A note on the equation $x^y+y^z=z^x$

Abstract

In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove that the exponential Diophantine equation $x^y+y^z=z^x$ has no positive integer solution $(x,y,z)$ with $2\mid y$.

MSC:11D61.

1 Introduction

Let ℤ, ℕ be the sets of all integers and positive integers, respectively. Recently, Zhang and Yuan [1] were interested in the equation

$$x^y+y^z=z^x,x,y,z\in \mathbb{N}.$$

(1.1)

Using the Gel’fond-Baker method, they proved that all solutions $(x,y,z)$ of (1.1) satisfy $max\{x,y,z\}<exp(exp(exp(5)))$. This upper bound is far beyond the computable scope at present. In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove the following result.

Theorem Equation (1.1) has no solution $(x,y,z)$ with $2\mid y$.

In addition, it is obvious that $(x,y,z)=(1,1,2)$ is a solution of (1.1). Because one have not found the other solutions, we propose a conjecture as follows:

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

Our theorem supports the above mentioned conjecture.

2 Preliminaries

Lemma 2.1 Let $f(X)=X/logX$, where X is a real number. Then $f(X)$ is an increasing function for $X>e$.

Proof Since $f^{\prime }(X)=(logX-1)/{(logX)}^2$, we have $f^{\prime }(X)>0$ for $X>e$. Thus, the lemma is proved. □

Lemma 2.2 Let $g(X)=\sqrt{X}-2(2+log(4X))/\pi$, where X is a real number. Then we have $g(X)>0$ for $X\ge 16$.

Proof Since $g^{\prime }(X)=1/2\sqrt{X}-2/\pi X>0$ for $X\ge 16$, g(X) is an increasing function satisfying $g(X)\ge g(16)>0$ for $X\ge 16$. The lemma is proved. □

Lemma 2.3 ([2, 3])

The equation

$$X^2+2^m=Y^n,X,Y,m,n\in \mathbb{N},gcd(X,Y)=1,n>2$$

(2.1)

has only the solutions $(X,Y,m,n)=(5,3,1,3)\mathit{\text{and}}(7,3,5,4)$.

Lemma 2.4 ([[4], Theorem 8.4])

The equation

$$X^2+Y^m=2^n,X,Y,m,n\in \mathbb{N},2\nmid Y,Y>1,m>1$$

(2.2)

has only the solution $(X,Y,m,n)=(13,7,3,9)$.

Lemma 2.5 ([[4], Theorem 8.4])

The equation

$$X^2-Y^m=2^n,X,Y,m,n\in \mathbb{N},2\nmid Y,Y>1,m>2,n>1$$

(2.3)

has only the solution $(X,Y,m,n)=(71,17,3,7)$.

Let D be a positive integer, and let $h(-4D)$ denote the class number of positive binary quadratic primitive forms of discriminant $-4D$.

Lemma 2.6 $h(-4D)\le D$.

Proof Notice that $h(-4)=h(-8)=h(-28)=1$, $h(-12)=h(-16)=h(-20)=h(-24)=h(-32)=h(-36)=h(-40)=h(-52)=h(-60)=2$, $h(-44)=3$, $h(-48)=h(-56)=4$. The lemma holds for $D\le 15$. By Theorems 11.4.3, 12.10.1, and 12.14.3 of [5], if $D\ge 1$, then

$$h(-4D)<\frac{2\sqrt{D}}{\pi }(2+log(4D)).$$

(2.4)

Therefore, if $h(-4D)>D$, then from (2.4) we get

$$\sqrt{D}<\frac{2}{\pi }(2+log(4D)).$$

(2.5)

But, by Lemma 2.2, (2.5) is impossible for $D\ge 16$. Thus, the lemma is proved. □

Lemma 2.7 Let k be a positive integer with $gcd(k,2D)=1$. Every solution $(X,Y,Z)$ of the equation

$$X^2+D{Y}^2=k^Z,X,Y,Z\in \mathbb{Z},gcd(X,Y)=1,Z>0,$$

(2.6)

can be expressed as

$$\begin{array}{c}Z=Z_{1}t,t\in \mathbb{N},\hfill \\ X+Y\sqrt{-D}={\lambda }_1{(X_1+{\lambda }_2{Y}_1\sqrt{-D})}^t,{\lambda }_1,{\lambda }_2\in \{\pm 1\},\hfill \end{array}$$

where $X_1$, $Y_1$, $Z_1$ are positive integers satisfying

$$X_1^2+D{Y}_1^2=k^{Z_1},gcd(X_1,Y_1)=1,Z_1\mid h(-4D).$$

Proof This lemma is the special case of [[6], Theorems 1 and 2] for $D_1=1$ and $D_2<0$.

