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?

Another method:
It's clear zero is a solution and moreover $xyz\ge0$. Let's prove there are no more solutions over the integers. By the generalized mean theorem:
$$\sqrt{\frac{x^2+y^2+z^2}{3}}>\sqrt[3]{xyz}$$
Which leads to:
$$x^2+y^2+z^2 > 3(xyz)^{2/3}$$
So now we know that whenever $3(xyz)^{2/3} > 2xyz$ there are no solutions. Simple algebra shows that this is true whenever $xyz< 27/8$, so the only possible integer solutions are:
$$(0,0,0), (0,a,b), (1, 1, 2), (1, 1, 3)$$
And their permutations / sign changes. Now verify these do not satisfy the equation. 
$\blacksquare$
