On the exponential Diophantine equation $(a(a-l)m^{2}+1)^{x}+(alm^{2}-1)^{y} = (am)^{z}$

1.   Introduction

Let $\mathbb{Z}$, $\mathbb{N}$ denote the sets of integers, positive integers, respectively. Let $a$, $b$, $c$ be fixed relatively prime positive integers greater than one. There have been many papers investigating the positive integer solutions of the exponential Diophantine equation

A lot of interesting results have been obtained. In recent years, T. Miyazaki, N. Terai and other scientific research workers have also made certain progress. For detailed historical research background, see ^[15,21].

In 1956, Sierpinski ^[20] considered the case of $(a, b, c)$ = $(3, 4, 5)$, and showed that $(x, y, z) = (2, 2, 2)$ is the only positive integer solution. Jesmanowicz ^[9] conjectured that if $a$, $b$, $c$ are Pythagorean numbers, then the Eq (1.1) has only the positive integer solution $(x, y, z) = (2, 2, 2)$. As an analogue of Jesmanowicz's conjecture, Terai ^[22] conjectured that the exponential Diophantine equation

has only the positive integer solution $(x, y, z) = (p, q, r)$ except for a handful of triples $(a, b, c).$ Exceptional cases are listed explicitly in ^[24] and ^[16]. This conjecture has been proved to be true in many special cases, see ^{[5,8,14,17,23]}, but is still unsolved in general.

Now take the Diophantine equation

where $a$, $b$, $c$ and $m$ are given positive integers such that $a+b = c^{2}$. Some authors studied Eq (1.3) for some special values, see ^{[1,10,18,25,26]}. Terai and Hibino ^[25] studied the equation

where $m$ be a positive integer and $p$ be a prime. As $m\equiv1\pmod4$, $m\equiv1, 2\pmod3$, $p\equiv1\pmod4$, and $p < 3784$, they proved that the only solution of Eq (1.4) is $(x, y, z) = (1, 1, 2)$. The proof of this result is based on elementary methods and Baker's method. In 2019, Deng, Wu and Yuan ^[7] studied the equation

where $a$ and $m$ be positive integers such that $am\equiv1, 2\pmod3$, $a\equiv1\pmod2$, and $a\geq5$. They showed that the exponential Diophantine Eq (1.5) has only the positive integer solution $(x, y, z)$ = $(1, 1, 2)$ by using some results on generalized Ramanujan-Nagell equations and some elementary methods. In 2020, Kizildere and Soydan ^[11] studied the equation

where $p$ be a prime with $p\geq5$, $p\equiv3\pmod 4$ and $m$ be a positive integer. They proved that the Diophantine Eq (1.6) has only the positive integer solution $(x, y, z) = (1, 1, 2)$ where $pm\equiv1, 4\pmod 5$ by the methods of ^[25]. In the same year, Kizildere, Le and Soydan ^[12] studied the equation

where $l$, $m$, $a$ be positive integers such that $a\equiv1\pmod2$, $am\equiv1, 2\pmod3$, $l\equiv0\pmod3$. They showed that the exponential Diophantine Eq (1.7) has only the positive integer solution $(x, y, z)$ = $(1, 1, 2)$ with $\min\{a(a-l)m^{2}+1, alm^{2}-1\} > 30$ by using the Bilu-Hanrot-Voutier Theorem on primitive divisors of Lehmer numbers.

In this paper, we consider the Diophantine equation

where $a\equiv1\pmod2$, $a > l\geq3$ and there exists a prime factor $p$ of $l$ satisfying $a^{2}m^{2}\equiv-2\pmod p$. The following results are the main theorems of this paper.

Theorem 1.1. Suppose that $a$, $l$ and $m$ are positive integers with $a\equiv1\pmod 2$, $a > l\geq3$ and there exists a prime factor $p$ of $l$ satisfying $a^{2}m^{2}\equiv-2\pmod p$. If $l\equiv1\pmod2$ or $l\equiv0\pmod2$ and $\min\{a(a-l)m^{2}+1, alm^{2}-1\} > 30$, then Diophantine Eq$(1.8)$ has only the positive integer solution $(x, y, z) = (1, 1, 2)$.

Theorem 1.2. Suppose that $a$ and $m$ are positive integers with $a = l$. Then Diophantine Eq $(1.8)$ has positive integer solutions $(x, y, z) = (n, 1, 2)$, $n\in\mathbb{N}$.

