1 Introduction

Let a, b, c be fixed relatively prime positive integers greater than one. The exponential Diophantine equation

$$\begin{aligned} a^x + b^y=c^z \end{aligned}$$
(1.1)

in positive integers x, y, z has been actively studied by a number of authors. This field has a rich history. Using elementary methods such as congruences, the quadratic reciprocity law and the arithmetic of quadratic (or cubic) fields, we can completely solve most of Eq. (1.1) for small values of a, b, c. (cf. Nagell [16], Hadano [7] and Uchiyama [22].) It is known that the number of solutions (x, y, z) is finite, and all solutions can be effectively determined by means of Baker’s method for linear forms in logarithms.

Jeśmanowicz [8] conjectured that if a,  b,  c are Pythagorean numbers, i.e., positive integers satisfying \(a^2 + b^2=c^2\), then (1.1) has only the positive integer solution \((x, y, z)=(2, 2, 2)\). As an analogue of Jeśmanowicz’ conjecture, the first author proposed that if a, b, c, p, q, r are fixed positive integers satisfying \(a^p + b^q=c^r\) with \(a,b,c, p, q,r \ge 2\) and \(\gcd (a,b)=1\), then (1.1) has only the positive integer solution \((x, y, z)=(p, q, r)\) except for a handful of triples (a, b, c). These conjectures have been proved to be true in many special cases. They however, are still unsolved in general. (cf. [6, 11, 12, 14, 15, 18, 20]).

In the previous paper Terai [19], the first author showed that if m is a positive integer with \(1 \le m \le 20\) or \(m \not \equiv 3~(\mathrm{mod}~6)\), then the Diophantine equation

$$\begin{aligned} (4m^2+1)^x+(5m^2-1)^y=(3m)^z \end{aligned}$$
(1.2)

has only the positive integer solution \((x, y, z)=(1, 1, 2).\) The proof is based on elementary methods and Baker’s method. Su-Li [17] proved that if \(m \ge 90\) and \(m \equiv 3~(\mathrm{mod}~6)\), then Eq. (1.2) has only the positive integer solution \((x, y, z)=(1, 1, 2)\) by means of a result of Bilu–Hanrot–Voutier [3] concerning the existence of primitive prime divisors in Lucas-numbers. Recently, Bertók [1] has completely solved the remaining cases \(20<m<90\) and \(m \equiv 3~(\mathrm{mod}~6)\) via the help of exponential congruences. (cf. Bertók-Hajdu [2].) In [13] and [21], we also showed that the Diophantine equations

$$\begin{aligned} (m^2+1)^x+(cm^2-1)^y= & {} (am)^z \quad \text{ with }\quad 1+c=a^2, \\ (12m^2 + 1)^x + (13m^2-1)^y= & {} (5m)^z \end{aligned}$$

have only the positive integer solution \((x, y, z)=(1, 1, 2)\) under some conditions, respectively.

In this paper, we consider the exponential Diophantine equation

$$\begin{aligned} (3pm^2-1)^x+(p(p-3)m^2+1)^y=(pm)^z \end{aligned}$$
(1.3)

with m positive integer and p prime . Our main result is the following:

Theorem 1.1

Let m be a positive integer with \(m \not \equiv 0~(\mathrm{mod}~3).\) Let p be a prime with \(p \equiv 1~(\mathrm{mod}~4).\) Moreover, suppose that if \(m \equiv 1~(\mathrm{mod}~4),\) then \(p<3784\). Then Eq. (1.3) has only the positive integer solution \((x, y, z)=(1, 1, 2).\)

In particular, for \(p=5\), we can completely solve Eq. (1.3) without any assumption on m. The proof is based on applying a result on linear forms in p-adic logarithms due to Bugeaud [5] to Eq. (1.3) with \(m \equiv 0~(\mathrm{mod}~3)\).

