# About the validity of a new conjecture about a diophantine equation

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.

• What do you know of the solutions? Can you show any properties or conclusions about them? Gerhard "Prefers Not Reinventing A Wheel" Paseman, 2017.12.05. Dec 5, 2017 at 16:29
• @GerhardPaseman: Unfortunately, the answer is No. I have no idea on that problem. Dec 5, 2017 at 16:32
• Wlog z <x,y and divide by z^(x-z) Then you can peove that z must be a divisor of x and also y. Then put x=az, y=bz and simplify. Dec 5, 2017 at 17:08
• By symmetry we can assume $x<y<z$ without loss of generality. Or alternatively we can assume $x^{y-z}<y^{z-x}<z^{x-y}$ without loss of generality; making $x^{y-z}<1$ and $z^{x-y}>1$ Dec 5, 2017 at 18:38
• That 2nd inequality gives us an ordering on $x,y,z$ Dec 5, 2017 at 18:45

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.