On the exponential Diophantine equation $\left(\frac{q^{2l}-p^{2k}}{2}n\right)^x+(p^kq^ln)^y = \left(\frac{q^{2l}+p^{2k}}{2}n\right)^z$

1.   Introduction

Let $a, b, c$ be fixed positive integers. Consider the exponential Diophantine equation

The following problem proposed by Jeśmanowicz ^[5] has been actively studied in the field of Eq (1.1).

Conjecture 1.1. Assume that $a^2+b^2 = c^2.$ Then the Eq (1.1) has no positive integer solution $(x, y, z)$ other than $x = y = z = 2.$

The pioneering work related to Conjecture 1.1 was given by Sierpinśki ^[11], he showed that $(2, 2, 2)$ is the unique positive solution of the equation $3^x +4^ y = 5^z.$ In the same journal, Jeśmanowicz ^[5] obtained the same conclusion for the following cases:

and furthermore he proposed Conjecture 1.1. After these works, Conjecture 1.1 has been proved to be true for various particular cases. For recent results, we only refer to the papers of Deng et al. ^[3], Hu and Le ^[4], Miyazaki ^[8,10], Miyazaki et al. ^[9], Terai ^[12], Yuan and Han ^[16] and the references given there. Most of the existing works on Conjecture 1.1 concern the coprimality case, that is, $\gcd(a, b) = 1.$ Indeed, all of the above mentioned results treat the coprimality case, and such a case is essential in the study of the Eq (1.1). Actually, the non-coprimalty case, i.e. $\gcd(a, b) > 1,$ is a degenerate one in the sense that Eq (1.1) can be often solved only by local arguments using the prime factors of $\gcd(a, b).$ Recently, several authors actively studied the non-coprimality case about Conjecture 1.1. For any primitive Pythagorean triple $(a, b, c)$, we can write

where N is a positive integer. Without loss of generality, we may assume $b$ is even. Then, (1.1) becomes

This equation has been solved for some special triples $(a, b, c),$ without assuming any conditions on $N$. For some results in this direction, we refer to the papers of Deng and Cohen ^[2], Yang and Tang ^[14,15], Deng ^[1], Ma and Chen ^[7], and the references given there. In particular, Tang and Weng ^[13] very recently solved Eq (1.2) for the case where $(a, b, c)$ is expressed as

where $r$ is any positive integer. Note that this is the first result dealing with (1.2) for infinitely many triples $(a, b, c)$. Miyazaki ^[10] extend this result as follows: If $b$ is a power of $2$, then Conjecture 1.1 is true. It is well known that any primitive Pythagorean triple $(a, b, c)$ is parameterized as follows:

where $u, v$ are co-prime positive integers of different parities with $u > v.$ In this notation, the mentioned result of Tang and Weng corresponds to $(u, v) = (2^{2^{r-1}}, 1)$ with $r \geq 1$, and the result of Miyazaki ^[10] corresponds to $(u, v) = (2^r, 1)$ with $r \geq 1.$ In this paper we consider the exponential Diophantine equation

where $k, l, n$ are positive integers and both $p$ and $q$ are odd primes such that $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$, where $a$ is odd prime with $a\equiv 5\pmod 8$ and $a\not\equiv 1\pmod 5$, $m_1$ and $m_2$ are positive integers. We obtain the following:

Theorem 1.1. Let $k, l, m_1, m_2$ be positive integers and let both $p$ and $q$ be odd primes such that $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$, where $a$ is odd prime with $a\equiv 5\pmod 8$ and $a\not\equiv 1\pmod 5$. Then the Eq (1.3) has only the positive integer solution $(x, y, z) = (2, 2, 2).$

This paper is organized as follows. First of all, in Section 2, we show some preliminary lemmas which are needed in the proof of Theorems 1.1. Then in Section 3, we give the proof of Theorem 1.1. Finally in Section 4, we give some examples of applications of Theorems 1.1.

2.   Preliminaries

In this section, we present some lemmas that will be used later.

Lemma 2.1. (^[6]) If $(x, y, z)$ is a solution of (1.2) with $(x, y, z)\neq (2, 2, 2)$, then one of the following conditions is satisfied

(i) $max\{x, y\}>min\{x, y\}>z$;

(ii) $x>z>y$;

(iii) $y>z>x$.

Lemma 2.2. (^{[2, 10]}) Assume that $n>1$, then (1.2) has no solution $(x, y, z)$ with $max\{x, y\}>min\{x, y\}>z$.

Lemma 2.3. Let $k, l, m_1, m_2$ be positive integers and let both $p$ and $q$ be odd primes such that $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$, where $a$ is odd prime with $a\equiv 5\pmod 8$. Then $m_2\equiv 1\pmod 2$.

