An application of the Baker method to a new conjecture on exponential Diophantine equations

1.   Introduction

Let $\mathbb{N}$ be the sets of all positive integers. Let $n$ be a positive integer and let $a, b$ be fixed coprime positive integers with $\min\{a, b\} > 2$. Recently, Yuan and Han ^[1] proposed the following conjecture:

Conjecture 1.1. For any positive integer $n$, if $\min\{a, b\}\geq 4$, then the equation

has only the solution $(x, y, z) = (1, 1, 1)$.

The above conjecture has been proved in many cases for $n = 1$ (see ^[2]). However, for general $n$, it is still widely open.

Let $A, B$ be coprime positive integers with $\min\{A, B\} > 1$ and $2|B$. In ^[1], Yuan and Han ^[1] deal with the solutions $(x, y, z)$ of $(1.1)$ for the case that $(a, b) = (A^2, B^2)$, then $(1.1)$ can be rewritten as

In this respect, they proved that, for any $n$, if $B\equiv2\pmod 4,$ then $(1.2)$ has no solutions $(x, y, z)$ with $y > z > x$; in particular, if $(a, b) = (A^2, B^2)$ and $B = 2$, then Conjecture 1.1 is true. Very recently, Le and Soydan ^[3] proved that, for any $n$, if $A > B^3/8$, then $(1.2)$ has no solutions $(x, y, z)$ with $x > z > y$. Thus, they deduce that, for any $n$, if $(a, b) = (A^2, B^2)$, $A > B^3/8$ and $B\equiv2\pmod4$, then Conjecture 1.1 is true. Their proof relies heavily on an upper bound for solutions of exponential Diophantine equations due to Scott and Styer ^[4].

In this paper, using the Baker method, we prove a general result as follows:

Theorem 1.2. For any $n > 1$, if $a > \max\{15064b, b^{3/2}\}$, then $(1.1)$ has no solutions $(x, y, z)$ with $x > z > y$.

Combining Theorem 1.2 with the above mentioned results of ^[1], we can obtain the following corollary:

Corollary 1.3. For any $n > 1$, if $(a, b) = (A^2, B^2)$, $A > \max\{123B, B^{3/2}\}$ and $B\equiv2\pmod4$, then Conjecture 1.1 is true.

Obviously, Theorem 1.2 and Corollary 1.3 improve the corresponding results in ^[3].

2.   Lemmas

Let $\mathbb{Z, Q, C}$ be the sets of all integers, rational numbers, and complex numbers, respectively. For any algebraic number $\alpha$ of degree $d$ over $\mathbb Q$, let $h(\alpha)$ denote the absolute logarithmic height of $\alpha$, then we have

where $a_0$ is the leading coefficient of the minimal polynomial of $\alpha$ over $\mathbb Z$ and $\alpha^{(i)}(i = 1, \dots, d)$ are the conjugates of $\alpha$ in $\mathbb C$. For $\alpha\not = 0$, let $\log\alpha$ be any determination of its logarithms.

Let $\alpha_1, \alpha_2$ be two algebraic numbers with $\min\{|\alpha_1|, |\alpha_2|\}>1$ and let $\beta_1, \beta_2$ be positive integers. Further, let

Lemma 2.1. If $\alpha_1$ and $\alpha_2$ are multiplicatively independent and $\alpha_1, \alpha_2, \log\alpha_1, \log\alpha_2$ are real and positive, then

where $D = [\mathbb Q(\alpha_1, \alpha_2):\mathbb Q]$,

Proof. This is the special case of Corollary 2 of ^[5] for $m = 10$. □

Lemma 2.2. (Theorem 1.1 of ^[6], Proposition 3.1 of ^[1]) Let $(x, y, z)$ be a solution of $(1.1)$ with $(x, y, z)\not = (1, 1, 1),$ then either $x>z>y$ or $y>z>x$. Moreover, if $n>1$, then either

or

where ${\rm rad}(n)$ is the product of all distinct prime divisors of $n$.

Lemma 2.3. Let $t$ be a real number. If $t>7600$, then $t>75.6(1.08+\log t)^2$.

Proof. Let $f(t) = t-75.6(1.08+\log t)^2$ for $t>1$, then we have $f'(t) = 1-151.2(1.08+\log t)/t$, where $f'(t)$ is the derivative of $f(t)$. Since $f'(t)>0$ for $t>1500$, $f(t)$ is an increasing function for $t>1500$. Therefore, since $f(7600)>0$, we get $f(t)>0$ for $t>7600$. Thus, the lemma is proved.□

3.   Proofs

Proof of Theorem 1.2. We now prove the first half of the theorem. Let $a > \max\{15064b, b^{3/2}\}$ and $(x, y, z)$ be a solution of $(1.1)$ with $x > z > y$. By Lemma $2.2$, we have

and

By $(3.2)$, we get

where

Further, by $(3.3)$, we have

Notice that $x > z > y, a > b$, and $b\geq b_2$ by $(3.1)$. We get $a^xn^{x-z}\geq a^x > a^y > b^y\geq b_2^y$. Hence, we see from $(3.2)$ that

Since $\log(1+t) < t$ for any $t > 0$, by $(3.4)$ and $(3.6)$, we have

Therefore, by $(3.7)$, we get

Let

By $(3.5)$ and $(3.9)$, $\Lambda$ can be rewritten as $(2.2)$. We see from $(3.9)$ that $\alpha_1$ and $\alpha_2$ are multiplicatively independent rational numbers with $\min\{\alpha_1, \alpha_2\} > 1$. By $(2.1)$ and $(3.9)$, we have

Since $\Lambda > 0$ by $(3.5)$, applying Lemma $2.1$, we get from $(3.5)$, $(3.9)$, $(3.10)$, and $(3.11)$ that

where

Therefore, by $(3.8)$ and $(3.12)$, we have

Hence, we obtain

Since $z > y$ and $a > b^{3/2}$, if $2b_2^y > (a+b)^{2z/3},$ then from $(3.1)$ we get $a^{2/3} > b$ and

which is a contradiction. So, we have $2b_2^y < (a+b)^{2z/3},$ which implies that

Hence, by $(3.14)$ and $(3.15)$, we get

When $10\geq0.38+\log K$ by $(3.13)$, we have

When $10 < 0.38+\log K$ by $(3.16)$, we get

Since

by $(3.5)$, we see from $(3.13)$ and $(3.19)$ that

Further, by $(3.18)$ and $(3.20)$, we have

Applying Lemma $2.3$ to $(3.21)$, we get

The combination of $(3.17)$ and $(3.22)$ yields

On the other hand, by $(3.5)$, we have

Therefore, we get

Hence, by $(3.23)$ and $(3.24)$, we obtain

However, since $a > 15064b$, $(3.25)$ is false. Thus, the theorem is proved. □

Proof of Corollary 1.3. Let $(a, b) = (A^2, B^2)$. By Theorem 1.2, if $A > \max\{123B, B^{3/2}\}$, then $(1.2)$ has no solutions $(x, y, z)$ with $x > z > y$. On the other hand, by ^[1], if $B\equiv2\pmod4$, then $(1.2)$ has no solutions $(x, y, z)$ with $y > z > x$. Therefore, by Lemma $2.2$, if $A > \max\{123B, B^{3/2}\}$ and $B\equiv2\pmod4$, then $(1.2)$ has no solutions $(x, y, z)$ with $(x, y, z)\not = (1, 1, 1)$. Thus, the corollary is proved.□

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Conflict of interest

The author declares there is no conflict of interest.