1 Introduction

In recent years, many papers investigated pure ternary exponential diophantine equations (see [6,7,8,9,10,11,12,13,14,15]).

Let \(a,\ b,\ c,\ m\) be positive integers such that

$$\begin{aligned} a+b=c^2, \ 2\mid a, \ 2\not \mid c,\ m>1. \end{aligned}$$
(1.1)

In this paper we discuss the equation

$$\begin{aligned} \left( am^2+1\right) ^x+\left( bm^2-1\right) ^y=(cm)^z,\ x, y, z \in {\mathbf {N}}. \end{aligned}$$
(1.2)

In 2012, Terai [13] proved that if \((a,\ b,\ c)=(4,\ 5,\ 3)\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\) under some conditions. Recently, Wang et al. [17] improved that for \((a,\ b,\ c)=(4,\ 5,\ 3)\) and \(3\not \mid m\). Then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\). In this paper we prove a general result as follows:

Theorem 1.1

If \(c\mid m \)  and  \(m>36c^3 \log c\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\).

As a direct consequence we get:

Corollary 1.2

If \((a,\ b,\ c)=(4,\ 5,\ 3)\) and \(3\mid m\) and \(m>1068\), then (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\).

Thus, combining the result of [17] and our corollary we get that (1.2) is basically solved for \((a,\ b,\ c)=(4,\ 5,\ 3)\).

2 Preliminaries

For any nonnegative integer n, let \(F_n\) and \(L_n\) be the n-th Fibonacci and Lucas number, respectively.

Lemma 2.1

([3]). The equation

$$\begin{aligned} F_n=X^2,\ n, X \in {\mathbf {N}}. \end{aligned}$$

has only the solutions \((n,\ X)=(1,\ 1),(2,\ 1)\) and \((12,\ 12)\).

For any positive integer D, let \(h(-4D)\) denote the class number of positive binary quadratic forms of discriminant \(-4D\).

Lemma 2.2

([4], Theorems 11.4.3, 12.10.1 and 12.14.3]).

$$\begin{aligned} h\left( -4D\right) <\frac{4}{\pi }\sqrt{D}\log \left( 2e\sqrt{D}\right) . \end{aligned}$$

Let \(D,\ D_1,\ D_2,\ k\) be positive integers such that \(\min \left\{ D,\ D_1,\ D_2\right\} >1, \gcd \left( D_1,\ D_2\right) =1,\ 2\not \mid k\) and \(\gcd (D,\ k)=\gcd \left( D_1,\ D_2,\ k\right) =1.\)

Lemma 2.3

([5], Theorems 1 and 2]). If the equation

$$\begin{aligned} X^2+DY^2=k^z, \ \gcd (X,\ Y)=1,\ Z>0,\ X, Y, Z \in {\mathbf {Z}} \end{aligned}$$
(2.1)

has solutions \((X,\ Y,\ Z)\), then every solution \((X,\ Y,\ Z)\) of (2.1) can be expressed as

$$\begin{aligned}&\displaystyle Z=Z_1t, \ t \in {\mathbf {N}},\\&\displaystyle X+Y\sqrt{-D}=\lambda _1\left( X_1+\lambda _2Y_1\sqrt{-D}\right) ^t, \ \lambda _1,\lambda _2 \in \left\{ 1, -1\right\} , \end{aligned}$$

where \(X_1, Y_1, Z_1\) are positive integers satisfying \(X_1^2+DY_1^2=k^{z_1},\ \gcd (X_1,\ Y_1)=1\) and \(h(-4D)\equiv 0(\bmod Z_1)\).

Lemma 2.4

([5], Lemma 1]). For a fixed solution \((X,\ Y,\ Z)\) of the equation

$$\begin{aligned} D_1X^2+D_2Y^2=k^z, \ \gcd (X,\ Y)=1,\ Z>0,\ X, Y, Z \in {\mathbf {Z}}, \end{aligned}$$
(2.2)

there exists a unique positive integer l such that

$$\begin{aligned} l=D_1\alpha X+D_2\beta Y, \ 0<l<k, \end{aligned}$$

where \(\alpha ,\ \beta \) are integers with \(\beta X-\alpha Y=1\).

