On the Exponential Diophantine Equation $(m^2+m+1)^x+m^y=(m+1)^z$

Abstract

Let $m \ge 1$ be a positive integer. We show that the exponential Diophantine equation $(m^2+m+1)^x+m^y=(m+1)^z$ has no positive integer solutions other than $(x,y,z)=(1,1,2)$ when $m \not \in \{1, 2, 3 \}$.

1 Introduction

Let u, v, w be relatively prime positive integers greater than one and assume that the exponential Diophantine equation

$$\begin{aligned} u^{x}+v^{y}=w^{z} \end{aligned}$$

(1.1)

in positive integers x, y, z has a solution $(x_0, y_0, z_0).$ Two famous conjectures related to uniqueness of this solution $(x_0, y_0, z_0)$ are due to Jeśmanowicz and Terai with some restriction on (1.1). In 1956, Jeśmanowicz conjectured that if u, v and w are any Pythagorean triples, i.e., positive integers satisfying $u^{2}+v^{2}=w^{2},$ then the solution $(x_0,y_0,z_0)=(2,2,2)$ is the unique solution of (1.1) [5]. Another similar conjecture is proposed by Terai which states that if u, v, w, p, q, r are fixed positive integers satisfying $u^{p}+v^{q}=w^{r}$ with $u,v,w,p,q,r\ge 2$, then the Eq.  (1.1) has unique positive integer solution $(x_0,y_0,z_0)=(p,q,r)$ [19, 20]. Exceptional cases are listed explicitly in [24]. Although both conjectures are proved to be true in many special cases, see for example [1, 3, 4, 6, 8, 10,11,12,13, 18, 21,22,23, 25], they are still remain an unsolved problem yet. We refer to [9, 17] for a detailed information on these two conjectures. In this note we study the exponential Diophantine equation

$$\begin{aligned} (m^2+m+1)^x+m^y=(m+1)^z \end{aligned}$$

(1.2)

where $m>1$ is a positive integer, and we prove the following.

Theorem 1.1

Let $m>1$ be a positive integer. If $m > 3$ then the Eq. (1.2) has only the positive integer solution $(x,y,z) =(1,1,2).$ For $m=2$ and $m=3$ the Eq. (1.1) has exactly two solutions, namely $(x,y,z) =(1,1,2), (2,5,4)$ and $(x,y,z) =(1,1,2), (1,5,4),$ respectively.

In the above theorem, we exclude the case $m=1$ just for preserving the exponent in the expression $m^y.$ In fact, it is easy to see that the equation $3^x+1=2^z$ has only the positive integer solution $(x, z)=(1, 2)$ by considering it modulo 8. For the next two values of m,  the Eq. (1.2) turns into the equations $7^x+2^y=3^z$ and $13^x+3^y=4^z,$ for which both of them have more than one solution [14, 26]. So the aim of this study is to give an answer to the question whether or not the Eq. (1.2) has any positive integer solutions other than $(x,y,z) =(1,1,2)$ when $m>3.$ The proof depends on elementary congruence considerations and some results on linear forms in two $m-$adic logarithms.

2 Proof of Theorem 1.1

Lemma 2.1

Let (x, y, z) be a positive integer solution of the Eq. (1.2). The following conditions hold.

• 1. y is odd.

• 2. There exists an integer t such that $|x-y |=(m+1)t.$

Proof

• 1. By reducing Eq. (1.2) modulo $m+1$ we get that $1+(-1)^y \equiv 0 \pmod {(m+1)}$ which implies that y is odd since $m>1.$

• 2. If $x=y$ then we may take $t=0.$ So assume that $|x-y |\ge 1.$ It is clear from (1.2) that $z \ge 2.$ So we have that

  $$\begin{aligned} (m^2+m+1)^x +m^y&\equiv 0 \pmod {(m+1)^2} \\ (-m)^x +m^y&\equiv 0 \pmod {(m+1)^2} \\ m^{|x-y |} + (-1)^x&\equiv 0 \pmod {(m+1)^2} \\ ((m+1)-1)^{|x-y |} + (-1)^x&\equiv 0 \pmod {(m+1)^2} \\ (-1)^{|x-y |}+(-1)^{|x-y |-1 }(m+1) |x-y |+ (-1)^x&\equiv 0 \pmod {(m+1)^2} . \end{aligned}$$

  Taking into account that y is odd we get more precisely

  $$\begin{aligned} |x-y |&\equiv 0 \pmod {(m+1)}, \\ \end{aligned}$$

  which means that $|x-y |=(m+1)t$ for some positive integer t.

