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?

We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation. 
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd then, $$x^2+y^2+z^2\equiv(0+0+1)\mod4 ; 2xyz \equiv 0 \mod4$$  So, the only possibility is that all are even. 
Let x = 2X , y = 2Y ,z  = 2Z. Then, $$4X^2 +4Y^2 +4Z^2 = 16XYZ $$ $$\implies X^2 + Y^2 + Z^2 = 4XYZ$$ The same argument goes show that X,Y,Z are even. The process can be continued indefinitely. 
This is possible only when x = y = z = 0. So ,there exist no other solution other than this.
