Let us consider the following conjecture:


**Conjecture**: There are no integer solutions of the equation $$x^{y-z}z^{x-y}=y^{x-z}$$ with $x,y,z$ distinct positive integers greater than or equal to $2$.


I came across this result when studying some diophantine equations. Several attempts were made to find a solution, but without any success. By this question I want to see if someone can give me a conterexample to this conjecture.



About the validity of a new conjecture about a diophantine equation
The conjecture is true, in fact the equation has no solution in distinct positive real numbers. To see this, let us write the equation in the more symmetric form
$$ x^y y^z z^x = x^z y^x z^y. \tag{$\ast$}$$
We get the same equation after interchanging $x$ and $y$, or $y$ and $z$, i.e., after permuting the variables arbitrarily. Hence we can assume without loss of generality that $x>y>z>0$. Then, with the notation $a:=x-y$ and $b:=y-z$, the original equation becomes
$$ (y+a)^b (y-b)^a = y^{a+b}, $$
where each factor and each exponent is positive. Equivalently,
$$ (1+a/y)^b (1-b/y)^a = 1, $$
where each factor and each exponent is positive. However, this is impossible, since
$$ (1+a/y)^b (1-b/y)^a < (e^{a/y})^b (e^{-b/y})^a = 1.$$


**Added on 22 January 2021.** Recently I posted the equation $(\ast)$ to a non-professional discussion board, and to my surprise two entirely new solutions arose. They are not mine, but I sketch them here as they are really nice and instructive. I will assume that $x,y,z>0$ are distinct and $(\ast)$ holds. I will derive a contradiction in two new ways.


**First new proof (sketch).** By assumption, $u:=y/x$ and $v:=z/x$ satisfy $u^{v-1}=v^{u-1}$. This contradicts (after some thought) the fact that the function $t\mapsto\frac{\ln t}{t-1}$ is strictly decreasing on the positive axis (the function is not defined at $t=1$, but it extends analytically there).


**Second new proof (sketch).** By assumption, the determinant
$$\begin{vmatrix}1&x&\ln x\\1&y&\ln y\\1&z&\ln z\\\end{vmatrix}$$
vanishes, hence its rows are linearly dependent. This contradicts (after some thought) the fact that the function $t\mapsto\ln t$ is strictly concave on the positive axis.


