17. Find all positive integers $(x, y, z, n)$, such that

$$
x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}
$$
17. Let $x \leq y \leq z, n$ be one solution. It is obvious that $z-x \geq y-x \geq 0$ and $z+x>y>0$. Hence $(z-x)(z+x) \geq y(y-x)$, i.e.

$$
z^{2} \geq x^{2}+y^{2}-x y=\frac{x^{3}+y^{3}}{x+y}
$$

or

$$
x+y \geq \frac{x^{3}+y^{3}}{z^{2}}
$$

Then it is easy to see that

$$
z=n x^{2} y^{2}-\frac{x^{3}+y^{3}}{z^{2}} \geq n x^{2} y^{2}-(x+y)>0
$$

except for $x=y=1, n=1,2$ (in this case there is no solution to the problem).

But $z^{2} \mid x^{3}+y^{3}$, and from the previous inequality we obtain

$$
x^{3}+y^{3} \geq z^{2} \geq\left(n x^{2} y^{2}-(x+y)\right)^{2}
$$

Therefore

$$
n^{2} x^{4} y^{4} \leq x^{3}+y^{3}+2 n x^{2} y^{2}(x+y)-(x+y)^{2}
$$

i.e.

$$
n^{2} x^{4} y^{4}<2 n x^{2} y^{2}(x+y)+x^{3}+y^{3}
$$

or

$$
n x y<2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}}
$$

If $x \geq 2$, then $y \geq x \geq 2$ and $n x y \geq 4$. But

$$
2\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{n x^{3}}+\frac{1}{n y^{3}} \leq 2\left(\frac{1}{2}+\frac{1}{2}\right)+\frac{1}{8}+\frac{1}{8}<3
$$

which is a contradiction. Hence $x=1$ and we have

$$
n y<2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}
$$

If $y \geq 4$, then $n y \geq 4$ and

$$
2+\frac{2}{y}+\frac{1}{n}+\frac{1}{n y^{3}}<2+\frac{2}{4}+1+\frac{1}{4}<4
$$

which is impossible. Therefore $y \leq 3$. Since $z^{2} \mid x^{3}+y^{3}$, i.e. $z^{2} \mid 1+y^{3}$ and $z \geq y$, we have several possibilities

(i) if $y=1$, then $1+y^{3}=2$ and $z=1$;

(ii) if $y=2$, then $1+y^{3}=9$ and $z=3$;

(iii) if $y=3$, then $1+y^{3}=28$ and $z$ does not exist.

If $x=y=z=1$, then $n=3$. If $x=1, y=2, z=3$, then $n=1$.

So all solutions of the problem are

$$
\begin{gathered}
(1,1,1,3) \quad(1,2,3,1) \quad(2,1,3,1) \quad(1,3,2,1) \\
(3,1,2,1) \quad(3,2,1,1) \quad(2,3,1,1)
\end{gathered}
$$