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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.

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  • $\begingroup$ 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. $\endgroup$
    – Gerhard Paseman
    Dec 5, 2017 at 16:29
  • $\begingroup$ @GerhardPaseman: Unfortunately, the answer is No. I have no idea on that problem. $\endgroup$
    – Safwane
    Dec 5, 2017 at 16:32
  • $\begingroup$ 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. $\endgroup$
    – user35593
    Dec 5, 2017 at 17:08
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    $\begingroup$ 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 $z^{x-y}>1$ $\endgroup$
    – it's a hire car baby
    Dec 5, 2017 at 18:38
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    $\begingroup$ That 2nd inequality gives us an ordering on $x,y,z$ $\endgroup$
    – it's a hire car baby
    Dec 5, 2017 at 18:45

1 Answer 1

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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.

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    $\begingroup$ That is a neat solution! $\endgroup$
    – Per Alexandersson
    Dec 5, 2017 at 19:38

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