Proof. On the contrary suppose that $m_2$ is even. If $m_1$ is also even, then we get from the condition

that

So

thus would give $k_2 = 0$ and $2^{\frac{m_1}{2}}-a^{\frac{m_2}{2}} = 1$. If $m_1 > 2$, then taking the equation $2^{\frac{m_1}{2}}-a^{\frac{m_2}{2}} = 1$ modulo $4$ yields $-1\equiv 1 \pmod 4$, which leads to a contradiction. Hence $m_1 = 2, m_2 = 0$, which contradicts the condition that $m_2$ is a positive integer. If $m_1$ is odd, then we get from the condition

that $q = 3$ since $q$ is prime. Taking modulo $4$ for the equation $3^l = 2^{m_1}+a^{m_2}$ would give $3^l\equiv 1\pmod 4$. It follows that $l$ is even and

which leads to a contradiction. This completes the proof.

Lemma 2.4. Assume that $n = 1$. If $(x, y, z)$ is a solution of the Eq (1.3) with $x\equiv y\equiv z \equiv 0 \pmod 2$, then $(x, y, z) = (2, 2, 2).$

Proof. It is easy to find that $m_1\geq 3$ by the condition $p^k = 2^{m_1}-a^{m_2}$. We may write $x = 2x_1, y = 2y_1, z = 2z_1$ by the assumption $x\equiv y\equiv z \equiv 0 \pmod 2$. It follows from (1.3) that

Since

then

or

or

If (2.1) holds, then taking modulo $4$ for the former equation we get that

It follows that $y_1$ is odd. On the other hand, subtracting the right equation from the left one yields

As

and

we get

Adding the two equations gives

We claim that $y_1 = 1$. On the contrary suppose $y_1 > 1$. Note that $y_1$ is odd, Eq (2.4) would give that

Thus $y_1a^{m_2(y_1-1)}\equiv 0 \pmod 2$, which is a contradiction. Therefore $y_1 = 1$ and $2^{(m_1+1)x_1} = 2^{m_1+1}$ yields that $x_1 = 1$. Substituting these values $x = y = 2$ and $n = 1$ into Eq (1.3) gives $z = 2$.

If (2.2) holds, then taking modulo $4$ for the former equation we get that

It follows that $y_1$ is odd. We then get taking modulo $a$ for the former equation that

It follows that $z_1$ is even. Finally adding the two equations and then dividing it by $2$ gives that

which is impossible. Hence the Eq (2.2) is not true.

The Eq (2.3) is obviously not true since

This completes the proof.

Lemma 2.5. Assume that $n = 1$. Then the Eq (1.3) has only the positive integer solution $(x, y, z) = (2, 2, 2).$

Proof. It is easy to find that is enough to prove that $x\equiv y\equiv z \equiv 0 \pmod 2$ by Lemma 2.4. Substituting the conditions $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$ into the Eq (1.3) gives

Taking modulo $4$ for the above equation we get that

It follows that $y\equiv 0 \pmod 2$. We now prove that $z$ is even. Taking modulo $a$ for the Eq (2.6) gives

It follows that $m_1z\equiv 0 \pmod 2$. We claim that $z\equiv 0 \pmod 2$. On the contrary suppose that $z$ is odd, then we must have that $m_1$ is an even. Hence $m_2$ is odd by Lemmas 2.3. On the other hand, since

then $(-1)^x = (-1)^z$ would give $x$ is odd. Again taking the Eq (2.6) modulo $3$ will lead to $a^{m_2x}\equiv 1 \pmod 3$. Since $m_2$ and $x$ are both odd, this means that $a\equiv 1\pmod 3$. Therefore, $q^l\equiv 0\pmod 3$, so that $q = 3$. This gives

a contradiction. Therefore $z$ is even. Finally we prove $x$ is also even. The congruence modulo $2^{m_1}-a^{m_2}$ of the Eq (2.6) gives

Notice that

we have that

This means that $x$ is also even. This completes the proof.

3.   Proof of Theorem 1.1

Assume that $(x, y, z)$ is a positive integer solution with $(x, y, z)\neq (2, 2, 2)$. Then we have by Lemmas 2.1, 2.2 and 2.5 that $n > 1$ and either $x > z > y$ or $y > z > x$. We shall discuss separately two cases.