The positive integer l defined as in Lemma 2.4 is called the characteristic number of the solution \((X,\ Y,\ Z)\) and is denoted by \(<X,\ Y,\ Z>\).

Lemma 2.5

([5], Lemma 6]). If \(<X,\ Y,\ Z>=l\), then \(D_1X\equiv -lY(\bmod k)\).

For a fixed positive integer \(l_0\), if (2.2) has a solution \((X_0,\ Y_0,\ Z_0)\) with \(<X_0,\ Y_0,\ Z_0>=l_0\), then the set of all solutions \((X,\ Y,\ Z)\) of (2.2) with \(<X,\ Y,\ Z>\equiv \pm l_0(\bmod k)\) is called a solution class of (2.2) and is denote by \(S(l_0)\).

Lemma 2.6

([5], Theorems 1 and 2]). For any fixed solution class \(S(l_0)\) of (2.2), there exists a unique solution \((X_1,\ Y_1,\ Z_1)\in S(l_0)\) such that \(X_1>0, Y_1>0\) and \(Z_1\le Z\), where Z runs through all solutions \((X,\ Y,\ Z)\in S(l_0)\). The solution \((X_1,\ Y_1,\ Z_1)\) is called the least solution of \(S(l_0)\). Every solution \((X,\ Y,\ Z)\in S(l_0)\) can be expressed as

$$\begin{aligned}&\displaystyle Z=Z_1t,\ 2\not \mid t, \ t\in {\mathbf {N}},\\&\displaystyle X\sqrt{D_1}+Y \sqrt{-D_2}=\lambda _1 \left( X_1 \sqrt{D_1}+\lambda _2Y_1 \sqrt{-D_2}\right) ^t, \ \lambda _1,\lambda _2 \in \{1, -1\}. \end{aligned}$$

Lemma 2.7

([2], Theorem 2]). Let \((X_1,\ Y_1,\ Z_1)\) be the least solution of \(S(l_0)\). If (2.2) has a solution \((X,\ Y,\ Z)\in S(l_0)\) satisfying \(X>0\) and \(Y=1\), then \(Y_1=1\). Further, if \((X,\ Z)\ne (X_1,\ Z_1)\), then one of the following conditions is satisfied:

  1. (i)

    \(D_1X^2_1=\frac{1}{4}(k^{Z_1}\pm 1),\ D_2=\frac{1}{4}(3k^{Z_1}\mp 1),\ (X,\ Z)=(X_1|D_1X^2_1-3D_2|,\ 3Z_1)\).

  2. (ii)

    \(D_1X^2_1=\frac{1}{4}F_{3r+3\varepsilon },\ D_2=\frac{1}{4}L_{3r},\ k^{Z_1}=F_{3r+\varepsilon },\ (X,\ Z)=(X_1|D^2_1X^4_1-10D_1D_2X^2_1+5D^2_2|,\ 5Z_1)\), where r is a positive integer, \(\varepsilon \in \{1, -1\}\).

Let \(\alpha ,\ \beta \) be algebraic integers. If \(\alpha +\beta \) and \(\alpha \beta \) are nonzero coprime integers and \(\frac{\alpha }{\beta }\) is not a root of unity, then \(\left( \alpha ,\ \beta \right) \) is called a Lucas pair. Further, let \(A=\alpha +\beta \) and \(C=\alpha \beta \). Then we have

$$\begin{aligned} \alpha =\frac{1}{2}\left( A+\lambda \sqrt{B}\right) , \ \beta =\frac{1}{2}\left( A-\lambda \sqrt{B}\right) ,\ \lambda \in \{1, -1\}, \end{aligned}$$
(2.3)

where \(B=A^2-4C\). We call \((A,\ B)\) the parameters of the Lucas pair \((\alpha ,\ \beta )\). Two Lucas pairs \((\alpha _1,\ \beta _1)\) and \((\alpha _2,\ \beta _2)\) are equivalent if \(\frac{\alpha _1}{\alpha _2}=\frac{\beta _1}{\beta _2}=\pm 1\). Given a lucas pair \((\alpha ,\ \beta )\), one defines the corresponding sequence of Lucas numbers by

$$\begin{aligned} L_n\left( \alpha ,\ \beta \right) =\frac{\alpha ^n-\beta ^n}{\alpha -\beta },\ n=0,\ 1,\ldots . \end{aligned}$$
(2.4)

