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Well, I have been trying to prove that:
$$x^2 + y^2 + z^2 = 2xyz \implies x = y = z = 0$$
and have made little progress. Till now, I have only been able to prove that if this is to happen then $x$, $y$ and $z$ must be even. I am explaining how I arrived to this result but don't know how to proceed further or even figuring this out would make sense.
If $x^2 + y^2 + z^2 = 2xyz$, then at least one of $x$, $y$ or $z$ must be a multiple of 2. Let that one number be x and now $2|x$. Also let $x = 2a$.Then, $4a^2 + y^2 + z^2 = 4ayz$.
Now assume that one of $y$ or $z$ is odd. So the other must be odd as well as odd and odd add upto even and we need an even number to make the equality true. But it can be proved that the sum of squares of two odd numbers is never a multiple of $4$. Then the LHS would no longer be a multiple of 4 ($4a^2$ is a multiple of $4$ but $y^2 + z^2$ isn't) while the RHS would be. This is a contradition. Hence, $y$ and $z$ both are even. That's what I have been able to prove till now and still don't know if that is useful.
Please help if you have an answer.