$\square$

Lemma 2.2

If $m \equiv 1 \pmod 4$ then $(x, y, z)=(1, 1,2)$ is the only solution of (1.2).

Proof

If $z \le 2$ then clearly $(x, y, z)=(1, 1,2)$ is the only solution of (1.2). Assume that $z \ge 3$ and $m=4k+1$ for some positive integer k. If x is even, then $(m^2+m+1)^x \equiv 1 \pmod 8$ and $m^{y-1} \equiv 1$, so from (1.2) we get that $1+m \equiv 0 \pmod 8$ which implies $2k+1 \equiv 0 \pmod 4,$ a contradiction. So x must be odd and hence $(m^2+m+1)^x \equiv m^2+m+1 \pmod 8 .$ Then again considering (1.2) modulo 8 we get that

$$\begin{aligned} (m+1)^2&\equiv 0 \pmod 8 \\ (4k+2)^2&\equiv 0 \pmod 8 \\ 4&\equiv 0 \pmod 8 \end{aligned}$$

which is a contradiction. Hence, $(x, y, z)=(1, 1,2)$ is the only solution of (1.2) when $m \equiv 1 \pmod 4.$ $\square$

Lemma 2.3

Let (x, y, z) be a positive integer solution of the Eq. (1.2). Then x and y are relatively prime integers. In particular, $x \ne y$ for $z >2.$

Proof

If $z \le 2$ then $x=y=1$ and hence the result is clear. So assume that $z \ge 3$ and that there exists an odd prime p such that $x=x_{1}p$ and $y=y_{1}p$ for some positive integers $x_{1}$ and $y_{1}$ since y is odd by Lemma 2.1. Let

$$\begin{aligned} K=(m^2+m+1)^{x_{1}}+m^{y_{1}}, \quad L= \frac{ (m^2+m+1)^{x_{1}p}+m^{y_{1}p} }{(m^2+m+1)^{x_{1}}+m^{y_{1}} } . \end{aligned}$$

So Eq. (1.2) is of the form

$$\begin{aligned} KL=(m+1)^z \end{aligned}$$

(2.1)

where $\gcd (K,L)=1$ or p. Note that $K \equiv 0 \pmod {m+1}.$ Hence if $\gcd (K,L)=1$ then $L=1$ which is clearly impossible for $p > 1.$ Thus, $\gcd (K,L)=p.$ Let $m+1=p^kq$ for some positive integer k such that $\gcd (p,q)=1.$ From (2.1) we have that either $K=p^{kz-1}q^z,$ $L=p$ or $K=pq^z,$ $L=p^{kz-1}.$ For $p>1$ it is easy to see that

$$\begin{aligned} p(m^2+m+1)^{x_{1}} < (m^2+m+1)^{x_{1}p} \end{aligned}$$

and

$$\begin{aligned} pm^{y_{1}} < m^{y_{1}p} . \end{aligned}$$

So the case $L=p$ leads to a contradiction. On the other hand it is known that $p^2 \not \mid L ,$ see, for example, [15, P1.2], thus for the case $K=pq^z,$ $L=p^{kz-1}$ we have the only possibility $kz-1=1$ which is also a contradiction since $z \ge 3.$ So, there do not exist such a prime p and hence x and y are relatively prime integers. $\square$

Let $m' > 1$ be an integer and let $m'=p_1^{t_1} \ldots p_k^{t_k}$ be the prime factorization of $m'$ for distinct primes $p_i.$ The proof of Theorem 1.1 mainly depends on a result due to Bugeaud [2] on linear forms in two $m'-$adic logarithms. Let $x_1 / y_1$ and $x_2 / y_2$ be two non-zero rational numbers with $x_1 / y_1 \ne \pm 1.$ In [2] Bugeaud provide an upper bound for the $m'-$adic valuation of

$$\begin{aligned} \Lambda = \left( x_1 / y_1 \right) ^{b_1} - \left( x_2 / y_2 \right) ^{b_2} \end{aligned}$$

whenever $v_{p_i} \left( x_1 / y_1 \right) = v_{p_i} \left( x_2 / y_2 \right) =0$ for all $1 \le i \le k$ where $b_1$ and $b_2$ are positive integers. Suppose that there exists a positive integer g which is coprime with $p_1, \ldots , p_k$ such that for all prime $p_i,$

$$\begin{aligned} v_{p_i} \left( \left( \dfrac{x_1}{y_1} \right) ^g -1 \right) \ge {t_i} , v_{p_i} \left( \left( \dfrac{x_2}{y_2} \right) ^g -1 \right) \ge 1, 1 \le i \le k \end{aligned}$$

