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A note on the equation ${x}^{y}+{y}^{z}={z}^{x}$
Journal of Inequalities and Applicationsvolume 2014, Article number: 170 (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.
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
Using the Gel’fondBaker 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 ${f}^{\prime}(X)=(logX1)/{(logX)}^{2}$, 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
Therefore, if $h(4D)>D$, then from (2.4) we get
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 ${D}_{2}<0$.
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
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
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 $p\nmid b{L}_{1}(\alpha ,\beta )\cdots {L}_{n1}(\alpha ,\beta )$. A Lucas pair $(\alpha ,\beta )$ such that ${L}_{n}(\alpha ,\beta )$ has no primitive divisor will be called an ndefective Lucas pair. Further, a positive integer n is called totally nondefective if no Lucas pair is ndefective. □
Lemma 2.8 ([7])
Let n satisfy $4<n\le 30$ and $n\ne 6$. Then, up to equivalence, all parameters of ndefective 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 nondefective.
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
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
or
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,
and
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
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
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
has the solution
Applying Lemma 2.7 to (4.1) and (4.2), we have
where ${X}_{1}$, ${Y}_{1}$, ${Z}_{1}$ are positive integers satisfying
and
Let
We see from (4.5) and (4.7) that $\alpha +\beta =2{X}_{1}$ and $\alpha \beta ={z}^{{Z}_{1}}$ are coprime nonzero integers, $\alpha /\beta =(({X}_{1}^{2}y{Y}_{1}^{2})+2{X}_{1}{Y}_{1}\sqrt{y})/{z}^{{Z}_{1}}$ is not a root of unity. Hence, $(\alpha ,\beta )$ is a Lucas pair with parameters $(2{X}_{1},4y{Y}_{1}^{2})$. Further, Let ${L}_{n}(\alpha ,\beta )$ ($n=0,1,2,\dots $) denote the corresponding Lucas numbers. By (4.4) and (4.7), we have
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
Let $d=gcd({Y}_{1},3{X}_{1}^{2}y{Y}_{1}^{2})$. 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), $gcd(y,3{X}_{1}^{2}y{Y}_{1}^{2})=1$ and
It implies that
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.
If $t=1$, then from (4.3) and (4.6) that $x={Z}_{1}$, $x\mid h(4y)$ and
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.
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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.
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The authors declare that they have no competing interests.
Authors’ contributions
YL obtained the theorems and completed the proof. XL corrected and improved the final version. Both authors read and approved the final manuscript.
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Keywords
 inequality
 exponential Diophantine equation
 primitive divisor of Lucas number
 existence