Solution of an equation with $3$ variables I've been thinking about a solution for the following equation for integers $x, y, z$:
$$6 = (x-y)^2 + (y-z)^2 + (x-z)^2$$
A possible approach would probably be to set $(x-y)^2 = a$, $(y-z)^2 = b$ and $(x-z)^2 = c$.
Now, the euqation woud look like this:
$$6 = a + b + c$$
Hence, the only possible solutions would be $$(a, b, c) = (4, 1, 1)$$ $$(a, b, c) = (1, 4, 1)$$ $$(a, b, c) = (1, 1, 4)$$
Now one could formulate equation for $x, y$ and $z$. However, this approach would be very time consuming. Is there a simpler solution to the problem?

Hint:
Let $x-y=a$ etc.
$\implies a+b+c=0\iff c=-?$
$$6=a^2+b^2+c^2=2a^2+2ab+2b^2\iff2a^2+a(2b)+2b^2-6=0$$
$$a=\dfrac{-2b\pm\sqrt{12(4-b^2)}}4=\dfrac{-b\pm\sqrt{3(4-b^2)}}2$$
For integer $a,$ $$b^2=4,1\implies b=\pm2,\pm1$$
We shall get the same set of values for $c,$ if we start with $b=-(c+a)$
So, $(a,b,c)\in(1,1,-2)$ or $\in(-1,-1,2)$ as $a+b+c=0$
