On a conjecture concerning the exponential Diophantine equation $(an^{2}+1)^{x}+(bn^{2}-1)^{y} = (cn)^{z}$

1.   Introduction

Let $\mathbb{N}$, $\mathbb{Z}$, and $\mathbb{Q}$ be the set of positive integers, the set of integers, and the set of rational numbers, respectively. Let $a$, $b$, $c$, and $n$ be positive integers such that

Under the assumption (1.1), Fujita and Le proposed the following conjecture:

Conjecture 1.1. [1, Conjecture 1.1] The exponential Diophantine equation

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

There is a more general conjecture that readers can refer to [2, Conjecture 1]. In 2012, Terai ^[3] verified that if $(a, b, c) = (4, 5, 3)$, then Conjecture 1.1 is true except for $n > 20$ and $n\equiv3\pmod6$. The proof of this result is based on elementary methods and Baker's method. In 2014, using some properties of exponential diophantine equations (see ^[4,5]) and some results on the existence of primitive divisors of Lucas numbers (see ^[6]), Su and Li ^[7] proved that if $n > 90$ and $n\equiv3\pmod 6$, then Conjecture 1.1 is true for $(a, b, c) = (4, 5, 3)$. Two years later, Bertók ^[8] showed Conjecture 1.1 is true when $(a, b, c) = (4, 5, 3)$ by completely solving Eq (1.2) for the remaining cases $20 < n < 90$ and $n\equiv3\pmod6$ with the help of exponential congruences. This is a nice application of Bertók and Hajdu ^[9]. On the other hand, Miyazaki and Terai ^[10] showed that if $a = 1$ and $c\equiv3, 5\pmod8$, under the condition $n\equiv\pm1\pmod c$, then Conjecture 1.1 is true except for the case $(n, b, c) = (1, 8, 3)$. Pan ^[11] proved that if $a\equiv4, 5\pmod8$, $\big(\frac{a+1}{c}\big) = -1$ and $n > 6c^{2}\log c$, under the condition $n\equiv\pm1\pmod c$, then Conjecture 1.1 is true, where $\big(\frac{a+1}{c}\big)$ is the Jacobi symbol. Fu and Yang ^[12] showed that if $a\equiv0\pmod2$ and $n > 36c^{3}\log c$, under the condition $c\mid n$, then Conjecture 1.1 is true. Kizildere et al. ^[13] proved that if $a = b+1$, $c\equiv11, 13\pmod{24}$ and $n > c^{2}$, under the condition $n\equiv\pm1\pmod c$, then Conjecture 1.1 is true. From these works, one can know that studying $c\mid n$ and $n\equiv\pm1\pmod c$ for Eq $(1.2)$ plays an important role in solving Conjecture 1.1.

In this paper, using an elementary approach and a deep result on linear forms in two $m$-adic logarithms due to Bugeaud ^[14], we investigate Conjecture 1.1 by handling the case $\gcd(c, n) = 1$ (which contains the case $n\equiv\pm1\pmod c$) for Eq $(1.2)$, and prove the following result:

Theorem 1.2. Let $a$, $b$, $c$, and $n$ be positive integers with $a+b = c^{2}$, $2\nmid c$ and $\gcd(an^{2}+1, bn^{2}-1) = 1$. If $\gcd(c, n) = 1$ and $n\geq 117.14c$, then Conjecture $1.1$ is true.

Evidently, Theorem 1.2 improves the result (see [1, Theorem 1.2]) of Fujita and Le when $\gcd(c, n) =$ $1$. Notice that $6c\log c > 117.14$ for any positive integer $c\geq9$, and the condition $\gcd(c, n) = 1$ contains $n\equiv\pm1\pmod c$. One can easily check that Theorem 1.2 extends the result of Pan ^[11] when $c\geq9$. We point out that the condition $\gcd(an^{2}+1, bn^{2}-1) = 1$ in Theorem 1.2 is equivalent to the existence of a positive integer $n$ such that $\gcd(an^{2}+1, bn^{2}-1) = 1$. Therefore, the condition $\gcd(an^{2}+1, bn^{2}-1) = 1$ is weaker than [1, Lemma 2.3].