For equivalent Lucas pairs \((\alpha _1,\ \beta _1)\) and \((\alpha _2,\ \beta _2)\), we have \(L_n(\alpha _1,\ \beta _1)=\pm L_n(\alpha _2,\ \beta _2) (n=0,\ 1,\ldots ).\) A prime p is called a primitive divisor of \(L_n(\alpha ,\ \beta )(n>1)\) if

$$\begin{aligned} p\mid L_n(\alpha ,\ \beta ),\ p\not \mid B L_1(\alpha ,\ \beta )\ldots L_{n-1}(\alpha ,\ \beta ). \end{aligned}$$

A Lucas pair \((\alpha ,\ \beta )\) such that \(L_n(\alpha ,\ \beta )\) has no primitive divisors will be called an n-defective Lucas pair. Further, a positive integer n is called totally non-defective if no Lucas pair is n-defective.

Lemma 2.8

([1], Theorem 1.4]). If \(n>30\), Then n is totally non-defective.

Lemma 2.9

([16]). Let n satisfy \(4<n\le 30\) and \(n\ne 6\). Then, up to equivalence, all parameters of n-defective Lucas pairs are given as follows:

  1. (i)

    \(n=5,\ (A,\ B)=(1,\ 5),\ (1,\ -7),\ (2,\ -40),\ (1,\ -11),\ (1,\ -15),\ (12,\ -76), (12,\ -1364)\).

  2. (ii)

    \(n=7,\ (A,\ B)=(1,\ -7),\ (1,\ -19)\).

  3. (iii)

    \(n=8,\ (A,\ B)=(2,\ -24),\ (1,\ -7)\).

  4. (iv)

    \(n=10,\ (A,\ B)=(2,\ -8),\ (5,\ -3),\ (5,\ -47)\).

  5. (v)

    \(n=12,\ (A,\ B)=(1,\ 5),\ (1,\ -7),\ (1,\ -11),\ (2,\ -56),\ (1,\ -15),\ (1,\ -19)\).

  6. (vi)

    \(n\in \{13,\ 18, \ 30 \},\ (A,\ B)=(1,\ -7)\).

3 Proof of theorem

We now assume that \((x,\ y,\ z)\) is a solution of (1.2) with \((x,\ y,\ z)\ne (1,\ 1,\ 2)\). Since \(c\mid m\), we have \(cm\mid m^2\), and by (1.2), we get

$$\begin{aligned} \gcd \left( am^2+1,\ cm\right) =\gcd \left( bm^2-1,\ cm\right) =\gcd \left( am^2+1,\ bm^2-1\right) =1. \end{aligned}$$
(3.1)

Since \(m>1\) and \(z>2\), by (1.2), we have \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y \equiv 1+(-1)^y(\bmod m^2)\) and

$$\begin{aligned} 2\not \mid y. \end{aligned}$$
(3.2)

Further, since \(z\ge 3\), by (1.2) and (3.2), we get \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y \equiv (ax+by)m^2(\bmod m^3)\) and

$$\begin{aligned} (ax+by)\equiv 0(\bmod m). \end{aligned}$$
(3.3)

Notice that \(2\mid a, \ 2\not \mid c\) and \(2\not \mid b\) by (1.1). We see from (3.2) and (3.3) that \(2\not \mid ax+by\) and

$$\begin{aligned} 2\not \mid m. \end{aligned}$$
(3.4)

So we have

$$\begin{aligned} 2\not \mid cm, \ 2\not \mid am^2+1, \ 2\mid bm^2-1. \end{aligned}$$
(3.5)

