Let $m \geq n$ $m^n | m^m+n^n \to m^n |n^n \to m=n$ $m^{2m}=2m^m$ there are not solutions.

Taking mod n on the diophane we have: $$m \equiv 0 \pmod n$$And taking mod m on the diophane we have: $$n \equiv 0 \pmod m$$Thus $n=m$ and now replacing on the diophane we have $n^n=2$ which is a contradiction, thus we have no sols. And we are done

MathLuis wrote: Taking mod n on the diophane we have: $$m \equiv 0 \pmod n$$And taking mod m on the diophane we have:

$$n \equiv 0 \pmod m$$...

No.

@above,why?

@above Because $m^m\equiv 0\mod n$ does not imply $m\equiv 0\mod n$.

$$where is N3$$

$m$ and $n$ have the same prime divisors. W.L.O.G $m\ge n$ If p is a prime number, $v_p(m)\ge v_p(n)+1$ $v_p(m^m + n^n)=v_p(n^n)=nv_p(n)=v_p(m^n * n^m)=mv_p(n)+nv_p(m)$ contradiction. $$m=n$$not solutions.