When $(a, b, c) = (12, 13, 5)$, Terai and Hibino ^[15] proved that $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$ except for $n\equiv17, 33\pmod{40}$. When $(a, b, c) = (18, 7, 5)$, Alan ^[16] showed that $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$ except for $n\equiv23, 47, 63, 87\pmod{120}$. Recently, Hasanalizade ^[17] proved that if $(a, b, c) = (44, 5, 7)$ and $n\equiv2\pmod5$ or $n\equiv0, \pm1, \pm3\pmod7$, then $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$. The proofs of these results are based on elementary methods and Baker's method. On the other hand, Miyazaki and Terai ^[10], Bertók ^[8], and Terai and Hibino ^[18] completely solve Eq $(1.2)$ when $(a, b, c) = (1, 8, 3)$, $(4, 5, 3)$, and $(10, 15, 5)$, respectively. With the help of Theorem 1.2, we can completely solve the Eq $(1.2)$ for $(a, b, c) = (12, 13, 5)$, $(18, 7, 5)$, and $(44, 5, 7)$ without any assumption on $n$. Namely, we show the following result:

Theorem 1.3. For any positive integer $n$, if $(a, b, c) = (44, 5, 7)$, $(12, 13, 5)$, or $(18, 7, 5)$, then $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$.

As an immediate result of Theorem 1.3, we can obtain the following corollary:

Corollary 1.4. If $(a, b, c) = (44, 5, 7)$, $(12, 13, 5)$, or $(18, 7, 5)$, then Conjecture $1.1$ is true.

This paper is organized as follows: In Section 2, we present several lemmas that will be useful for the proofs of the main results. In Sections 3 and 4, we provide the proofs of Theorems 1.2 and 1.3, respectively.

2.   Preliminary results

For a prime number $p$ and a nonzero integer $x$, we write $v_{p}(x)$ for the largest power of $p$ dividing $x$, and, for nonzero rational $\frac{r}{t}$, set $v_{p}(\frac{r}{t}) = v_{p}(r)-v_{p}(t)$. First of all, we need the following known result:

Lemma 2.1. [19, Lemma 2.8] Let $a$ and $b$ be distinct coprime rational integers, and let $q$ be an odd prime.

$({\rm i})$. $\gcd(a+b, \frac{a^{q}+b^{q}}{a+b}) = 1 \ {\rm or} \ q$.

$({\rm ii})$. If $q\mid (a+b)$, then $v_{q}(\frac{a^{q}+b^{q}}{a+b}) = 1$.

Definition 2.2. Two nonzero complex numbers $\alpha$ and $\beta$ are called multiplicatively independent if the only solution of the equation $\alpha^{X}\beta^{Y} = 1$ in $\mathbb{Z}$ is $X = Y = 0$.

Next, we quote a result on linear forms in two $m$-adic logarithms due to Bugeaud ^[14], which is crucial to the proof of Theorem 1.2. In order to describe this result, we introduce a related notation. If $r$ is a nonzero rational number with $r = \frac{s}{t}$, and $s$ and $t$ being coprime integers, we define the logarithmic height of $r$ as $h(r): = \max \{\log|s|, \log|t|, 1\}$. Note that if $m = p^{j_{1}}_{1}\cdots p^{j_{k}}_{k}$, where the $p_{i}$'s are distinct primes and $j_{i}\in \mathbb{N}$, for any nonzero integer $x$, we define the arithmetic function $v_{m}$ by

where $[\cdot]$ denotes the integer part.

Proposition 2.3. [14, Theorem 2] Let $\alpha_{1}$ and $\alpha_{2}$ be two nonzero rational numbers with $\alpha_{1}\neq\pm1$, $b_{1}$, and $b_{2}$ being positive integers, and set

For any set of distinct primes $p_{1}, \cdots, p_{k}$ and positive integers $j_{1}, \cdots, j_{k}$, we set $m = p^{j_{1}}_{1}\cdots p^{j_{k}}_{k}$ and suppose that there exists a positive integer $g$ such that for each $i$, we have

and also

Then, if $m$, $b_{1}$, and $b_{2}$ are relatively prime, we have

where

and

The assumptions of Proposition 2.3 (which is a consequence of [14, Theorem 2] with $(y_{1}, y_{2}) = (1, 1)$ and $\mu = 4$) are very restrictive, but are satisfied (and easy to check) in our context. Proposition 2.3 has many applications in studying the Diophantine equation and related Diophantine problems; readers can refer to ^{[14,20,21,22,23,24]}. We stress that $\alpha_{1}$ and $\alpha_{2}$ need not be multiplicatively independent in Proposition 2.3. Indeed, under the additional assumption that $\alpha_{1}$ and $\alpha_{2}$ are multiplicatively independent, we have the following result:

Remark 2.4. ^[14] The constant $66.8$ in the upper bound for $v_{m}(\Lambda)$ may be improved to $53.6$.

