Show that equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$ While solving some old  INMO problems I found that one, and I am completely stuck at it. The problem is:

Show that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$.

I shall be thankful if you can provide me any hints or suggestions. Thanks.
 A: Instead of solving 
$$x^2+y^2+z^2 = (x-y)(y-z)(z-x) \tag{*1}$$
one look at a simpler problem first. Let's say you have found a non-trivial integer solution for 
$$(u-v)(v-w)(w-u)|u^2+v^2+w^2\tag{*2}$$ 
one can set $\lambda$ to the integer $\displaystyle\;\frac{u^2+v^2+w^2}{(u-v)(v-w)(w-u)}\;$ and $(x,y,z) = (\lambda u,\lambda v, \lambda w)$ will give us a solution for $(*1)$.
It turns out it isn't that hard to find solutions for $(*2)$. One just take any two non-zero integers $p, q$, set $(u,v,w) = (v-p,v,v+q)$ and looks for expression of $v$ which makes $$pq(p+q) \;|\; (v-p)^2 + v^2 + (v+q)^2 = 3v^2 -2(p-q)v + (p^2+q^2)$$
For $p = q = 1$, we find
$$(u,v,w) = (2t-1,2t,2t+1)
\quad\implies\quad \lambda = \frac{(2t-1)^2 + (2t)^2 + (2t+1)^2}{(-1)(-1)(2)} = 6t^2+1
$$
This will give us a parametrized family of solution of $(*1)$
$$(x,y,z) = ((2t-1)(6t^2+1), 2t(6t^2+1), (2t+1)(6t^2+1))$$
This demonstrate the original equation $(*1)$ does have infinitely many solutions.
Others solutions can be constructed in similar manner. For example,
take $p = 1, q = 2$, we find $v = 6t-1$ give us another family of solutions:
$$(x,y,z) =  ((6t-2)(18t^2-4t+1),(6t-1)(18t^2-4t+1),(6t+1)(18t^2-4t+1))$$
The more interesting question is whether there are some ways to systematically exhaust all solutions of $(*1)$ and I've no idea on that.
A: In this type of problems when you have to prove that there are infinitely many solutions it's convenient to look for possible values in P.A. So, suppose that $z-y=y-x=k$, then $z-x=2k$ and our equation becomes $$x^2+(x+k)^2+(x+2k)^2=2k^3.$$
So, after a little calculus we get the equation $3x^2+(6k)x+(5k^2-2k^3)=0$. Using the formula for the quadratic equation gives us $$x=\frac{-6k\pm \sqrt{(6k)^2-12(5k^2-2k^3)}}{6}=-k\pm \frac{k\sqrt{6(k-1)}}{3}.$$
Now, since we want $x\in \mathbb{Z}$ we need that $6(k-1)=u^2$ for some $u\in \mathbb{Z}$. Then $6\mid u^2$, so $6\mid u$ (why?). Set $u=6t$, thus we get $k=6t^2+1$ and hence we get $x=-(6t^2+1)\pm (6t^2+1)(2t)$. If we take the plus sign we get $x=12t^3-6t^2+2t-1$. Now, since $y=x+k$ and $z=x+2k$, replacing $x$ gives us $y=12t^3+2t$ and $z=12t^3+6t^2+2t+1$.
Finally, you can check that $$(12t^3-6t^2+2t-1)^2+(12t^3+2t)^2+(12t^3+6t^2+2t+1)^2=2(6t^2+1)^3.$$  
Hence, the equation has infintely many integer solutions.