We first consider the case of \(2\mid x\). Then, by (3.2), the equation

$$\begin{aligned} X^2+(bm^2-1)Y^2=(cm)^Z, \ \gcd (X,\ Y)=1, \ Z>0,\ X, Y, Z \in {\mathbf {Z}} \end{aligned}$$
(3.6)

has the solution

$$\begin{aligned} \left( X,\ Y,\ Z\right) =\left( (am^2+1)^{\frac{x}{2}},\ (bm^2-1)^{\frac{y-1}{2}},\ z\right) . \end{aligned}$$
(3.7)

By (3.1) and (3.5), applying Lemma 2.3 to (3.6) and (3.7), we have

$$\begin{aligned}&\displaystyle z=Z_1t,\ t\in {\mathbf {N}}, \end{aligned}$$
(3.8)
$$\begin{aligned} \left( am^2+1\right) ^{\frac{x}{2}}+\left( bm^2-1\right) ^{\frac{y-1}{2}}\sqrt{1-bm^2}= & {} \lambda _1\left( X_1+\lambda _2Y_1\sqrt{1-bm^2}\right) ^t,\nonumber \\&\displaystyle \quad \lambda _1, \lambda _2\in \{1, -1\}, \end{aligned}$$
(3.9)

where \(X_1,\ Y_1,\ Z_1\) are positive integers satisfying

$$\begin{aligned}&\displaystyle X^2_1+\left( bm^2-1\right) Y^2_1=(cm)^{Z_1},\ \gcd \left( X_1,\ Y_1\right) =1, \end{aligned}$$
(3.10)
$$\begin{aligned}&\displaystyle h\left( -4(bm^2-1)\right) \equiv 0\left( \bmod Z_1\right) . \end{aligned}$$
(3.11)

If \(2\mid t\), let

$$\begin{aligned} X_2+Y_2\sqrt{1-bm^2}=\left( X_1+\lambda _2Y_1\sqrt{1-bm^2}\right) ^{\frac{t}{2}}. \end{aligned}$$
(3.12)

By Lemma 2.3, \(X_2\) and \(Y_2\) are integers satisfying

$$\begin{aligned} X^2_2+\left( bm^2-1\right) Y^2_2=(cm)^\frac{{Z_1t}}{2}=(cm)^{\frac{z}{2}},\ \gcd \left( X_2,\ Y_2\right) =1. \end{aligned}$$
(3.13)

Substitute (3.12) into (3.9), we have \((am^2+1)^{\frac{x}{2}}+(bm^2-1)^{\frac{y-1}{2}}\sqrt{1-bm^2}=\lambda _1(X_2+Y_2\sqrt{1-bm^2})^2\) and

$$\begin{aligned} \left( bm^2-1\right) ^{\frac{y-1}{2}}=2\lambda _1X_2Y_2. \end{aligned}$$
(3.14)

By (3.1) and (3.13), we get \(\gcd \left( X_2,\ bm^2-1\right) =1\). Therefore, we see from (3.14) that

$$\begin{aligned} |X_2|=1,\ |Y_2|=\frac{1}{2}(bm^2-1)^{\frac{y-1}{2}}. \end{aligned}$$
(3.15)

Substitute (3.15) into (3.13), we get

$$\begin{aligned} 1+\frac{1}{4}(bm^2-1)^y=(cm)^{\frac{{z}}{2}}. \end{aligned}$$
(3.16)

Since \(z>2\), we have \(\frac{{z}}{2}\ge 2\). By (3.2) and (3.16), we get \(0\equiv (cm)^\frac{{z}}{2}\equiv 1+\frac{1}{4}(bm^2-1)^y\equiv 1-\frac{1}{4}\equiv \frac{3}{4}(\bmod m^2)\) and \(m^2\mid 3\), a contradiction. So we have

$$\begin{aligned} 2\not \mid t. \end{aligned}$$
(3.17)

Let

$$\begin{aligned} \alpha =X_1+Y_1\sqrt{1-bm^2},\ \beta =X_1-Y_1\sqrt{1-bm^2}. \end{aligned}$$
(3.18)