Corollary 1.2

The exponential Diophantine equation

$$\begin{aligned} (15m^2-1)^x+(10m^2+1)^y=(5m)^z \end{aligned}$$
(1.4)

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

2 Preliminaries

In order to obtain an upper bound for a solution y of Pillai’s equation \(c^z-b^y=a\) under some conditions, we need a result on lower bounds for linear forms in the logarithms of two algebraic numbers. We will introduce here some notations. Let \(\alpha _1\) and \(\alpha _2\) be real algebraic numbers with \(|\alpha _1| \ge 1\) and \(|\alpha _2| \ge 1\). We consider the linear form

$$\begin{aligned} \Lambda = b_2 \log \alpha _2 - b_1\log \alpha _1, \end{aligned}$$

where \(b_1\) and \(b_2\) are positive integers. As usual, the logarithmic height of an algebraic number \(\alpha \) of degree n is defined as

$$\begin{aligned} h(\alpha ) = \frac{1}{n} \left( \log |a_0| + \sum _{j=1}^{n}\, \log \max \left\{ 1, |\alpha ^{(j)}| \right\} \right) , \end{aligned}$$

where \(a_0\) is the leading coefficient of the minimal polynomial of \(\alpha \) (over \(\mathbb {Z}\)) and \((\alpha ^{(j)})_{1 \le j \le n}\) are the conjugates of \(\alpha \). Let \(A_1\) and \(A_2\) be real numbers greater than 1 with

$$\begin{aligned} \log A_i \ge \max \left\{ h(\alpha _i), \frac{|\log \alpha _i|}{D}, \frac{1}{~D~}\right\} , \end{aligned}$$

for \(i \in \{1,2\}\), where D is the degree of the number field \(\mathbb {Q}(\alpha _1,~\alpha _2)\)  over \(\mathbb {Q}\). Define

$$\begin{aligned} b^{\prime } =\frac{b_1}{~D\log A_2~}+ \frac{b_2}{~D\log A_1~}. \end{aligned}$$

We choose to use a result due to Laurent [10], Corollary 2] with \(m = 10\) and \(C_2 = 25.2\).

Proposition 2.1

(Laurent [10])  Let \(\Lambda \) be given as above, with \(\alpha _1 > 1\) and \(\alpha _2 > 1\). Suppose that \(\alpha _1\) and \(\alpha _2\) are multiplicatively independent. Then

$$\begin{aligned} \log |\Lambda | \ge -25.2\, D^4 \left( \max \left\{ \log b^{\prime }+0.38 , \frac{~10~}{D}\right\} \right) ^2 \log A_1 \log \log A_2. \end{aligned}$$

Next, we shall quote a result on linear forms in p-adic logarithms due to Bugeaud [5]. Here we consider the case where \(y_{1}=y_{2}=1\) in the notation from [5, p. 375].

Let p be an odd prime. Let \(a_{1}\) and \(a_{2}\) be non-zero integers prime to p. Let g be the least positive integer such that

$$\begin{aligned} \mathrm{ord}_{p}(a_{1}^{\,g}-1) \ge 1, \quad \mathrm{ord}_{p}(a_{2}^{\,g}-1) \ge 1, \end{aligned}$$

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

$$\begin{aligned} 1/(p -1)<E \le \mathrm{ord}_{p}(a_{1}^{\,g}-1). \end{aligned}$$

We consider the integer

$$\begin{aligned} \Lambda =a_{1}^{b_{1}}-a_{2}^{b_{2}}, \end{aligned}$$

where \(b_{1}\) and \(b_{2}\) are positive integers. We let \(A_1\) and \(A_2\) be real numbers greater than 1 with

$$\begin{aligned} \log A_{i} \ge \max \{\log |a_{i}|,\,E\log p\} \quad (i=1,2), \end{aligned}$$

and we put \(b'=b_{1}/\log A_{2}+b_{2}/\log A_{1}\).

