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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.

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    Also for other values of $x$ as well, continuing from those solutions. Here is an animation showing the solutions in 0, 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. http://www.math.ubc.ca/~israel/problems/Fanim.gif2012-04-16

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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.

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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.

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    @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)2012-04-16