The case $x > z > y$. Then dividing Eq (1.3) by $n^y$ yields

Since $\gcd\left((p^kq^l)^y, \left(\frac{q^{2l}+p^{2k}}{2}\right)^z\right) = 1$, we can observe that the two factors on the right-hand side are co-prime. Hence then Eq (3.1) yields $n = p^u$ for some positive integer $u$ and

or $n = q^v$ for some positive integer $v$ and

or $n = p^uq^v$ for some positive integers $u$ and $v$ and

If (3.2) holds, then substituting the conditions $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$ into Eq (3.2) would give

Taking modulo $8$ for Eq (3.5) leads to $a^{m_2y}\equiv 1 \pmod 8$. It follows that $y$ is even since $m_2$ is odd by Lemma 2.3. Taking modulo $8$ for equation $p^k = 2^{m_1}-a^{m_2}$ leads to $p\equiv -a\equiv 3 \pmod 8$. Again taking modulo $p$ for Eq (3.5) leads to

It follows that

Therefore $z$ is even. Then we get from Eq (3.5) that

Since

so it follows that

however, this is impossible since

If (3.3) holds, then substituting the conditions $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$ into Eq (3.3) would give

Then taking modulo $4$ for Eq (3.6) leads to $(-1)^{y}\equiv 1 \pmod 4$. It follows that $y$ is even. Taking modulo $8$ for equation $q^k = 2^{m_1}+a^{m_2}$ leads to $q\equiv a\equiv 5 \pmod 8$. Again taking modulo $q$ for Eq (3.6) leads to

It follows that

Therefore $z$ is even. Then similarly we get from Eq (3.6) that

which is impossible by the above result that has been proved.

If (3.4) holds, then substituting the conditions $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}$ into Eq (3.4) would give

Taking modulo $8$ for equation $q^k = 2^{m_1}+a^{m_2}$ leads to $q\equiv a\equiv 5 \pmod 8$. Again taking modulo $q$ for Eq (3.6) leads to

It follows that $(-1)^z = \left(\frac{2}{q}\right)^z = \left(\frac{1}{q}\right) = 1$. Therefore $z$ is even. Then similarly we get from Eq (3.7) that

which is impossible by the above result that has been proved.

The case $y > z > x$. Then dividing Eq (1.3) by $n^x$ yields

It is easy to see that the two factors on the right-hand side are co-prime. Thus, Eq (3.8) yields $n = a^s$ for some positive integer $s$ and

or $n = 2^r$ for some positive integer $r$ with $(m_1+1)x = r(z-x)$ and

or $n = 2^ra^s$ for some positive integers $r$ and $s$ and

If (3.9) holds, taking modulo $4$ for Eq (3.9) leads to $(-1)^{y}\equiv 1 \pmod 4$. It follows that $y$ is even. Taking modulo $a$ for Eq (3.9) leads to

It follows that

It follows that $(m_1+1)x$ is even. Then we get from Eq (3.9) that

It follows either

or

Hence

which is impossible since

If (3.10) holds, we first prove that $r(y-z) = 2$. In fact, taking modulo $4$ for Eq (3.10) yields $2^{r(y-z)}\equiv 0 \pmod 4$. Therefore $r(y-z)\geq 2$. On the other hand, if $r(y-z)\geq 3$, then taking modulo $8$ for Eq (3.10) leads to $a^{m_2x}\equiv 1\pmod 8$. It follows that $m_2x$ is even. Taking modulo $3$ for Eq (3.10) leads to $1\equiv a^{m_2x}\equiv 2^z\equiv (-1)^z \pmod 3,$ which implies that $z$ is even. Then we get from Eq (3.10) that

It follows either

or

Hence

which is impossible since

So $r(y-z)\leq 2$ and $r(y-z) = 2$. We now prove that

First taking modulo $8$ for Eq (3.10) we get that

It follows that $m_2x$ is odd. Again taking modulo $2^{m_1}-a^{m_2}$ for Eq (3.10) leads to

It follows that

since

Thus $m_1m_2x\equiv z\pmod 2$. If $z$ is even, then we get that $m_1$ is also even. Finally taking modulo $a$ for Eq (3.10) leads to $2^{2m_1z}\equiv 2^{2m_1y+2}\pmod a$. It follows that

which leads to a contradiction. Therefore we must have

Notice that $(m_1+1)x = r(z-x)$ and $r(y-z) = 2$, we must have that

Taking modulo $3$ for Eq (3.10) leads to $a\equiv a^{m_2x}\equiv 2^z\equiv 2 \pmod 3.$ Taking modulo $3$ for equation $p^k = 2^{m_1}-a^{m_2}$ leads to $p^k\equiv 0\pmod 3$. It follows that $p = 3$ and taking modulo $4$ for equation $3^k = 2^{m_1}-a^{m_2}$ we get that $k$ is odd. If $k\equiv 3\pmod 4$, we then get taking modulo $5$ for equation $3^k = 2^{m_1}-a^{m_2}$ that $a\equiv 1\pmod 5$, which contradicts to the condition $a\not\equiv 1\pmod 5$. If $k\equiv 1\pmod 4$, then taking modulo $5$ for equation $3^k = 2^{m_1}-a^{m_2}$ leads to