Proposition 2.2

(Bugeaud [5]) With the above notation, if \(a_{1}\) and \(a_{2}\) are multiplicatively independent, then we have the upper estimate

$$\begin{aligned} \mathrm{ord}_{p}(\Lambda ) \le \frac{36.1 g}{\,E^{3} (\log p)^{4}} \big (\max \{ \log b' + \log (E \log p) +0.4, \, 6 E \log p, \, 5\} \big )^{ 2} \log A_{1} \log A_{2}. \end{aligned}$$

3 Proof of Theorem 1.1

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

Let (x, y, z) be a solution of (1.3). Taking (1.3) modulo p implies that \( (-1)^x+1 \equiv 0~(\mathrm{mod}~p).\) Hence x is odd.

3.1 The case where m is even

Using a congruence method, we ca easily show that if m is even, then Eq. (1.3) has only the positive integer solution \((x, y, z)=(1, 1, 2)\).

Lemma 3.1

If m is even, then Eq. (1.3) has only the positive integer solution \((x, y, z)=(1, 1, 2).\)

Proof

If \(z \le 2\), then \((x, y, z)=(1, 1, 2)\) from (1.3). Hence we may suppose that \(z \ge 3\). Taking (1.3) modulo \(m^3\) implies that

$$\begin{aligned} -1+3pm^2x+1+p(p-3)m^2y \equiv 0~(\mathrm{mod}~m^3), \end{aligned}$$

so

$$\begin{aligned} 3px+p(p-3)y \equiv 0~(\mathrm{mod}~m), \end{aligned}$$

which is impossible, since x is odd and m is even. We therefore conclude that if m is even, then Eq. (1.3) has only the positive integer solution \((x, y, z)=(1, 1, 2).\) \(\square \)

3.2 The case where m is odd with \(m \not \equiv 0~(\mathrm{mod}~3)\)

Lemma 3.2

If m is odd with \(m \not \equiv 0~(\mathrm{mod}~3)\), then \(x=1.\)

Proof

Suppose that \(x \ge 2\). We show that this will lead to a contradiction. The proof is devided into two cases: Case 1: \(m \equiv 1~(\mathrm{mod}~4)\),    Case 2: \(m \equiv 3~(\mathrm{mod}~4)\)

Case 1: \(m \equiv 1~(\mathrm{mod}~4)\). Then, taking (1.3) modulo 4 implies that \(3^y \equiv 1~(\mathrm{mod}~4)\), so y is even.

On the other hand, taking (1.3) modulo 3, together with our assumption \(m \not \equiv 0~(\mathrm{mod}~3)\), implies that

$$\begin{aligned} (-1)^x +(-1)^y \equiv (pm)^z \not \equiv 0~(\mathrm{mod}~3), \end{aligned}$$
(3.1)

which contradicts the fact that x is odd and y is even. Hence we obtain \(x=1\).

Case 2: \(m \equiv 3~(\mathrm{mod}~4)\). Then \(\displaystyle \left( \frac{3pm^2-1}{p(p-3)m^2+1}\right) =1 \) and \(\displaystyle \left( \frac{pm}{p(p-3)m^2+1}\right) =-1,~ \) where \(\displaystyle {\left( \frac{*}{*}\right) }\) denotes the Jacobi symbol. Indeed,

$$\begin{aligned} \left( \frac{3pm^2-1}{p(p-3)m^2+1}\right) = \left( \frac{3pm^2+p(p-3)m^2}{p(p-3)m^2+1}\right) = \left( \frac{p^2m^2}{p(p-3)m^2+1}\right) =1 \end{aligned}$$

and

$$\begin{aligned} \left( \frac{pm}{p(p-3)m^2+1}\right)= & {} \left( \frac{p}{p(p-3)m^2+1}\right) \left( \frac{m}{p(p-3)m^2+1}\right) \\= & {} -\left( \frac{p(p-3)m^2+1}{p}\right) \left( \frac{p(p-3)m^2+1}{m}\right) \\= & {} -1, \end{aligned}$$

since \(m \equiv 3~(\mathrm{mod}~4)\) and \(p \equiv 1~(\mathrm{mod}~4)\). In view of these, z is even from (1.3).