If $z\leq2$, then it is clear that Conjecture $1.1$ is true. Thus, we will assume that $z\geq3$ in what follows.

Lemma 2.5. If $(x, y, z)$ is a solution of Eq $(1.2)$, then $y$ is odd.

Proof. Taking $(1.2)$ modulo $n^{2}$ implies that

Hence, $y$ is odd since $n\geq2$.

Lemma 2.6. If $2\mid a$ and $2\mid n$, then $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$.

Proof. Taking $(1.2)$ modulo $n^{3}$ implies that

Therefore, $ax+by\equiv0\pmod {n}$, which is impossible since $2\mid a$, $2\mid n$, $y$ is odd and $c$ is odd.

Lemma 2.7. Let $a$, $b$, $c$, and $n$ be positive integers such that $2\nmid c$ and $a+b = c^{2}$. If $n\geq 3c$, then each of the following is true.

$({\rm i})$

$({\rm ii})$

$({\rm iii})$

Proof. $({\rm i})$ We divide the proof into the following two cases:

Case 1: $b\geq2$. Since $\log(an^{2}+1)\geq\log(n^{2}+1) > 2\log n$ and $\log(bn^{2}-1)\geq2\log n$, we get $\delta_{1}(n)\leq0$.

Case 2: $b = 1$. Since $c$ is odd, we have $a\geq2$ and $\log(an^{2}+1)\geq\log(2n^{2}+1)$. Then for any integer $n\geq9$, we can deduce that

Part (i) is proven.

$({\rm ii})$ Since $a+b = c^{2}$, we have

From $c\leq n/3$ and $n\geq3c\geq9$, we get that

as desired. Part (ii) is proven.

$({\rm iii})$ By (2.4), we have

as desired. Part (iii) is proven.

This completes the proof of Lemma 2.7.

Lemma 2.8. Let $a$, $b$, $c$, and $n$ be positive integers such that $2\nmid c$ and $a+b = c^{2}$, and let $(x, y, z)$ be a solution of $(1.2)$. If $n > c$, then

where $N = \max\{x, y\}$ and $\beta(n): = \min\{an^{2}+1, bn^{2}-1\}$.

Proof. From $(1.2)$, we have

A direct computation gives us that

Notice that

From (2.6) and (2.7), one can deduce that

as required.

This completes the proof of Lemma 2.8.

To establish an upper bound on $z$, we need to prove the following result:

Lemma 2.9. Let $\gcd(c, n) = 1$, and let $(x, y, z)$ be a solution of $(1.2)$. If $n\geq\lambda c$, then $\gcd(x, y, n) = 1$, where $\lambda$ is any constant with $\lambda > 2^{\frac{5}{8}}$.

Proof. If $z\leq2$, by $(1.2)$, one can easily check that $x = y = 1$. Hence,

as required.

If $z\geq3$ and $x$ is a power of 2, then by Lemma 2.5, one can derive that

Next, we may consider only the case where $z\geq3$ and $x$ has an odd prime factor. Let us assume that $\gcd(x, y, n) > 1$. Then there exists an odd prime $p$ such that $x = px_{1}$, $y = py_{1}$, and $n = pn_{1}$, since $y$ is odd by Lemma 2.5. By $(1.2)$, we can deduce that

where $C = (an^{2}+1)^{x_{1}}+(bn^{2}-1)^{y_{1}}$ and $D = \frac{(an^{2}+1)^{px_{1}}+(bn^{2}-1)^{py_{1}}}{(an^{2}+1)^{x_{1}}+(bn^{2}-1)^{y_{1}}}$. Therefore, we have $(an^{2}+1)^{x_{1}}-(bn^{2}-1)^{y_{1}}\equiv2\pmod{n^{2}}$. Since $n^{2}\geq p^{2}\geq9$, we can get that $\big((an^{2}+1)^{x_{1}}-(bn^{2}-1)^{y_{1}}\big)^{2}\geq4$ and