By (3.10) and (3.18), we have \(\alpha +\beta =2X_1,\ \alpha -\beta =2Y_1\sqrt{1-bm^2},\ \alpha \beta =(cm)^{Z_1}\) and \(\frac{\alpha }{\beta }\) satisfies \((cm)^{Z_1}(\frac{\alpha }{\beta })^2-2\left( X^2_1-(bm^2-1)Y^2_1\right) (\frac{\alpha }{\beta })+(cm)^{Z_1}=0\). It implies that \((\alpha ,\ \beta )\) is a Lucas pair with parameters

$$\begin{aligned} (A,\ B)=\left( 2X_1,\ -4(bm^2-1)Y^2_1\right) . \end{aligned}$$
(3.19)

Further, let \(L_n(\alpha ,\ \beta )\ (n=0,\ 1,\ldots )\) be the corresponding Lucas numbers. By (2.3), (3.9) and (3.18), we have

$$\begin{aligned} (bm^2-1)^{\frac{y-1}{2}}=Y_1\left| \frac{\alpha ^t-\beta ^t}{\alpha -\beta }\right| =Y_1|L_t(\alpha ,\ \beta )|. \end{aligned}$$
(3.20)

We see from (3.19) and (3.20) that the Lucas number \(L_t(\alpha ,\ \beta )\) has no primitive divisors. Therefore, by Lemmas 2.8 and 2.9, we get from (3.17) and (3.19) that

$$\begin{aligned} t\le 3. \end{aligned}$$
(3.21)

By (3.8), (3.11) and (3.21), we have

$$\begin{aligned} z\le 3h\left( -4(bm^2-1)\right) . \end{aligned}$$
(3.22)

Applying Lemma 2.2 to (3.22), we get

$$\begin{aligned} z< \frac{12}{\pi }\sqrt{bm^2-1}\log \left( 2e\sqrt{bm^2-1}\right) . \end{aligned}$$
(3.23)

Further, since \(b<a+b=c^2\), by (3.23), we have

$$\begin{aligned} z< \frac{12}{\pi }cm\log (2ecm). \end{aligned}$$
(3.24)

On the other hand, since \(2\mid x\), if \(z=3\), then \((cm)^3>(am^2+1)^x\ge (am^2+1)^2>a^2m^4\), whence we get \(c^3>a^2m>m>36c^3\log c\), a contradiction. It implies that \(z\ge 4\) and \(0\equiv (cm)^z\equiv (am^2+1)^x+(bm^2-1)^y\equiv (ax+by)m^2(\bmod m^4)\), whence we obtain \(ax+by\equiv 0(\bmod m^2)\) and

$$\begin{aligned} ax+by\ge m^2. \end{aligned}$$
(3.25)

Since \(m>36c^3\log c\), by (1.2), we have

$$\begin{aligned} z>x\frac{\log (am^2+1)}{\log (cm)}>x,\ z>y\frac{\log (bm^2-1)}{\log (cm)}>y. \end{aligned}$$
(3.26)

Hence, by (3.25) and (3.26), we get

$$\begin{aligned} c^2z=(a+b)z>ax+by\ge m^2. \end{aligned}$$
(3.27)

The combination of (3.24) and (3.27), we have

$$\begin{aligned} m<\frac{12}{\pi }c^3\log (2ecm). \end{aligned}$$
(3.28)

But, since \(m>36c^3\log c\), (3.28) is false. Thus, (1.2) has no solutions \((x,\ y,\ z)\) with \(2\mid x\).