Taking (1.3) modulo 4 implies that \( 3^y \equiv (pm)^z \equiv 3^z \equiv 1~(\mathrm{mod}~4), \) since z is even. Hence y is even. Similarly, (3.1) also leads to a contradiction. We therefore obtain \(x=1\). \(\square \)

3.3 Pillai’s equation \(c^z-b^y=a\)

From Lemma 3.2, it follows that \(x=1\) in (1.3), provided that m is odd with \(m \not \equiv 0~(\mathrm{mod}~3)\). If \(y \le 2\), then we obtain \(y=1\) and \(z=2\) from (1.3). From now on, we may suppose that \(y \ge 3\). Hence our theorem is reduced to solving Pillai’s equation

$$\begin{aligned} c^z-b^y=a \end{aligned}$$
(3.2)

with \(y \ge 3\), where \(~a=3pm^2-1,~~b=p(p-3)m^2+1\) and \(c=pm.\)

We now want to obtain a lower bound for y.

Lemma 3.3

\(\displaystyle y > m^2-2.\)

Proof

Since \(y \ge 3\), Eq. (3.2) yields the following inequality:

$$\begin{aligned} (pm)^z \ge 3pm^2-1+(p(p-3)m^2+1)^3 > (pm)^3. \end{aligned}$$

Hence \(z \ge 4.\) Taking (3.2) modulo \(p^2m^4\) implies that

$$\begin{aligned} 3pm^2-1+1+p(p-3)ym^2 \equiv 0~(\mathrm{mod}~p^2m^4), \end{aligned}$$

so \(3+(p-3)y \equiv 0~(\mathrm{mod}~pm^2)\). Hence we have

$$\begin{aligned} y \ge \frac{~1~}{~p-3~}(pm^2-3)=\frac{~p~}{~p-3~}m^2-\frac{~3~}{~p-3~}>m^2-2, \end{aligned}$$

as desired. \(\square \)

We next want to obtain an upper bound for y.

Lemma 3.4

\(y < 2521 \log c.\)

Proof

From (3.2), we now consider the following linear form in two logarithms:

$$\begin{aligned} \Lambda = z\log c - y\log b \quad ({>}0). \end{aligned}$$

Using the inequality \(\log (1+t)<t\) for \(t>0\), we have

$$\begin{aligned} 0<\Lambda = \log \left( \frac{c^z}{b^y}\right) =\log \left( 1+\frac{a}{b^y}\right) <\frac{a}{b^y}. \end{aligned}$$
(3.3)

Hence we obtain

$$\begin{aligned} \log \Lambda < \log a - y\log b. \end{aligned}$$
(3.4)

On the other hand, we use Proposition 2.1 to obtain a lower bound for \(\Lambda \). It follows from Proposition 2.1 that

$$\begin{aligned} \log \Lambda ~\ge ~ -25.2~ \left( \max \left\{ \log b^{\prime }+0.38 , 10\right\} \right) ^2~ (\log b)~(\log c), \end{aligned}$$
(3.5)

where \(b^{\prime }=\displaystyle {\frac{y}{\log c} + \frac{z}{\log b}}.\)

We note that \(b^{y+1}>c^z\). Indeed,

$$\begin{aligned} b^{y+1}-c^z = b(c^z-a)-c^z= & {} (b-1)c^z-ab \ge p(p-3)m^2 \cdot p^2m^2 \\&-(3pm^2-1)(p(p-3)m^2+1)>0. \end{aligned}$$

Hence \(b^{\prime }<\displaystyle \frac{~2y+1~}{~\log c~}.\)

