Determine all integers $x,\ y,\ z$ that satisfy $x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$ 
Determine all integers $x,\ y,\ z$ that satisfy
  $$x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$$

We expand the RHS and sort the equation by $x$, which yields a quadratic equation with the discriminant:
$$D=-12y^{2}-12z^{2}+24yz+12y+12z+1$$
$x$ is an integer, thus $D$ must be square of an integer. We can write similar equalities by solving the original equation for $y$ and $z$. I couldn't get any further.
My other approaches include considering the greatest common divisor of $x,\ y,\ z$ and write $$x=da,\ y=db,\ z=dc$$
where $d=\gcd(x,y,z)$ and $\gcd(a,b,c)=1$ which wasn't much useful.

Let $y=u+x$ and $z=v+x$, then $y-z = u-v$ and hence
$$
\begin{align}
(y-u) + y + (y-u+v) &= u^2 + (u-v)^2 + v^2\\
3y &= 2u^2 + 2v^2 - 2uv +2u - v 
\end{align}
$$
So we can almost freely choose $u,v$, but we need RHS to be divisible by $3$. By considering the equation modulo $3$, we see that solutions are
$$
(u,v) \equiv (0,0),(0,2),(1,1),(1,2),(2,0),(2,1) \pmod 3
$$
For example, taking $(u,v)\equiv (0,0)\pmod 3$, this gives parametrizations:
$$
\begin{align}
u = 3m, v = 3n &\implies y= 2 m + 6 m^2 - n - 6 m n + 6 n^2\\
&\implies x = -m + 6 m^2 - n - 6 m n + 6 n^2\\
&\implies z = -m + 6 m^2 + 2n - 6 m n + 6 n^2
\end{align}
$$
where we can freely choose $m,n\in\mathbb Z$. The other cases are the same.  
Edit 1: One more case: let $(u,v)\equiv (2,1)\pmod 3$, so that
$$
u = 3m+2,\quad v = 3n+1
$$
then
$$
\begin{align}
y &= 3 + 8 m + 6 m^2 - n - 6 m n + 6 n^2\\
x = y-u &= 1 + 5 m + 6 m^2 - n - 6 m n + 6 n^2\\
z =v+x &= 2 + 5 m + 6 m^2 + 2 n - 6 m n + 6 n^2
\end{align}
$$
