On the generalized Ramanujan-Nagell equation $x^2+(2k-1)^y = k^z$ with $k\equiv 3$ (mod 4)

1.   Introduction

Let $\mathbb{N}$ be the set of all positive integers. Let $k$ be a fixed positive integer with $k > 1$. In this paper, we deal with an exponential generalized Ramanujan-Nagell equation with the form

In 2014, N. Terai ^[6] proposed the following conjecture:

Conjecture 1.1. (1.1) has only the solution $(x, y, z) = (k-1, 1, 2)$.

Obviously, if $k\equiv 2$(mod 4), then $2k-1\equiv 3$(mod 8). Since $z > 1$, by (1.1), we have $2\nmid x, 2\nmid y$ and $k^z\equiv x^2 +(2k-1)^y \equiv 1+3 \equiv 4$(mod 8). It implies that $z = 2$. Therefore, Conjecture 1.1 is true for $k\equiv 2$(mod 4). However, in addition to this case, it is only proved in some special cases (see ^{[1,2,3,4,6,7]}). For example, M. J. Deng, J. Guo and A. J. Xu ^[3] gave some conditions for (1.1) to have solutions $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$. So they proved that if $k\equiv 3$ (mod 4) and $k < 500$, then Conjecture 1.1 is true.

For any fixed positive $r$, there exist unique positive integers $d$ and $s$ such that $r = ds^2$ and $d$ is square-free. Such $d$ is call the square-free part of $r$, and denoted by $Q(r)$. In this paper, using some elementary methods, we prove a general result as follows:

Theorem 1.1. If $k\equiv 3$(mod 4) and $Q(k-1)\geq 2.65$ log $k$, then (1.1) has only the solution $(x, y, z) = (k-1, 1, 2)$.

By the above theorem, we can deduce the following corollaries:

Corollary 1.1. If $k\equiv 3$(mod 4) and $k-1$ is square-free, then Conjecture 1.1 is true.

Corollary 1.2. If $k\equiv 3$(mod 4) and $500 < k < 1000$, then Conjecture 1.1 is true.

Corollary 1.3. Conjecture 1.1 is true for almost all positive integers $k$ with $k\equiv3$(mod 4).

2.   Preliminaries

Lemma 2.1. (^[5]) Let $n$ be an odd integer with $n > 1$, and let $X, Y$ be coprime positive integers. Further, let $p$ be an odd prime with $p\nmid XY$. If $p\mid X+Y$ and $p^r\mid(X^n+Y^n)/(X+Y)$, where $r$ is a positive integer, then $p^r\mid n$.

Here and below, we assume that $k\equiv 3$(mod 4) and (1.1) has a solution $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$.

Lemma 2.2. ((i) of Lemma 2.5 of ^[3]) $2\nmid y$ and $2\mid z$.

Lemma 2.3. ((ii) of Lemma 2.5 of ^[3]) $y > 3$.

Lemma 2.4. (Lemma 2.6 of ^[3]) There exist positive integers $a$ and $b$ such that

and

Lemma 2.5. ((ii) of Theorem 1.1 of ^[3]) $a\equiv b\equiv 1$ (mod 4).

Lemma 2.6. (Theorem 1.2 of ^[3]) $k$ is not an odd prime power.

Lemma 2.7. ((i) of Theorem 1.1 of ^[3]) $2k-1$ is not an odd prime power.

Lemma 2.8. (Corollary 1.4 of ^[3]) $k > 500$.

Lemma 2.9. $y < z < 2y$.

Proof. We see from (1.1) that $k^z > (2k-1)^y > k^y$. So we have $y < z$. On the other hand, by Lemma 2.4, we get from (2.1) and (2.2) that $b\geq 3$ and

Therefore, by (2.3), we obtain $z < 2y$. The lemma is proved.

Lemma 2.10. $(a+b)y\geq 2.3^{z/2}$.

Proof. Notice that $2\nmid y$, $2\mid (a+b)$, $a+b > 2$ and $(a^{y}+b^{y})/(a+b)$ is an odd positive integer. By (2.2), we have

where $k_0$ is square free, neither $f$ nor $g$ has $z/2$-th power divisors.

If $k_0 = 1$, then from (2.4) we get $f = 1$, $k_1 > 1$ and $a+b = 2k_1^{z/2}\geq 2\cdot 3^{z/2}$. Therefore, the lemma holds for this case.

If $k_0 > 1$, then $k_0$ has an odd prime divisor $p$. Since neither $f$ nor $g$ has $z/2$-th power divisors, by (2.4), there exists a positive integer $s$ such that

Hence, applying Lemma 2.1 to (2.4) and (2.5), we have

Therefore, by (2.4) and (2.6), we get $(a+b)y\geq 2p^{z/2}\geq 2\cdot 3^{z/2}$. The lemma is proved.

Lemma 2.11. $y < 2.65$ log $k$, where log is the Napierian logarithm.