(2.2)

and

$$\begin{aligned} v_2 \left( \left( \dfrac{x_1}{y_1} \right) ^g -1 \right) \ge 2, v_2 \left( \left( \dfrac{x_2}{y_2} \right) ^g -1 \right) \ge 2 \text { if } 2\mid m'. \end{aligned}$$

(2.3)

Theorem 2.4

([2], Theorem 2]) Let $A_{1}>1, A_{2}>1$ be real numbers such that

$$\begin{aligned} \log A_{i} \ge \max \{ \log |x_{i} |, \log |y_{i} |, \log m' \}, \quad i=1,2 \end{aligned}$$

and put

$$\begin{aligned} b'= \frac{b_{1}}{\log A_{2}}+\frac{b_{2}}{\log A_{1}}. \end{aligned}$$

Under the hypotheses (2.2) and (2.3) assume that $x_1 / y_1$ and $x_2 / y_2$ are multiplicatively independent. If $m',$ $b_1$ and $b_2$ are relatively prime then we have the upper estimate

$$\begin{aligned} v_{m'} (\Lambda ) \le \dfrac{53.6g}{ \left( \log m' \right) ^4 } \left( \max \{ \log b' +\log \log m' +0.64 , 4 \log m' \} \right) ^2 \log A_{1} \log A_{2}. \end{aligned}$$

Now we apply the above theorem to the Eq. (1.2) by considering the $(m+1)-$adic valuation.

Lemma 2.5

Let $m>7$. If $m \equiv 3 \pmod 4$ or $2 \mid m$ then the Eq. (1.2) has only the positive integer solution $(x, y, z)=(1, 1,2).$

Proof

If $z \le 2$ then then the assertion is trivially true. Assume that $z \ge 3 .$ Since y is odd, we rewrite the Eq. (1.2) as

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

and consider the $(m+1)-$adic valuation of $(m^2+m+1)^x - (-m)^y.$ Since ${(m+1)} \mid {m^2+m}$ , ${(m+1)} \mid {-m-1}$, and also $4 \mid {m^2+m} ,$ $4 \mid {(-m-1)}$ if $m+1$ is even. So, by Lemma 2.3, the hypotheses of Theorem 2.4 are satisfied for $g=1$ by taking $x_{1}:=m^2+m+1$ and $x_{2}:=-m.$ Thus, from Theorem 2.4 we have the estimate

$$\begin{aligned} z\le & {} \dfrac{53.6}{ \left( \log {(m+1)} \right) ^4 } \left( \max \{ \log b' +\log \log {(m+1)} +0.64 , 4 \log {(m+1)} \} \right) ^2 \nonumber \\&\times \log {(m^2+m+1)} \log {m} \end{aligned}$$

(2.4)

where $b'=\dfrac{x}{\log m} + \dfrac{y}{\log m^2+m+1}.$

First assume that $\log b' +\log \log {(m+1)} +0.64 > 4 \log {(m+1)}.$ We will show that this is not possible. Put $M=\max \{x, y\}.$ Then

$$\begin{aligned} M \left( \dfrac{1}{\log m} +\dfrac{1}{\log (m^2+m+1)} \right) \ge b'> \dfrac{(m+1)^4}{ e^{0.64} \log {(m+1)}} \end{aligned}$$

(2.5)

and it follows that $M > 2205$ since $m \ge 8.$ On the other hand, from the Eq. (1.2) we see that

$$\begin{aligned} x \dfrac{\log (m^2+m+1)}{\log (m+1)}< z \quad \text {and} \quad y \dfrac{\log m}{\log (m+1)} < z. \end{aligned}$$

Thus $M \dfrac{\log m}{\log (m+1)} < z.$ Combining this inequality and (2.4) together with (2.5) we get that

$$\begin{aligned} M \le 53.6 \left( \log M +U(m) +0.64 \right) ^2 V(m). \end{aligned}$$

