), then there are only finitely many integers c such that $c=U_n-V_m$ has more than one positive integer solution $(n,m)$. Given $(U,V):=(\{U_n\},\{V_m\})$, we write $m_{U,V}(c)$ for the “multiplicity” of c as an element of the form $U_n-V_m$; that is, as the number of pairs $(n,m)$ of positive integers such that $c=U_n-V_m$. With this notation, we have that $m_{U,V}(c)\le 1$ for all but finitely many c.

Let $d>1$ be an integer which is not a square and let $(X_n,Y_n)$ be the nth solution of the Pell equation

$$X^2-d{Y}^2=\pm 1.$$

Putting $\alpha :=X_1+\sqrt{d}{Y}_1$ and $\beta :=X_1-\sqrt{d}{Y}_1$, it is known that $X_n=({\alpha }^n+{\beta }^n)/2$ for all $n\ge 0$. Furthermore, $\beta =\epsilon {\alpha }^{-1}$, where $\epsilon =X_1^2-d{Y}_1^2\in \{\pm 1\}$. Note that the sequence ${\{U_n\}}_{n\ge 0}$ of general term $U_n=X_n$ is a binary recurrent sequence satisfying $U_{n+2}=(2X_1)U_{n+1}-\epsilon U_n$ for all $n\ge 1$ and has α as its dominant root. Let $V_m=2^m$ for all $m\ge 0$. Then the sequence ${\{V_m\}}_{m\ge 1}$ is also a linear recurrence (in fact, a geometric progression) which has 2 as its dominant root. Since α and 2 are multiplicatively independent, it follows by the result from [1], that there are only finitely manly c such that $m_{U,V}(c)>1$.

The purpose of this paper is to give an upper bound on $m_{U,V}(c)$ which is uniform in c and d, for our case $U=\{X_n\}$ and $V=\{2^m\}$. We prove the following theorem:

Theorem 1

If $U:={\{U_n\}}_{n\ge 0}$ is the sequence of X–coordinates of the positive integer solutions $(X,Y)$ of the Pell equation $X^2-d{Y}^2=\pm 1$ and $V:={\{V_m\}}_{m\ge 0}$ is the sequence of powers of 2; that is, $V_m=2^m$ for all $m\ge 0$. Then for all integers c we have

$$m_{U,V}(c)\le 3.$$

Note that for $d=2$, the sequence ${\{X_n\}}_{n\ge 0}$ starts as

$$1,3,7,17,41,99,\dots ,$$

and $-1=1-2^1=3-2^2=7-2^3$. Hence, the equation $c=X_n-2^m$ when $c=-1$ has three positive integer solutions $(n,m)$, namely $(1,1),(2,2),(3,3)$. In particular, our result is best possible. We offer the following conjecture.

Conjecture 1

There are only finitely many pairs of integers $(c,d)$ with $d>1$ such that $m_{U,V}(c)>2$.

Of course, the conjecture is true by the main result from [1] for a fixed d so one only has to show that there are only finitely many d such that $m_{U,V}(c)>2$ for some c. One may ask whether the above conjecture holds if we replace the lower bound $m_{U,V}(c)>2$ by $m_{U,V}(c)>1$. We show that this is not the case because of the following family of “counterexamples”.

Take $ϵ=1$. We then have the identity

$$X_3-X_1=2X_1{X}_2-2X_1=2X_1(2X_1^2-1)-2X_1=4X_1^3-4X_1.$$

So, if we choose $X_1=2^n$, we get the identity

$$X_3-2^{3n+2}=X_1-2^{n+2}=c,c=c_n:=-3\cdot 2^n.$$

Writing $2^{2n}-1=d_n{Y}^2$ with a square–free integer $d_n$ and a positive integer Y depending on n, we get $m_{U,V}(c)\ge 2$ for the pair $(c,d)=(c_n,d_n)$ and for all $n\ge 1$. While not needed, let us note that

$$\underset{n\to \infty }{\lim }{d}_n=\infty .$$

Indeed, let us write again $2^{2n}-1=d_n{Y}^2$. We omit the index n on d for simplicity. Then $(2^n,Y)=(X_k,Y_k)$ for some k, where this is the sequence of positive integer solutions to $X^2-d{Y}^2=1$. Since $X_k=2^n$, it follows, by a result from [7], that $k=1$. In particular, $2^n=X_1<e^{3\sqrt{d}\log d}$, where the last inequality is Lemma 1 in [5]. This shows that

$$\sqrt{d}\log d>\frac{n\log 2}{3}>\frac{n}{5},$$

so

$$n<5\sqrt{d}\log d=10\sqrt{d}\log \sqrt{d}.$$

The above inequality implies right away that

$$\sqrt{d}>\frac{n}{10\log n},$$

so, in particular,

$$d_n>{\left(\frac{n}{10\log n}\right)}^2,$$

which implies that the limit (3) holds in a stronger form.