# 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. – Gerhard Paseman Dec 5 '17 at 16:29
• @GerhardPaseman: Unfortunately, the answer is No. I have no idea on that problem. – Germany Dec 5 '17 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. – user35593 Dec 5 '17 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 $z^{x-y}>1$ – samerivertwice Dec 5 '17 at 18:38
• That 2nd inequality gives us an ordering on $x,y,z$ – samerivertwice Dec 5 '17 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 $$x^y y^z z^x = x^z y^x z^y.$$ 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 $$(1+a/y)^b (1-b/y)^a < (e^{a/y})^b (e^{-b/y})^a = 1.$$