It follows that $a = 5$ since $a$ is prime. Substituting the Eq (3.12) and $a = 5$ into the Eq (3.10), we get that

If $m_1\equiv m_2 \pmod 3$, say $m_1\equiv m_2\equiv \lambda \pmod 3$. Then applying Fermat's little theorem to Eq (3.13) yields

which implies that $5^{m_2x}\equiv {2^{(2\lambda +1)z}} \pmod 7$. This leads to $-1 = \left(\frac{5}{7}\right) = \left(\frac{2}{7}\right) = 1$, a contradiction. So in the following discussion we will assume that $m_1\not\equiv m_2 \pmod 3$. We now distinguish three cases.

Case 1: $m_1\equiv 0 \pmod 3$. Then $m_1\equiv 3 \pmod {12}$.

Subcase 1.1: $m_2\equiv 1 \pmod 3$. Then $m_2\equiv 1 \pmod 6$. Applying Fermat's little theorem to Eq (3.13), we get that $2^x\equiv 3 \pmod 7$. It follows that

which is a contradiction.

Subcase 1.2: $m_2\equiv 2 \pmod 3$. Then $m_2\equiv 5 \pmod 6$. Applying Fermat's little theorem to Eq (3.13), we get that $4^x\equiv 5 \pmod 7$. It follows that

which is a contradiction.

Case 2: $m_1\equiv 1 \pmod 3$. Then $m_1\equiv 7 \pmod {12}$.

Subcase 2.1: $m_2\equiv 0 \pmod 3$. Then $m_2\equiv 3 \pmod 6$. Applying Fermat's little theorem to Eq (3.13), we get that $2^{x+1}+2^{2x}\equiv 1 \pmod 7$. If $x\equiv 0 \pmod 3$, then $x\equiv 3 \pmod 6$, which yields to $3\equiv 1 \pmod 7,$ a contradiction. If $x\equiv 2 \pmod 3$, then $x\equiv 5 \pmod 6$, which yields to $3\equiv 1 \pmod 7,$ a contradiction again. Therefore $x\equiv 1 \pmod 3$, then $x\equiv 1 \pmod 6$. Again applying Euler's theorem to Eq (3.13), we have that $-1\equiv {5^5}\equiv -7 \pmod 9$, which is also impossible.

Subcase 2.2: $m_2\equiv 2 \pmod 3$. Then $m_2\equiv 5 \pmod 6$. Applying Fermat's little theorem to Eq (3.13), we get that $-1\equiv 0 \pmod 7$, which is a contradiction.

Case 3: $m_1\equiv 2 \pmod 3$. Then $\frac{m_1+3}{2}\equiv {7} \pmod {12}$ or $\frac{m_1+3}{2}\equiv {1} \pmod {12}$.

Subcase 3.1: $m_2\equiv 0 \pmod 3$. Then $m_2\equiv 3 \pmod {6}$. Applying Fermat's little theorem to Eq (3.13), we get that $3^{x-1}\equiv 1 \pmod 7$. It follows that $x\equiv 1 \pmod 6$, which implies either $x\equiv 1 \pmod {12}$ or $x\equiv 7 \pmod {12}$. If the first case holds, then applying Fermat's little theorem to Eq (3.13), we get either

or

which leads to a contradiction. If the last case holds, then using Fermat's little theorem to Eq (3.13), we get either

or

which leads to a contradiction again.

Subcase 3.2: $m_2\equiv 1 \pmod 3$. Then $m_2\equiv 1 \pmod 6$. Using Fermat's little theorem to Eq (3.13), we get that

This leads to $0\equiv 1 \pmod 7$, a contradiction.

If (3.11) holds, taking modulo $2^{m_1}-a^{m_2}$ for Eq (3.11) leads to

It follows that

Thus $z$ is even. Then we get from Eq (3.11) that

It follows either

or

Hence

which is impossible since

This completes the proof.

4.   Applications

In this section, we give some examples of applications of the result (see Table 1).

From the Table 1, one can easily see that the Conjecture 1.1 is true for the following cases:

5.   Conclusions

Jeśmanowicz' conjecture is true for the following set of Pythagorean numbers:

where $p$ and $q$ are odd primes such that $p^k = 2^{m_1}-a^{m_2}$ and $q^l = 2^{m_1}+a^{m_2}, a$ is odd prime with $a\equiv 5\pmod 8$ and $a\not\equiv 1\pmod 5$.

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 (N0. 202118).

Conflict of interest

All authors declare no conflicts of interest in this paper.