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

Question

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.

Answer 1

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.

Answer 2

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.