(2.6)

where

$$\begin{aligned} U(m)= \log \left( \dfrac{\log {(m+1)}}{\log m} +\dfrac{\log {(m+1)}}{\log (m^2+m+1)} \right) \end{aligned}$$

and

$$\begin{aligned} V(m)= \dfrac{\log {(m^2+m+1)} }{ \left( \log {(m+1)} \right) ^3 }. \end{aligned}$$

For $m \ge 8 ,$ both U(m) and V(m) are decreasing and $U(m)\le U(8)<0.46 ,$ $V(m) \le V(8)< 0.41 .$ Thus from (2.6) we get

$$\begin{aligned} M < 22 \left( \log M +1.1 \right) ^2 , \end{aligned}$$

which implies $M<1576,$ a contradiction. So $\log b' +\log \log {(m+1)} +0.64 \le 4 \log {(m+1)}.$ In this case from (2.4) we have that

$$\begin{aligned} z < 53.6 \times 16 \times W(m), \end{aligned}$$

where $W(m)= \dfrac{\log m \log {(m^2+m+1)}}{( \log {(m+1)})^2}.$ In the above one can see that $W(m)< 2$ for all positive m, and hence we get that $z<1716.$ Therefore, y is also bounded as $0.95 y<y\dfrac{\log m}{\log (m+1)}<z < 1716$ and hence $y < 1807$ for $m \ge 8.$ Similarly, from the inequality

$$\begin{aligned} 1.96x< x\dfrac{\log m^2+m+1}{\log (m+1)}<z<1716 , \end{aligned}$$

we get $x<876$ for $m \ge 8.$ Thus all x, y and z are bounded. Moreover, from Lemma 2.1m is also bounded with $m+1<M<1807.$ As a final step we checked with a short computer program in Maple that the equation (1.2) has no solution other than $(x, y, z)=(1, 1,2)$ with these restrictions and those of Lemma 2.1 when m is in the range $8 \le m \le 1807.$ This completes the proof.

$\square$

Proof of Theorem 1.1

From Lemmas 2.2 and 2.5 it remains to check the Eq. (1.2) only for $m \in \{2, 3, 4 , 6, 7 \} .$ The results for the equations $7^x+2^y=3^z,$ $13^x+3^y=4^z,$ and $57^x+7^y=8^z$ which corresponds to the case $m=2,$ $m=3$ and $m=7$ in the Eq. (1.2) have already been established by a number of authors, at least [14, 26] and [16, Theorem 6] respectively. For $m=4,$ the equation (1.2) turns into the equation $21^x+4^y=5^z.$ If x is even then by [7] this equation has no solution in positive integers whereas if x is odd then by [16, Lemma 6] it has only one solution, namely $(x, y, z)=(1, 1,2).$ Finally we consider the equation $43^x+6^y=7^z$ for $m=6.$ If x is odd then the equation has only one solution $(x, y, z)=(1, 1,2)$ by [16, Lemma 6]. Suppose that x is even, say $x=2X$. If $y>1$ then from the congruence $1 \equiv 7^z \pmod 8$ we see that z is also even, say $z=2Z.$ Thus we write

$$\begin{aligned} 2^y3^y=(7^Z-43^X)(7^Z+43^Z). \end{aligned}$$

Note that only one of the factors in the right hand side is divisible by 4 and $3 \not \mid 7^Z+43^Z.$ So we have two possibilities

$$\begin{aligned}&7^Z-43^X=2^{y-1}3^y \\&7^Z+43^Z=2 \end{aligned}$$

or

$$\begin{aligned}&7^Z-43^X=2 \cdot 3^y \\&7^Z+43^Z=2^{y-1} \end{aligned}$$

Clearly the first one is impossible. From the second one, we get that $7^Z=2^{y-2} +3^y.$ Reducing this equation modulo 3,  we find that y is even, a contradiction. Therefore, we conclude that $y=1.$ Assume that x is even for otherwise the equation $43^x+6=7^z$ has only one solution $(x, y, z)=(1, 1,2)$ from [16, Lemma 6]. Let $x=2X .$ By reducing modulo 4 we see that z is odd, but it is easy to see that the equation $43^{2X}+6=7^{z}$ has no solution in positive integers when z is odd by considering it modulo 43. This completes the proof. $\square$