# Very interesting equation $x^y + y^z + z^x = x^z + y^x + z^y$

How to solve equation of the form $x^y + y^z + z^x = x^z + y^x + z^y$. I grouped like this: $(x^y-y^x) + (y^z -z^y) + (z^x - x^z) = 0$ one of the case is $x^y-y^x = 0; y^z - z^y = 0$ and $z^x - x^z = 0$. Can you discuss either trivial or non-trivial solutions of $(x, y , z)$ in $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{R}$ and $\mathbb{C}$? Thanks in advance.

-
In $\mathbb R$, you get lots of solutions by setting $x=0$, which leads you to solve $y^z=z^y$ or equivalently $f(y)=f(z)$ where $f(t) = (\log t)/t$. This equation has infinitely many solutions $(y,z)$ where $1\le y<e$ and $e<z$. –  Greg Martin Apr 15 '12 at 18:39
Also for other values of $x$ as well, continuing from those solutions. Here is an animation showing the solutions in $0<y\le 6$, $0 < z \le 6$ for $0 \le x \le 2$. Regions where $F(x,y,z) = x^y + y^z + z^x - (x^z + y^x + z^y) > 0$ are blue, $< 0$ are red. math.ubc.ca/~israel/problems/Fanim.gif –  Robert Israel Apr 16 '12 at 5:17

Note that switching any two of $(x,y,z)$ interchanges the two sides of the equation. In particular, the set of solutions is invariant under permutations, and any $x,y,z$ that are not all distinct is a solution (which I will consider a "trivial" solution).

Nontrivial integer solutions include $(0,2,4)$ and $(1,2,3)$ and their permutations. I don't know if there are any other integer solutions, but a computer search didn't find any small ones.

-

At first thought:

I could only think of the trivial triples $(x,y,z)$

$$(0,n,n), (1,n,n), (n,n,0), (n,n,1), (n,0,n), (n,1,n)$$ for any $n \in \mathbb{Z}$

But then further trying $x=2,3,4 \dots$ and $y=z$ also seem to work.

Therefore $$\{ (x,y,z) | x,y,z\in \mathbb{Z}, y=z \}$$ which also includes those when $x=y=z$

By symmetry

$$\{ (x,y,z) | x,y,z\in \mathbb{Z}, x=z \}$$

$$\{ (x,y,z) | x,y,z\in \mathbb{Z}, x=y \}$$

also should work.

-
Raman can you generalize the each case in others (R and C) –  gandhi Apr 16 '12 at 6:05
@gandhi, yes this idea can be applied to $\mathbb{R}$ and $\mathbb{C}$ as well, but I remember that Diophantine equations allow varibles to be $\mathbb{Z}$ only. (But then I studied this three decades ago) –  Kirthi Raman Apr 16 '12 at 13:00