Now, using Lemma 2.1$({\rm i})$, we divide the proof into the following two cases:

Case 1: $\gcd(C, D) = 1$. Since $C\equiv0\pmod n$ and $\gcd(c, n) = 1$, one can easily get that $C = n^{z}c^{z}_{1}$ and $D = c_{2}^{z}$, where $c = c_{1}c_{2}$ and $\gcd(c_{1}, c_{2}) = 1$. By condition $n\geq\lambda c$ and $c_{2}^{z} = D\geq2C = 2(c_{1}n)^{z}$, one has $4\leq2(\lambda c^{2}_{1})^{z}\leq1$, which is impossible. Thus, $\gcd(x, y, n) = 1$ in this case.

Case 2: $\gcd(C, D) = p$. Since $C\equiv0\pmod n$ and $n = pn_{1}$, by Lemma 2.1$({\rm ii})$, one can easily get that $C = p^{-1}\cdot n^{z}\cdot c_{1}^{z}$ and $D = p\cdot c^{z}_{2}$, where $c = c_{1}c_{2}$ and $\gcd(c_{1}, c_{2}) = 1$. Notice that $p\cdot c^{z}\geq D\geq$ $2C\geq$ $2p^{-1}n^{z}\geq 2p^{-1}(\lambda c)^{z}$. Therefore $p\geq2^{\frac{1}{2}}\lambda^{\frac{z}{2}}$. Putting $z > \max\{x, y\}$ of Lemma 2.8 into $p\geq2^{\frac{1}{2}}\lambda^{\frac{z}{2}}$ and using our assumption yields that

which is impossible for $z\geq3$. So one arrives at $\gcd(x, y, n) = 1$ in this case.

This concludes the proof of Lemma 2.9.

Lemma 2.10. Let $\gcd(c, n) = 1$, and let $(x, y, z)$ be a solution of $(1.2)$. If $n\geq c+\frac{1}{n}$ and $z\leq3$, then $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$.

Proof. If $x\geq3$ or $y\geq3$, then

which is impossible. Therefore, Lemma 2.5 tells us that $y = 1$ and $x = 2$ or $x = 1$.

If $x = y = 1$, then $(1.2)$ gives us that $(an^{2}+1)+(bn^{2}-1) = (cn)^{2}$, which is impossible since $z = 3$.

If $x = 2$ and $y = 1$, then by $(1.2)$, one can deduce that

A simple transformation gives us that

By the condition $\gcd(an^{2}+1, bn^{2}-1) = 1$ of Theorem 1.2, we have $\gcd(an^{2}+1, cn) = 1$. Further, $\gcd(an^{2}+1, c) = 1$. On the other hand, $c^{2}\mid a(an^{2}+1)$, thus we have $c^{2}\mid a$, which contradicts the assumption $a < a+b = c^{2}$.

This concludes the proof of Lemma 2.10.

3.   Proof of Theorem 1.2

In this section, we present the proof of Theorem 1.2.

Proof of Theorem 1.2. First of all, suppose that $(x, y, z)$ is a solution of $(1.2)$. We apply Proposition 2.3 and Remark 2.4 to get an upper bound for $z$. For this, we set

and

Evidently, one has $\alpha_{1} = an^{2}+1\neq\pm1$. Let $n = p^{j_{1}}_{1}\cdots p^{j_{k}}_{k}$, and let $g = 1$. One can easily check that

and $A_{1} = \alpha_{1}$, $A_{2} = -\alpha_{2}$ satisfy the assumption of Proposition 2.3. Lemma 2.9 gives us that $n$, $x$, and $y$ are relatively prime, and we know from Definition 2.2 that $an^{2}+1$ and $bn^{2}-1$ are multiplicatively independent. Thus, by Proposition 2.3 and Remark 2.4, we have

where

Let $N = \max\{x, y\}$. Assume that

By (3.1), Lemma 2.8, and the definition of $\delta_{2}(n)$, we have

implies that

From the proof of Lemma 2.7$({\rm ii})$ and $({\rm i})$, one can get $\delta_{2}(n)\leq \frac{9}{8(\log3)^2}$ and

