A month without replies, anyone?

let's assume that $x \geq y \geq z$, thus $3x^3 \geq x^3+y^3+z^3=nx^2y^2z^2$, so $x \geq \frac{ny^2z^2}{3}$. Because $x^3+y^3+z^3=nx^2y^2z^2$ thus $x^2|y^3+z^3$, so $2y^3 \geq y^3+z^3 \geq x^2 \geq \frac{n^2y^4z^4}{9}$. If $z>1$ then $y \leq \frac{18}{16n^2}$ so $y=1$ but $y \geq z$ - contradiction. If $z=1$ then $y^3+1 \geq \frac{n^2y^4}{9}$, so $9+\frac{9}{y^3} \geq n^2y$. -If $y=1$ then $x^3+2=nx^2$ so the only solution is $(x,y,z,n)=(1,1,1,3)$ $(x^2|2=>x=1)$ so $n=3$ satisfies. -If $y>1$ then $10 \geq n^2y$ so a) $n=1$ then for instance $(x,y,z,n)=(3,2,1,1)$ is a solution so $n=1$ also satisfies b) $n=2$ then $y=2$ or $y=1$ then there are no solutions-easy to check. c) $n=3$ and $y=1$ that gives $(1,1,1,3)$ what we had before. (q.e.d.) Answer : $n=1,3$

Actually, isn't this infinite descent? After reducing to one of x,y,z being 1, then solve n and get n=1 and n=3 only.

at last the sith has won with the latex code...