Let α, β be algebraic integers. If $\alpha +\beta$ and αβ are nonzero coprime integers and $\alpha /\beta$ is not a root of unity, then $(\alpha ,\beta )$ is called a Lucas pair. Further, let $a=\alpha +\beta$ and $c=\alpha \beta$. Then we have

$$\alpha =\frac{1}{2}(a+\lambda \sqrt{b}),\beta =\frac{1}{2}(a-\lambda \sqrt{b}),\lambda \in \{\pm 1\},$$

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 ${\alpha }_1/{\alpha }_2={\beta }_1/{\beta }_2=\pm 1$. Given a Lucas pair $(\alpha ,\beta )$, one defines the corresponding sequence of Lucas numbers by

$$L_n(\alpha ,\beta )=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },n=0,1,2,\dots .$$

For equivalent Lucas pairs $({\alpha }_1,{\beta }_1)$ and $({\alpha }_2,{\beta }_2)$, we have $L_n({\alpha }_1,{\beta }_1)=\pm L_n({\alpha }_2,{\beta }_2)$ for any $n\ge 0$. A prime p is called a primitive divisor of $L_n(\alpha ,\beta )$ ($n>1$) if $p\mid L_n(\alpha ,\beta )$ and $p\nmid b{L}_1(\alpha ,\beta )\cdots L_{n-1}(\alpha ,\beta )$. A Lucas pair $(\alpha ,\beta )$ such that $L_n(\alpha ,\beta )$ has no primitive divisor 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 ([7])

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:

• (i) $n=5$, $(a,b)=(1,5),(1,-7),(2,-40),(1,-11),(1,-15),(12,-76),(12,-1,364)$.

• (ii) $n=7$, $(a,b)=(1,-7),(1,-19)$.

• (iii) $n=8$, $(a,b)=(2,-24),(1,-7)$.

• (iv) $n=10$, $(a,b)=(2,-8),(5,-3),(5,-47)$.

• (v) $n=12$, $(a,b)=(1,5),(1,-7),(1,-11),(2,-56),(1,-15),(1,-19)$.

• (vi) $n\in \{13,18,30\}$, $(a,b)=(1,-7)$.

Lemma 2.9 ([8])

If $n>30$, then n is totally non-defective.

3 Further lemmas on the solutions of (1.1)

Throughout this section, we assume that $(x,y,z)$ is a solution of (1.1) with $(x,y,z)\ne (1,1,2)$.

Lemma 3.1 ([1])

x, y and z are coprime.

Lemma 3.2 $min\{x,y,z\}\ge 3$.

Proof Since $z^x=x^y+y^z>1$, we have $z>1$. If $x=1$, since $(x,y,z)\ne (1,1,2)$, then $y>1$ and $z=1+y^z\ge 1+2^z\ge z+3$, a contradiction. Similarly, if $y=1$, then $x>1$ and $x+1=z^x\ge 2^x\ge x+2$, a contradiction. Therefore, we have $min\{x,y,z\}\ge 2$.

If $x=2$, then

$$2^y+y^z=z^2.$$

(3.1)

Further, by Lemma 3.1, y and z are odd integers with $min\{y,z\}\ge 3$. Hence, we see from (3.1) that (2.3) has the solution $(X,Y,m,n)=(z,y,z,y)$. But, by Lemma 2.5, it is impossible.

Similarly, if $y=2$ or $z=2$, then we have

$$x^2+2^z=z^x,2\nmid xz,min\{x,z\}\ge 3$$

(3.2)

or

$$x^y+y^2=2^x,2\nmid xy,min\{x,y\}\ge 3.$$

(3.3)

But, by Lemmas 2.3 and 2.4, (3.2) and (3.3) are impossible. Thus, we get $min\{x,y,z\}\ge 3$. The lemma is proved. □

Lemma 3.3 $y<x$.

Proof By (1.1), we have $z^x>x^y$ and $z^x>y^z$. Hence,

$$\frac{x}{logx}>\frac{y}{logz}$$

(3.4)

and

$$\frac{x}{logy}>\frac{z}{logz}.$$

(3.5)

In addition, by Lemmas 3.1 and 3.2, x, y and z are distinct.

