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

- Yulin Lu
^{1}and - Xiaoxue Li
^{2}Email author

**2014**:170

https://doi.org/10.1186/1029-242X-2014-170

© Lu and Li; licensee Springer. 2014

**Received: **11 March 2014

**Accepted: **7 April 2014

**Published: **7 May 2014

## 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.

## Keywords

## 1 Introduction

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. □

*The equation*

*has only the solutions* $(X,Y,m,n)=(5,3,1,3)\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{0.25em}{0ex}}(7,3,5,4)$.

**Lemma 2.4** ([[4], Theorem 8.4])

*The equation*

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

**Lemma 2.5** ([[4], Theorem 8.4])

*The equation*

*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

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*

*can be expressed as*

*where*${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$

*are positive integers satisfying*

*Proof* This lemma is the special case of [[6], Theorems 1 and 2] for ${D}_{1}=1$ and ${D}_{2}<0$.

*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

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 $min\{x,y,z\}\ge 2$.

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.

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,

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

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

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 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 $t\in \{1,3\}$.

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

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.

## Declarations

### Acknowledgements

The authors would like to thank the referee for his very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P. S. F. (2013JZ001) and N.S.F. (11371291) of P.R. China.

## Authors’ Affiliations

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