As a corollary to Theorem 1.1, taking $l = 3\pmod6$, we get a general result as follows :

Corollary 1.3. Suppose that $a$ and $m$ are positive integers with $a\geq5$, $a\equiv1\pmod2$, and $3\not|am$. Then Diophantine equation

has only the positive integer solution $(x, y, z) = (1, 1, 2)$.

As another corollary to Theorem 1.1, taking $l\equiv0\pmod3$, we give a general result as follows:

Corollary 1.4. ^[12] Suppose that $a$, $m$ and $l$ are positive integers with $a\geq5$, $a\equiv1\pmod2$, $3\not|am$ and $l\equiv0\pmod3$. Then Diophantine equation

has only the positive integer solution $(x, y, z) = (1, 1, 2)$ with $\min\{a(a-l)m^{2}+1, alm^{2}-1\} > 30$.

By Corollary 1.3, we know that Eq (1.8) has only the positive integer solution $(x, y, z)$ = $(1, 1, 2)$ when $a = 11$ and $3\not|m$. By using a result on linear forms in $p$-adic logarithms due to Bugeaud ^[2], we give a general result as follows:

Corollary 1.5. If $m > 2360$ or $3\not|m$, then the exponential Diophantine equation

has only the positive integer solution $(x, y, z) = (1, 1, 2)$.

It's easy to see, letting $a$ a prime and $l = 3$ in Theorem 1.1 gives us the result of ^[25]. Picking $l = 3$ in Theorem 1.1 yields the result of Deng, Wu and Yuan ^[7] and taking $l\equiv0\pmod3$ in Theorem 1.1 yields the result of Kizildere, Le and Soydan ^[12], respectively.

This paper is organized as follows. First of all, in Section 2, we show some preliminary lemmas which are needed in the proofs of our main results. Then in Section 3, we give the proof of Theorem 1.1. Section 4 is devoted to the proof of Theorem 1.2. Finally, Section 5 is devoted to the proof of Corollary 1.5.

2.   Preliminaries

In this section, we present several auxiliary lemmas that are needed in the proof of Theorem 1.1 and Theorem 1.2. Firstly, we will introduce here some definitions.

For any positive integer $n$, let $F_{n}$ and $L_{n}$ be the $n$-th Fibonacci number and Lucas number, respectively.

Next, we recall some useful lemmas.

Lemma 2.1. ^[13] For a fixed solution $(X, Y, z)$ of the equation

there exists a unique positive integer $L$ such that

where $\alpha$, $\beta$ are integers with $\beta X-\alpha Y = 1$ and $\{D_{1}, D_{2}, k\}\in\mathbb{Z}$, $\min\{D_{1}, D_{2}, k\} > 1$, $\gcd(D_{1}, D_{2}) = 1$, $2\not|k$.

The positive integer $L$ defined as in Lemma 2.1 is called the characteristic number of the solution $(X, Y, z)$ and is denoted by $\langle X, Y, z \rangle.$

Lemma 2.2. ^[13] If $\langle X, Y, z\rangle = L$, then $D_{1}X\equiv-LY\pmod k.$

For a fixed positive integer $L_{1}$, if Eq (2.1) has a solution $(X_{1}, Y_{1}, z_{1})$ with $(X_{1}, Y_{1}, z_{1}) = L_{1}$, then the set of all solutions $(X, Y, z)$ of Eq (2.1) with $(X, Y, z)\equiv\pm L_{1}\pmod k$ is called a solution class of (2.1) and is denoted by $S(L_{1})$.

Lemma 2.3. ^[13] For any fixed solution class $S(L_{1})$ of Eq $(2.1)$, there exists a unique solution $(X_{1}, Y_{1}, z_{1})\in S(L_{1})$ such that $X_{1}\geq1$, $Y_{1}\geq1$, and $z_{1}\leq z$, where $z$ runs through all solution $(X, Y, z)\in S(L_{1})$.The solution$(X_{1}, Y_{1}, z_{1})$ is called the least solution of $S(L_{1})$. Every solution $(X, Y, z)\in S(L_{1})$ can be expressed as