Proof. By Lemmas 2.9 and 2.10, we have

whence we get

Futher, by Lemma 2.5, we see from (2.1) that

Hence, by (2.7) and (2.8), we get

Let

Subtitute (2.10) into (2.9), we have

Since $k\geq 503$ by Lemma 2.8, we get

Therefore, by (2.11) and (2.12), we have

Let

Then we have

where $f'(t)$ is the derivative of $f(t)$. We see from (2.15) that $f'(t) > 0$ for $t > 2$. Hence, $f(t)$ is an increasing function for $t > 2$. Notice that $f(2.65) > 0$. We have $f(t) > 0$ for $t > 2.65$. Thus, by (2.13) and (2.14), we get that $t < 2.65$, and by (2.10), the lemma is proved.

3.   Proofs

In this section, we assume that $k\equiv 3$(mod 4) and $(x, y, z)$ is a solution of (1.1) with $(x, y, z)\neq(k-1, 1, 2)$.

3.1.   Proof of Theorem 1.1

Proof. By (1.1), we have

whence we get

By the definition of the square-free part, we have

$Q(k-1)$ is square free. Hence, we see from (3.1) and (3.2) that $x\equiv0$(mod $Q(k-1)m)$. Further, since $x^2\equiv 0$(mod$\left(Q(k-1)\right)^2m^2$) and $(k-1)^2\equiv 0$(mod$\left(Q(k-1)\right)^2m^2)$, by (3.1) and (3.2), we get

Furthermore, by Lemma 2.9, we find from (3.3) that $z-2y\neq 0$ and

Therefore, combining (3.4) with Lemma 2.11, we get $2.65$ log$k > Q(k-1)$. This implies that if $Q(k-1)\geq 2.65$ log $k$, then(1.1) has no solution $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$. The theorem is proved.

3.2.   Proof of Corollary 1.1

Proof. Since $k-1$ is square free, we have $Q(k-1) = k-1$. Therefore, by Theorem 1.1, if (1.1) has solution $(x, y, z)\neq (k-1, 1, 2)$, then

But, since $k > 500$ by Lemma 2.8, (3.5) is false. The corollary is proved.

3.3.   Proof of Corollary 1.2

Proof. We now assume that $k\equiv3$ (mod 4), $500 < k < 1000$ and (1.1) has solutions $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$. By Theorem 1.1, we have

Since $Q(k-1)$ is square free with $2\mid Q(k-1)$, by (3.6), we get

Further, by (3.2), we have

Therefore, by (3.7) and (3.8), we just have to consider the following cases:

Since $727$ and $883$ are odd primes, by Lemma 2.6, Conjecture 1.1 is true for $k \in \{727, \; 883\}$. Similarly, since $1373$ and $1801$ are odd primes, by Lemma 2.7, Conjecture 1.1 is true for $k\in\{687, \; 901\}$.

When $k = 579$, we have $2k-1 = 1157 = 13\times 89$, where $13$ and $89$ are odd primes. Hence, by Lemma 2.4, if (1.1) has solutions $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$ for $k = 579$, then we have

But, since $2\nmid y$, $89+13 = 102 = 2\times 3\times17$ and $17\nmid579$, (3.10) is false. Therefore, Conjecture 1.1 is true for $k = 579$.

Similarly, when $k = 723$, we have $2k-1 = 1445 = 5\times17^2$, $17^2+5 = 294 = 2\times3\times7^2$ and $7\nmid723$. Therefore, $17^{2y}+5^y\neq 2.723^{x/2}$ and Conjecture 1.1 is true for $k = 723$. Thus, by (3.9), the corollary is proved.

3.4.   Proof of Corollary 1.3

Proof. For any positive integer $K$, let $F(K)$ denote the number of positive integer $k$ with $k\leq K$ and $k\equiv 3$ (mod 4). Then we have

where $[(k+1)/4]$ is the integer part of $(K+1)/4$. Further, let $G(K)$ denote the number of positive integers $k$ such that $k\leq K$, $k\equiv3$ (mod 4) and $k$ can make (1.1) has solutions $(x, y, z)$ with $(x, y, z)\neq (k-1, 1, 2)$. By Theorem 1.1, we see from (3.2) that

where $d$ through all positive integers with $d\leq2.11$ log $K$ and $d$ is square free. Therefore, by (3.11) and (3.12), we get

This implies that Conjecture 1.1 is true for almost all positive integers $k$ with $k\equiv 3$ (mod 4). The corollary is proved.

Acknowledgments

The authors would like to thank the referee for their very helpful and detailed comments. This work is supported by Young talent-training plan for college teachers in Henan province (2019GGJS241), Startup Foundation for Introducing Talent (HYRC2019007) of Hetao College(CN) and N. S. F. (2021MS01003) of Inner Mongolia(CN).

Conflict of interest

The authors declare that they have no competing interests.