<s>[INST] 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.
 [/INST] 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.
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