Put \(M=\displaystyle \frac{y}{~\log c~}.\) Combining (3.4) and (3.5) leads to

$$\begin{aligned} y\log b < \log a + 25.2~ \left( \max \left\{ \log \left( 2M+\frac{1}{~\log c~}\right) +0.38,~ 10 \right\} \right) ^2~(\log b)~(\log c), \end{aligned}$$

so

$$\begin{aligned} M < 1+ 25.2~ \left( \max \left\{ \log \left( 2M+1 \right) +0.38,~ 10 \right\} \right) ^2, \end{aligned}$$

since \(\log c=\log (pm) \ge \log 5>1\). We therefore obtain \( M < 2521\). This completes the proof of Lemma 3.4. \(\square \)

We are now in a position to prove Theorem 1.1. It follows from Lemmas 3.3, 3.4 that

$$\begin{aligned} m^2 -2< 2521\log (pm). \end{aligned}$$
(3.6)

We want to obtain an upper bound for p and then one for m. We first show that if \(m \equiv 3~(\mathrm{mod}~4),\) then \(p<3784\). Recall that z is even for the case \(m \equiv 3~(\mathrm{mod}~4)\), as seen in the proof of Lemma 3.2. Put \(z=2Z\) with Z positive integer. Now Eq. (3.2) can be written as

$$\begin{aligned} (c^2)^Z-b^y= c^2-b. \end{aligned}$$

Then \(y \ge Z\). If \(y=Z\), then we obtain \(y=Z=1\). If \(y>Z\), then we consider a “gap” between the trivial solution \((y,Z)=(1,1)\) and (possibly) another solution (y, Z).

From \(a+b=c^2\) and \(a+b^y=c^{2Z}\), consider the following two linear forms in two logarithms:

$$\begin{aligned} \Lambda _0 = 2\log c - \log b ~(>0),~~~ \Lambda = 2Z\log c - y\log b ~(>0). \end{aligned}$$

Then

$$\begin{aligned} y\Lambda _0-\Lambda =2(y-Z)\log c \ge 2\log c, \end{aligned}$$

so

$$\begin{aligned} y> \frac{2}{~\Lambda _0 ~}\log c. \end{aligned}$$

By Lemma 3.4, we have \(\displaystyle \frac{2}{~\Lambda _0 ~}\log c <2512 \log c\). Hence

$$\begin{aligned} \frac{2}{~2521~}< \Lambda _0= & {} \log \left( \frac{c^2}{b}\right) =\log \left( 1+\frac{\,a\,}{b}\right)<\frac{\,a\,}{b} =\frac{~3pm^2-1~}{~p(p-3)m^2+1~}< \frac{~3pm^2~}{~p(p-3)m^2~}\\= & {} \frac{~3~}{~p-3~}. \end{aligned}$$

Consequently we obtain \(p < 3784\). When \(m \equiv 1~(\mathrm{mod}~4)\),we could not prove that z is even in Lemma 3.2. We therefore suppose that if \(m \equiv 1~(\mathrm{mod}~4)\), then \(p<3784\). In any case, (3.6) yields \(m \le 183.\)

From (3.3), we have the inequality

$$\begin{aligned} \left| \frac{\log b}{\log c}-\frac{z}{y}\right| < \frac{a}{~yb^y\log c~}, \end{aligned}$$

which implies that \(\displaystyle \left| \frac{\log b}{\log c}-\frac{z}{y}\right| < \frac{1}{2y^2}\) , since \(y \ge 3\). Thus \(\displaystyle \frac{z}{y}\) is a convergent in the simple continued fraction expansion to \(\displaystyle \frac{\log b}{\log c}\).

