<s>[INST] Integer solutions of the equation $x^2+y^2+z^2 = 2xyz$ 
Calculate all integer solutions $(x,y,z)\in\mathbb{Z}^3$ of the equation $x^2+y^2+z^2 = 2xyz$.

My Attempt:
We will calculate for $x,y,z>0$. Then, using the AM-GM Inequality, we have
$$
\begin{cases}
x^2+y^2\geq 2xy\\
y^2+z^2\geq 2yz\\
z^2+x^2\geq 2zx\\
\end{cases}
$$
So $x^2+y^2+z^2\geq xy+yz+zx$. How can I solve for $(x,y,z)$ after this?
 [/INST] Suppose that $(x,y,z)$ is a solution. An even number of these must be odd. If two are odd, say $x$ and $y$, then $x^2+y^2$ has shape $4k+2$, and therefore so does $x^2+y^2+z^2$, since $z^2$ is divisible by $4$. But $2xyz$ has shape $4k$. 
So $x,y,z$ are all even, say $2u,2v,2w$. Substituting we get $u^2+v^2+w^2=4uvw$.
Again, $u,v,w$ must be all even. 
Continue, forever. We conclude that $x$, $y$, and $z$ are divisible by every power of $2$. 
It follows that $x=y=z=0$.
Remark: The same argument can be used for $x^2+y^2+z^2=2axyz$. 
This is an instance of Fermat's Method of Infinite Descent, aka induction. 
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