If $x<y<z$, by Lemma 3.2, then $3\le x<y<z$. Hence, by Lemma 2.1, we get

$$\frac{z}{logz}>\frac{x}{logx}>\frac{x}{logy},$$

(3.6)

which contradicts (3.5). Similarly, we can remove the case that $x<z<y$.

If $z<x<y$, then $3\le z<x<y$ and

$$\frac{y}{logz}>\frac{y}{logy}>\frac{x}{logx},$$

(3.7)

which contradicts (3.4). Thus, we get $y<x$. The lemma is proved. □

4 Proof of theorem

We now assume that $(x,y,z)$ is a solution of (1.1) with $2\mid y$. Since $(x,y,z)\ne (1,1,2)$, by Lemmas 3.1, 3.2 and 3.3, we have $2\nmid xz$, $gcd(y,z)=1$, $min\{x,y,z\}\ge 3$ and $x>y$.

We see from (1.1) that the equation

$$X^2+y{Y}^2=z^Z,X,Y,Z\in \mathbb{Z},gcd(X,Y)=1,Z>0$$

(4.1)

has the solution

$$(X,Y,Z)=(x^{y/2},y^{(z-1)/2},x).$$

(4.2)

Applying Lemma 2.7 to (4.1) and (4.2), we have

$$x=Z_{1}t,t\in \mathbb{N},$$

(4.3)

$$x^{y/2}+y^{(z-1)/2}\sqrt{-y}={\lambda }_1{(X_1+{\lambda }_2{Y}_1\sqrt{-y})}^t,{\lambda }_1,{\lambda }_2\in \{\pm 1\},$$

(4.4)

where $X_1$, $Y_1$, $Z_1$ are positive integers satisfying

$$X_1^2+y{Y}_1^2=z^{Z_1},gcd(X_1,Y_1)=1$$

(4.5)

and

$$Z_1\mid h(-4y).$$

(4.6)

Let

$$\alpha =X_1+Y_1\sqrt{-y},\beta =X_1-Y_1\sqrt{-y}.$$

(4.7)

We see from (4.5) and (4.7) that $\alpha +\beta =2X_1$ and $\alpha \beta =z^{Z_1}$ are coprime nonzero integers, $\alpha /\beta =((X_1^2-y{Y}_1^2)+2X_1{Y}_1\sqrt{-y})/z^{Z_1}$ is not a root of unity. Hence, $(\alpha ,\beta )$ is a Lucas pair with parameters $(2X_1,-4y{Y}_1^2)$. Further, Let $L_n(\alpha ,\beta )$ ($n=0,1,2,\dots$) denote the corresponding Lucas numbers. By (4.4) and (4.7), we have

$$y^{(z-1)/2}=|L_t(\alpha ,\beta )|.$$

(4.8)

We find from (4.7) and (4.8) that the Lucas number $L_t(\alpha ,\beta )$ has no primitive divisor. Therefore, by Lemma 2.9, we have $t\le 30$. Further, since $2\nmid x$ and $2\nmid t$ by (4.3), it is easy to remove all cases in Lemma 2.8 and conclude that $t\in \{1,3\}$.

If $t=3$, then from (4.4) we get

$$y^{(z-1)/2}={\lambda }_1{\lambda }_2{Y}_1(3X_1^2-y{Y}_1^2).$$

(4.9)

Let $d=gcd(Y_1,3X_1^2-y{Y}_1^2)$. Since $gcd(X_1,Y_1)=1$, we have $d\mid 3$ and $d\in \{1,3\}$. Further, since $t\mid x$, we get $3\mid x$, $3\nmid y$ and $d\ne 3$ by (4.9). Therefore, we have $d=1$ and, by (4.9), $gcd(y,3X_1^2-y{Y}_1^2)=1$ and

$$Y_1=y^{(z-1)/2},3X_1^2-y{Y}_1^2=\pm 1.$$

(4.10)

It implies that

$$3X_1^2\mp 1=y^z.$$

(4.11)

But, since $2\mid y$ and $z\ge 3$, we get from (4.11) that $2\nmid X_1$ and $0\equiv y^z\equiv 3X_1^2\mp 1\equiv 3\mp 1\not\equiv 0(mod8)$, a contradiction.

If $t=1$, then from (4.3) and (4.6) that $x=Z_1$, $x\mid h(-4y)$ and

$$x\le h(-4y).$$

(4.12)

But recall that $x>y$, by Lemma 2.6, (4.12) is impossible. Thus, (1.1) has no solution $(x,y,z)$ with $2\mid y$. The theorem is proved.