Lemma 2.4. ^[3]Let $(X_{1}, Y_{1}, z_{1})$ be the least solution of $S(L_{1})$. If Eq $(2.1)$ has solution $(X, Y, z)\in S(L_{1})$ satisfying$X\geq1$ and $Y = 1$, then $Y_{1} = 1$. Furthermore, if $(X, z)\neq(X_{1}, z_{1})$, then one of the following conditions issatisfied:

$(1)$ $D_{1}X_{1}^{2} = \frac{1}{4}(k^{z_{1}}\pm1), D_{2} = \frac{1}{4}(3k^{z_{1}}\mp1), (X, z) = (X_{1}|D_{1}X^{2}_{1}-3D_{2}|, 3z_{1}).$

$(2)$ $D_{1}X_{1}^{2} = \frac{1}{4}F_{3r+3\varepsilon}, D_{2} = \frac{1}{4}L_{3r}, k^{z_{1}} = F_{3r+\varepsilon}, (X, z) = (X_{1}|D_{1}^{2}X^{4}_{1}-10D_{1}D_{2}X^{2}_{1}+5D_{2}^{2}|, 5z_{1}),$where $r$ is a positive integer, $\varepsilon\in\{1, -1\}$.

Lemma 2.5. ^[6] For any positive integer $n$, let $F_{n}$ be the n-th Fibonacci number. The equation

has only the solutions $(n, u) = (1, 1), (2, 1)$, and $(12, 12)$.

Lemma 2.6. ^[19] Suppose that $x$, $y$, $m$, $n$ are positive integers greater than one. Then the equation

has only the positive integer solution $(x, y, m, n) = (3, 2, 2, 3).$

Finally, we need a result on primitive divisors of Lehmer numbers due to Bilu, Hanrot, Voutier ^[4].

Let $\alpha$, $\beta$ be algebraic integers. If $(\alpha+\beta)^{2}$, $\alpha\beta$ are nonzero coprime integers, and $\frac{\alpha}{\beta}$ is not a root of unity, then $(\alpha, \beta)$ is called a Lehmer pair. Let $E = (\alpha+\beta)^{2}$ and $G = \alpha\beta$. Then we have

where $F = E-4G$. Further, one defines the corresponding sequence of Lehmer numbers by

Obviously, $L_{n}(\alpha, \beta), (n\geq1)$ are nonzero integers.

A prime $q$ is called a primitive divisor of the Lehmer number $L_{n}(\alpha, \beta), (n\geq1)$ if $q|L_{n}(\alpha, \beta)$ and $q\not|FL_{1}(\alpha, \beta)...L_{n-1}(\alpha, \beta).$

Lemma 2.7. ^[4] If $n > 30$, then $L_{n}(\alpha, \beta)$ has primitive divisors.

Lemma 2.8. Let $m\geq2100$ be a positive integer and $(x, y, z)$ be a positive integer solution of Eq $(1.11)$. Suppose that $N = \max\{x, y\}$, then $z > 1.87N.$

Proof. Let $(x, y, z)$ be a positive integer solution of Eq $(1.11)$, then we derive that

and

By (2.5) and (2.6), we get that

and

Notice that $f(m)$ and $g(m)$ are increasing and $m\geq2100$, we get that

and

If $N = \max\{x, y\} = x$, then

If $N = \max\{x, y\} = y$, then

Therefore, we conclude that if $m\geq2100$ and $N = \max\{x, y\}$, then

This completes the proof of Lemma 2.8.

3.   Proof of Theorem 1.1

In this section, we give a proof of Theorem 1.1.

Lemma 3.1. Let $(x, y, z)$ be a positive integer solution of Eq $(1.8)$. Suppose there exists a prime factor $p$ of $l$ satisfying $a^{2}m^{2}\equiv-2\pmod p$. Then both $x$ and $y$ are odd.

Proof. Since $a > l\geq3$, we see from Eq (1.8) that

and $y$ is odd.

Since $a^{2}m^{2}\equiv-2\pmod p$, by Eq (1.8) and $y$ is odd, we have

Hence, we obtain from (3.1) that $x$ is odd.

Lemma 3.1 is proved.

3.1.   The case $m\equiv0\pmod2$

Lemma 3.2. Suppose that $m\equiv0\pmod2$. Then Eq $(1.8)$ has only the positive integer solution $(x, y, z) = (1, 1, 2)$.

