4. 176 Find all positive integer solutions \(x, y, z\) that satisfy the equation \(5(xy + yz + zx) = 4xyz\).

 untranslated text:
4. 176 求出所有满足方程 $5(x y+y z+z x)=4 x y z$ 的正整数 $x, y, z$.

translated text:
4. 176 Find all positive integer solutions \(x, y, z\) that satisfy the equation \(5(xy + yz + zx) = 4xyz\).
［Solution］The original equation can be transformed into
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5} \text {. }
$$

Assume without loss of generality that $x \leqslant y \leqslant z$, then we have
$$
\frac{3}{x} \geqslant \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5} \text {, }
$$

which implies $x1$. Hence, $1<x<4$.
We will discuss two cases:
Case 1 When $x=2$, we have
$$
\frac{2}{y} \geqslant \frac{1}{y}+\frac{1}{z}=\frac{4}{5}-\frac{1}{2}=\frac{3}{10} \text {. }
$$

This implies $3<y<7$.
If $y=4$, then $z=20$;
If $y=5$, then $z=10$;
If $y=6$, then $z$ is not an integer.
Case 2 When $x=3$, similar to Case 1, we get
$$
\frac{2}{y} \geqslant \frac{4}{5}-\frac{1}{3}=\frac{7}{15} \text {, and } \quad \frac{1}{y}<\frac{7}{15} \text {. }
$$

Therefore, $2<y<5$.
When $y=3$ or $y=4$, $z$ has no integer solutions.
Thus, the original equation has 12 solutions:
$$
\begin{array}{l}
(2,4,20) ;(2,20,4) ;(4,2,20) ;(20,2,4) ; \\
(4,20,2) ;(20,4,2) ;(2,5,10) ;(2,10,5) ; \\
(5,2,10) ;(10,2,5) ;(5,10,2) ;(10,5,2)
\end{array}
$$