Applying Lemma 2.7$({\rm i})$, (3.3), and (3.4) gives us that

which implies $N < 3985$. Therefore

a contradiction. The claim is proven. Hence, one must have

Putting (3.6) into (3.1), we can deduce that

Now, applying $({\rm iii})$ of Lemma 2.7, one has

Suppose that $z\geq4$. Taking Eq (1.2) modulo $n^{4}$, one can arrive at

so

This implies that

Further, applying Lemma 2.8 and (3.8), we have

Therefore, using (3.7) and (3.9), one can deduce that

which contradicts the assumption $n\geq 117.14c$.

Finally, we conclude that $z\leq3$. By Lemma 2.10, one can easily know that $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$.

This concludes the proof of Theorem 1.2.

4.   Proof of Theorem 1.3

In this section, we present the proof of Theorem 1.3.

Proof of Theorem 1.3 (i) $(a, b, c) = (44, 5, 7)$. Suppose that $(x, y, z)$ is a solution of the equation

First of all, we show that $\gcd(44n^{2}+1, 5n^{2}-1) = 1$. In fact, from $5\cdot(44n^{2}+1)-44\cdot(5n^{2}-1) = 49$, it follows that $\gcd(44n^{2}+1, 5n^{2}-1)\mid 49$. Since $44n^{2}+1\not\equiv0\pmod7$ for any positive integer $n$, we get $\gcd(44n^{2}+1, 5n^{2}-1) = 1$. By Theorem 1.2, [17, Example 1 and Lemma 12], and Lemma 2.6, we only solve the remaining cases of $19\leq n\leq820$ and $n\not\equiv0\pmod2$ and $n\not\equiv0\pmod7$. Since $n\geq18\geq\lambda c$, Lemma 2.9 tells us that $n$, $x$, and $y$ are relatively prime. Using (3.1) of Theorem 1.2, one can immediately deduce that

where

For brevity, we let $N = \max\{x, y\}$ and

Then we have

Subsequently, suppose that

Because $n\geq19$, from $(4.2)$, we derive

On the other hand, because $n\geq19$, according to Lemma 2.8, one must have

Putting (4.5) into $(4.1)$ gives us that

where $\vartheta_{1}(n)$ and $\vartheta_{2}(n)$ are given by

and

Notice that $\vartheta_{1}(n)$ and $\vartheta_{2}(n)$ decrease as $n$ increases in the interval $[19, \infty)$. Then

By (4.6), one can immediately deduce that

which implies that $N < 2392$. This contradicts $N > 98575$.

Therefore, one must have

Because $n\geq19$, one arrives at

Thus, from (4.1), (4.8), and (4.9), one can immediately deduce that

Further, by (4.5) and (4.10), we see that

Hence, all of $x$, $y$, and $z$ are bounded. Using program search, we now show that $z\leq3$ by following two steps:

Step 1: Under the hypotheses $19\leq n\leq820$, $n\not\equiv0\pmod2$, $n\not\equiv0\pmod7$, $N\leq 4680$, and $2\nmid y$, one can check that $z\leq5$.

Step 2: Under the hypotheses $19\leq n\leq820$, $n\not\equiv0\pmod2$, $n\not\equiv0\pmod7$, $\max\{x, y\} < z$, and $4\leq z\leq5$, one can deduce Eq (1.2) has no positive integer solution $(x, y, z)$.

Finally, we conclude that $z\leq3$. By Lemma 2.10, one can easily know that $(1.2)$ has only the solution $(x, y, z) = (1, 1, 2)$.

When $(a, b, c) = (12, 13, 5)$ and $(18, 7, 5)$, using a similar way as in the proof of Theorem 1.3(i), the desired result follows immediately.

The proof of Theorem 1.3 is complete.

Use of AI tools declaration

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

Acknowledgments

This work is supported in part by the Southwest Minzu University Research Startup Funds (Grant No. RQD2021100), and also supported in part by Natural Science Foundation of Sichuan Province (No. 2022NSFSC1830).

Conflict of interest

The authors declare there is no conflict of interest.