Proof. If $z\leq2$, then $(x, y, z) = (1, 1, 2)$ from Eq (1.8). Hence, we may assume that $z\geq3$. Taking Eq (1.8) modulo $m^{3}$ implies that

That is,

which is impossible, since both $x$ and $a$ are odd. Therefore, we conclude that if $m$ is even, then Eq (1.8) has only the positive integer solution $(x, y, z) = (1, 1, 2).$ Lemma 3.2 is proved.

3.2.   The case $m\equiv1\pmod2$

Lemma 3.3. Suppose that $m$ is odd and there exists a prime factor $p$ of $l$ satisfying $a^{2}m^{2}\equiv-2\pmod p$. Then Eq $(1.8)$ has only the positive integer solution $(x, y, z) = (1, 1, 2)$.

Proof. We assume that Eq $(1.8)$ has another positive integer solution $(x, y, z)\neq(1, 1, 2)$. By Lemma 3.1, we get that both $x$ and $y$ are odd positive integers. Thus, Eq $(1.8)$ yields the following equation:

where

Hence, $(X, Y, z) = ((a(a-l)m^{2}+1)^{\frac{x-1}{2}}, (alm^{2}-1)^{\frac{y-1}{2}}, z)\neq(1, 1, 2)$ is a positive integer solution of Eq $(3.2)$ with $\gcd(X, Y) = 1.$

Let $L = \langle(a(a-l)m^{2}+1)^{\frac{x-1}{2}}, (alm^{2}-1)^{\frac{y-1}{2}}, z)\rangle$. By Lemma 2.2 and Eq $(3.2)$, we know that $L$ satisfies

On the other hand, let $L_{1} = \langle1, 1, 2\rangle$, since $(X_{1}, Y_{1}, z_{1}) = (1, 1, 2)$ is a positive integer solution of Eq $(3.2)$. Then by Lemma 2.2 we have

It implies that $L\equiv\pm L_{1}\pmod {am},$ and it is obvious that $(X_{1}, Y_{1}, z_{1}) = (1, 1, 2)$ is the least solution of $S(L_{1})$, since $z\geq2$. Hence, $(X, Y, z)\in S(L_{1})$. Therefore, using Lemma 2.3, we get that

By Eq (3.4), we get

and

Now, we consider the following two cases.

Case 1. $l\equiv1\pmod2.$ Since $2|(alm^{2}-1)$ and $2\not|(a(a-l)m^{2}+1)$, we see from Eq $(3.5)$ that $y = 1$ and $(alm^{2}-1)^{\frac{y-1}{2}} = 1$. It implies that $(X, Y, z)$ is a solution of $S(L_{1})$ satisfying $Y = 1$ and $(X, z)\neq(X_{1}, z_{1}) = (1, 2)$.

Therefore, by Lemma 2.4, we get either

or

By Eq $(3.7)$, we can obtain

and

If Eq $(3.9)$ holds, taking modulo 4, we get $6\equiv0\pmod4$, which is impossible.

If Eq $(3.10)$ holds, we get $am^{2}\mid5$, that is, $a = 5$, $m = 1$ and $l = 4$, which is impossible.

By Lemma 2.5, we know that Eq $(3.8)$ has no positive integer solution, since $am\equiv1\pmod2$.

Hence, Eq $(1.8)$ has only the positive integer solution $(x, y, z) = (1, 1, 2)$ with $l\equiv1\pmod2.$

Case 2. $l\equiv0\pmod2$ and $\min\{a(a-l)m^{2}+1, alm^{2}-1\} > 30.$

By Eqs (3.5) and (3.6), we get either $a(a-l)m^{2}+1|t$ or $alm^{2}-1|t$. This implies that

Further, since $\min\{a(a-l)m^{2}+1, alm^{2}-1\} > 30$, we see from (3.11) that

Let

Then $(\alpha+\beta)^{2} = 4(alm^{2}-1), \alpha\beta = (am)^{2}$ are nonzero coprime integers, and $\frac{\alpha}{\beta}$ is not a root of unity. Hence, $(\alpha, \beta)$ is a Lehmer pair. By Eqs (3.5) and (3.13), we have

We see from (3.14) that the Lehmer number $L_{t}(\alpha, \beta)$ has no primitive divisors. But, by Lemma 2.7, we find from (3.14) that this is false.

Thus, under the assumption, Eq (1.8) has only the positive integer solution $(x, y, z) = (1, 1, 2)$ with $l\equiv0\pmod2.$

This completes the proof of Theorem 1.1.

