$[$ Arithmetic of residues (miscellaneous).]

Solve the equation $x^{2}+y^{2}+z^{2}=2 x y z$ in integers.
If there is a non-zero solution to this equation in integers, then there is also a solution in natural numbers. Let's prove that there are no solutions in natural numbers.

Suppose this is not the case. Then consider the smallest natural number \(a\), for which there exist such natural numbers \(b, c, n\) that \(a^{2} + b^{2} + c^{2} = 2^{n} a b c\). Note that all terms on the left side are even (otherwise, the left side is not divisible by 4, while the right side is). Substituting \(a = 2u, b = 2v, c = 2w\) we get \(4(u^{2} + v^{2} + w^{2}) = 2^{n+3} u v w\), or \(u^{2} + v^{2} + w^{2} = 2^{n+1} u v w\). Since \(u < a\), this contradicts the choice of the number \(a\).

## Answer

\((0,0,0)\).