12-146 Determine all positive integers $x, y$ that satisfy the equation $x+y^{2}+z^{3}=x y z$, where $z$ is the greatest common divisor of $x, y$.
(36th International Mathematical Olympiad Preliminary Question, 1995)
[Solution] Since $(x, y)=z$. We can set $x=z c, y=z b$, where $(c, b)=1$.
Thus, the given equation transforms to $z c+z^{2} b^{2}+z^{3}=z^{3} b c$
which simplifies to $c+z b^{2}+z^{2}=z^{2} b c$.
Therefore, $z \mid c$, meaning there exists an integer $a$ such that $c=z a$, thus (1) further simplifies to $a+b^{2}+z=z^{2} a b$,
or $a=\frac{b^{2}+z}{z^{2} b-1}$,
(1) If $z=1$, then
$$
a=\frac{b^{2}+1}{b-1}=b+1+\frac{2}{b-1}.
$$

Thus, $b-1 \mid 2$, so $b=2$ or 3. Correspondingly, $a=5$.
$$
x=z c=z(z a)=5.
$$

Therefore, the solutions are $(x, y)=(5,2)$ or $(5,3)$.
(2) If $z=2$, then $a=\frac{b^{2}+2}{4 b-1}$,

Thus, $16 a=\frac{16 b^{2}+3^{2}}{4 b-1}=4 b+1+\frac{33}{4 b-1}$,
$4 b-1 \mid 33$, so $b=1$ or 3. Correspondingly, $a=1$.
$x=z^{2} a=4 \cdot y=z b=2$ or 6.
Therefore, the solutions are $(x, y)=(4,2)$ or $(4,6)$.
(3) If $z^{2} \geqslant 3$, then
$$
z^{2} a=\frac{z^{2} b^{2}+z^{3}}{z^{2} b-1}=b+\frac{b+z^{3}}{z^{2} b-1}.
$$

Since $x=z^{2} a$ is a positive integer, we must have $\frac{b+z^{3}}{z^{2} b-1} \geqslant 1$,
which implies $b \leqslant \frac{z^{2}-z+1}{z-1}$.
Since $z \geqslant 3$, then $\frac{z^{2}-z+1}{z-1}<z+1$,
thus $b<z+1$, so $b \leqslant z$.
This leads to $a=\frac{b^{2}+z}{z^{2} b-1}<\frac{z^{2}+z}{z^{2}-1}=1+\frac{z+1}{z^{2}-1}<2$.
Therefore, $a=1$.
From (2), we have $1+b^{2}+z=b z^{2}$,
so $b$ is an integer solution to the equation $t^{2}-z^{2} t+z+1=0$. Considering the discriminant $z^{4}-4 z-4$ must be a perfect square, but
$$
\left(z^{2}-1\right)^{2}<z^{4}-4 z-4<z^{2},
$$

which means $z^{4}-4 z-4$ lies between two consecutive squares, and thus is not a square.
Therefore, when $z \geqslant 3$, the equation has no positive integer solutions.
In summary, the solutions to the equation are
$$
(x, y)=(4,2),(4,6),(5,2),(5,3).
$$