On the other hand, if \(\displaystyle \frac{p_r}{q_r}\) is the r-th such convergent, then

$$\begin{aligned} \left| \frac{\log b}{\log c} -\frac{p_r}{q_r}\right| > \frac{1}{~(a_{r+1} + 2)q_r^2~}, \end{aligned}$$

where \(a_{r+1}\) is the \((r + 1)\)-st partial quotient to \(\displaystyle \frac{\log b}{\log c}\) (see e.g. Khinchin [9]). Put \(\displaystyle \frac{z}{y}=\frac{p_r}{q_r}\). Note that \(q_r \le y\). It follows, then, that

$$\begin{aligned} a_{r+1} > \frac{~b^y \log c~}{ay}-2 \ge \frac{~b^{q_r}\log c~}{aq_r}-2. \end{aligned}$$
(3.7)

Finally, we checked by Magma [4] that for each \(p <3784\) with \(p \equiv 1~(\mathrm{mod}~4)\), inequality (3.7) does not hold for any r with \(q_r < 2521\log (pm)\) in the range \(3 \le m \le 183\). This completes the proof of Theorem 1.1.

4 Proof of Corollary 1.2

Let (x, y, z) be a solution of (1.4). By Theorem 1.1, we may suppose that \(m \equiv 0~(\mathrm{mod}~3)\). Recall that x is odd. Here, we apply Proposition 2.2. For this, we set \(p:=3, a_1:=10m^2+1, a_2:=1-15m^2, b_1:=y, b_2:=x\), and

$$\begin{aligned} \Lambda :=(10m^2+1)^y-(1-15m^2)^x. \end{aligned}$$

Then we may take \(g=1, E=2, A_1=10m^2+1, A_2:=15m^2-1\). Hence we have

$$\begin{aligned} z \le \frac{36.1}{8 (\log 3)^{4}} \big (\max \{ \log b' + \log (2 \log 3) +0.4, \, 12 \log 3\} \big )^{ 2} \log (10m^2+1) \log (15m^2-1), \end{aligned}$$

where \(\displaystyle b':=\frac{y}{\log (15m^2-1)}+\frac{x}{\log (10m^2+1)}\). Suppose that \(z \ge 4\). We will observe that this leads to a contradiction. Taking (1.4) modulo \(m^4\), we find

$$\begin{aligned} 15x+10y \equiv 0~(\mathrm{mod}~m^2). \end{aligned}$$

In particular, we find \(M:=\max \{x,y\} \ge m^2/25\). Therefore, since \(z \ge M\) and \(b'\le \frac{M}{\log m}\), we find

$$\begin{aligned} M \le 3.1 \left( \max \left\{ \log \left( \frac{M}{\log m}\right) + \log (2 \log 3) +0.4, \, 12 \log 3\right\} \right) ^{ 2} \log (10m^2+1) \log (15m^2-1). \end{aligned}$$
(4.1)

If \(m \ge 3450\), then

$$\begin{aligned} M \le 3.1 \left( \log \left( \frac{M}{\log m}\right) + \log (2 \log 3) +0.4\right) ^{ 2} \log (10m^2+1) \log (15m^2-1). \end{aligned}$$

Since \(m^2 \le 25M\), the above inequality gives

$$\begin{aligned} M \le 3.1 \left( \log {M} -\log (\log 3450)+1.19\right) ^{ 2} \log (250M+1) \log (375M-1). \end{aligned}$$

We therefore obtain \(M \le 105186\), which contradicts the fact that \(M \ge m^2/25 \ge 476100.\)

If \(m<3450\), then inequality (4.1) gives

$$\begin{aligned} \frac{~m^2~}{~25~} \le 539 \log (10m^2+1) \log (15m^2-1). \end{aligned}$$

This implies \(m \le 2062\). Hence all x, y and z are also bounded. It is not hard to verify by Magma [4] that there is no (m, x, y, z) under consideration satisfying (1.4). We conclude \(z \le 3\). In this case, one can easily show that \((x,y,z)=(1,1,2)\). This completes the proof of Corollary 1.2.

Remark

In the same way as in the proof of Corollary 1.2, we can completely solve “the remaining case” \(m \equiv 0~(\mathrm{mod}~3)\) of Eq. (1.2), which is shown by Su-Li [17] and Bertók [1].