Number of ordered sets of integers How many ordered sets of integers $(x,y,z)$  satisfying $$x,y,z \in [-10,10]$$  are solutions to the following system of equations:
                           $$x^2y^2+y^2z^2=5xyz$$
                          $$y^2z^2+z^2x^2=17xyz$$
                           $$z^2x^2+x^2y^2=20xyz$$
By a quite cumbersome method, which is too long for me to type down here, I arrived at the answer 5: the ordered sets being $(0,0,0), (8,2,4), (-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$. But this does not seem to be the right answer. What I basically did was divide the system of equations with $(xyz)$, then by a process of elimination of the variables, found the possible values of x, y and z. However, it seems I've somehow missed out other solutions as well. Can anyone help me out with this? 
Thanks!

Let $(x,y,z)$ be a solution. Then the three numbers
$$
u_1=x^2y^2+y^2z^2-5xyz, \ u_2=y^2z^2+z^2x^2-17xyz, \ u_3=z^2x^2+x^2y^2=20xyz
$$
are all zero. 
Now 
$$
\begin{array}{lcl}
(1) \ -u_1+u_2+u_3 &=& 2x^2z^2-32xyz=32xz\bigg( \frac{xz}{16}-y\bigg) \\
(2) \ u_1-u_2+u_3 &=& 2x^2y^2-8xyz=8xy\bigg( \frac{xy}{4}-z\bigg) \\
(3) \ u_1+u_2-u_3 &=& 2y^2z^2-2xyz=2yz\bigg( yz-x\bigg)
\end{array}
$$
If $x,y$ and $z$ are all nonzero, we see that $x=yz, y=\frac{xz}{16}$ and $z=\frac{xy}{4}$. Multiplying, we deduce $xyz=\frac{(xyz)^2}{64}$, so $xyz=64$. Using 
$x^2=xyz, y^2=\frac{xyz}{16}$ and $z^2=\frac{xyz}{4}$, we deduce $x^2=64,y^2=4,z^2=16$. 
This produces four solutions : $(8\varepsilon_1,2\varepsilon_2,4\varepsilon_1\varepsilon_2)$ where each of $\varepsilon_1$ and $\varepsilon_2$ can be $+1$ or $-1$.
If $z=0$, then by (2) we have $2(xy)^2=0$ and hence $x=0$ or $y=0$. Similarly when $x=0$ or $y=0$ : this will produce the solutions found by Johan.