4.   Proof of Theorem 1.2

Proof. Since $l = a$, Eq (1.8) can be written as

By Lemma 2.6, we obtain that Eq (4.1) has only positive integer solutions $(x, y, z) = (n, 1, 2)$, $n\in\mathbb{N}$.

This completes the proof of Theorem 1.2.

5.   Proof of Corollary 1.5

In order to obtain an upper bound for $m$, we shall quote a result on linear forms in $p$-adic logarithms due to Bugeaud ^[2]. Here we consider the case where $y_{1} = y_{2} = 1$ in the notation of ^[2], page 375.

Let $p$ be an odd prime. Let $a_{1}$ and $a_{2}$ be nonzero integers prime to $p$. Let $g$ be the least positive integer such that

where we denote the $p$-adic valuation by $v_{p}(\cdot)$. Assume that there exists a real number $E$ such that

We consider the integer

where $b_{1}$ and $b_{2}$ are positive integers. We let $A_{1}$ and $A_{2}$ be real numbers greater than one with

and we put

Proposition 5.1. ^[2] With the above notation, if $a_{1}$ and $a_{2}$ are multiplicatively independent, then we have the upper estimates

Proof. Now let $(x, y, z)$ be a solution of Eq (1.11). By Corollary 1.3 and Lemma 3.2, we know that Eq (1.11) has only the positive integer solution $(x, y, z) = (1, 1, 2)$ when $3\not|m$ or $2|m$. Thus, the proof of Corollary 1.5 suffices to prove the case $m > 2360$, $3|m$, $2\not| m$. Recall that $y$ is odd. Here, we apply Proposition 5.1. For this, we set $p = 11$, $a_{1} = 88m^{2}+1$, $a_{2} = 1-33m^{2}$, $b_{1} = x$, $b_{2} = y$, and

Then we may take $g = 1$, $E = 1$, $A_{1} = 88m^{2}+1$, $A_{2} = 33m^{2}-1$. Hence by (5.1), we have

where

Suppose that $z\geq4$. Taking Eq (1.11) modulo $m^{4}$ we have

In particular, we put $N = \max\{x, y\}$, then $N\geq \frac{m^{2}}{121}$. Since $z > N$ and $B\leq\frac{N}{\log5m}$ we have that

Let

Note that $6\log11 > 5$, then

If

Then the inequality $\log N > 6\log11-\log(\log 11)-0.4$ implies that $N > 495231$. On the other hand, from (5.3) we have that

which implies that $N < 66033$, a contradiction. Hence $H(N, m) = 6\log11$ and therefore from (5.3) we have the inequality

This implies that $m\leq3342$.

If $2100\leq m\leq3342$, by Lemma 2.8 we have

Then $m\leq2359$, which contradicts the fact that $m > 2360$.

We conclude $z\leq3$. If $z = 3,$ let $a = 88m^2+1, b = 33m^2-1$ we obtain that

If $y = 1$, we get from (5.6) that $a|b^2(b-1)$. It follows that $a = 88m^2+1|b-1 = 33m^2-2$ since $\gcd(a, b) = 1$, which is impossible. If $x = 1$, we get from (5.6) that $b|a^2(a-1)$. It follows that $b = 33m^2-1|a-1 = 88m^2$, so $33m^2-1|11m^2-3$, which is impossible. Thus we have $x > 1$ and $y > 1$. This will lead to

a contradiction.

Hence $z\leq2$. In this case, one can easily show that $(x, y, z) = (1, 1, 2)$.

This completes the proof of Corollary 1.5.

Remark. It is easy to see that the Diophantine equation $(a(a-l)m^2+1)^x+(alm^2-1)^y = (am)^z$ has solution $(x, y, z) = (1, 1, 2)$ if we remove all the conditions in Theorem 1.1. How can prove it? On the other hand, it is worth noting that from Theorem 1.1, the condition $\min\{a(a-l)m^{2}+1, alm^{2}-1\}\leq30$ still has the value of further research.

Acknowledgments

The Authors express their gratitude to the anonymous referee for carefully examining this paper and providing a number of important comments and suggestions. This research was supported by the Major Project of Education Department in Sichuan (No. 16ZA0173) and NSF of China (No. 11871058)and Nation project cultivation project of China West Normal University.

Conflict of interest

All authors declare no conflicts of interest in this paper.