Finally, we consider the case of \(2\not \mid x\). Then, by (1.2) and (3.2), the equation

$$\begin{aligned} \left( am^2+1\right) X^2+(bm^2-1)Y^2=(cm)^Z,\ \gcd (X,\ Y)=1,\ Z>0,\ X, Y, Z\in {\mathbf {Z}} \end{aligned}$$
(3.29)

has the solution

$$\begin{aligned} \left( X,\ Y,\ Z\right) =\left( (am^2+1)^{\frac{x-1}{2}},\ (bm^2-1)^{\frac{y-1}{2}},\ z\right) . \end{aligned}$$
(3.30)

Let \(l=\langle (am^2+1)^{\frac{x-1}{2}},\ (bm^2-1)^{\frac{y-1}{2}},\ z\rangle \). Since \(cm\mid m^2\), by Lemma 2.5, l satisfies

$$\begin{aligned} am^2+1\equiv (am^2+1)^{\frac{x+1}{2}}\equiv -l (bm^2-1)^{\frac{y-1}{2}}\equiv (-1)^{\frac{y+1}{2}}l (\bmod cm). \end{aligned}$$
(3.31)

On the other hand, since \((x,\ y,\ z)\ne (1,\ 1,\ 2)\), (3.29) has an other solution

$$\begin{aligned} \left( X,\ Y,\ Z\right) \ne \left( 1,\ 1,\ 2\right) . \end{aligned}$$
(3.32)

Let \(l_0=\langle 1,\ 1,\ 2\rangle \). Then we have

$$\begin{aligned} am^2+1\equiv -l_0 (\bmod cm). \end{aligned}$$
(3.33)

Obviously, since \(z\ge 2\) for any solution \((x,\ y,\ z)\) of (3.29), the least solution of \(S(l_0)\) is

$$\begin{aligned} \left( X_1,\ Y_1,\ Z_1\right) =\left( 1,\ 1,\ 2\right) . \end{aligned}$$
(3.34)

Compare (3.31) and (3.33), we have \(l\equiv \pm l_0 (\bmod cm)\). It implies that the solution (3.30) belongs to \(S(l_0)\). Therefore, using Lemma 2.6, we get from (3.30) and (3.32) that

$$\begin{aligned} z= & {} 2t,\ 2\not \mid t,\ t\in {\mathbf {N}},\nonumber \\&(am^2+1)^{\frac{x-1}{2}}\sqrt{am^2+1}+(bm^2-1)^{\frac{y-1}{2}}\sqrt{1-bm^2}\nonumber \\= & {} \lambda _1(\sqrt{am^2+1}+\lambda _2\sqrt{1-bm^2})^t,\ \lambda _1, \lambda _2\in \{1, -1\}. \end{aligned}$$
(3.35)

By (3.35), we have

$$\begin{aligned} (bm^2-1)^{\frac{y-1}{2}}=\lambda _1\lambda _2\sum \limits _{i=0}^{\frac{t-1}{2}}\left( \begin{array}{c} t\\ 2i+1 \end{array}\right) (am^2+1)^{\frac{t-1}{2}-i}(1-bm^2)^i. \end{aligned}$$
(3.36)

Further, since \(2\mid bm^2-1\) and \(2\not \mid (am^2+1)t\), we see from (3.36) that \(y=1\) and \((bm^2-1)^{\frac{y-1}{2}}=1\). It implies that (3.30) is a solution of \(S(l_0)\) satisfying \(X>0, \ Y=1\) and \((X,\ Z)\ne (X_1,\ Z_1)=(1,\ 2)\). Therefore, by Lemma 2.7, we get either

$$\begin{aligned} am^2+1=(am^2+1)X_1^2=\frac{1}{4}\left( (cm)^2\pm 1\right) \end{aligned}$$
(3.37)

or

$$\begin{aligned} (cm)^2=(cm)^{Z_1}=F_{3r+\varepsilon }. \end{aligned}$$
(3.38)

When (3.37) holds, since \(c\mid m\), we have \(1\equiv am^2+1\equiv \frac{1}{4}\left( (cm)^2\pm 1\right) \equiv \pm \frac{1}{4}(\bmod c^2)\). But, since \(c^2\ge 9\), it is impossible. On the other hand, since \(cm>1\) and \(2\not \mid cm\), by Lemma 2.1, (3.38) is false. Thus, (1.2) has only the solution \((x,\ y,\ z)=(1,\ 1,\ 2)\) with \(2\not \mid x\). the theorem is proved.