18. N4 (BUL) Find all positive integers $x$ and $y$ such that $x+y^{2}+z^{3}=x y z$. where $z$ is the greatest common divisor of $x$ and $y$ -
18. Let $x=z a$ and $y=z b$, where $a$ and $b$ are relatively prime. The given Diophantine equation becomes $a+z b^{2}+z^{2}=z^{2} a b$, so $a=z c$ for some $c \in \mathbb{Z}$. We obtain $c+b^{2}+z=z^{2} c b$, or $c=\frac{h^{2}+2}{2^{2} b-1}$.
(i) If $z=1$, then $c=\frac{h^{2}+1}{b-1}=b+1+\frac{2}{b-1}$, so $b=2$ or $b=3$. These values yield two solutions: $(x, y)=(5,2)$ and $(x, y)=(5,3)$.
(ii) If $z=2$, then $16 c=\frac{1 \cos ^{2}+32}{40-1}=4 b+1+\frac{33}{40-1}$, so $b=1$ or $b=3$. In this case $(x, y)=(4,2)$ or $(x, y)=(4,6)$.
(iii) Let $z \geq 3$. First, we see that $z^{2} c=\frac{z^{2} b^{2}+z^{2}}{z^{2} b-1}=b+\frac{b+z^{a}}{z^{2} b-1}$. Thus $\frac{b+z^{a}}{z^{2} b-1}$ must be a positive integer, so $b+z^{3} \geq z^{2} b-1$, which implies $b \leq$ $\frac{a^{2}-z+1}{a-1}$. It follows that $b \leq z$. But then $b^{2}+z \leq z^{2}+b2$. Therefore $c=\frac{b^{2}+z}{x^{2} b-1}<1$, a contradiction.
The only solutions for $(x, y)$ are $(4,2),(4,6),(5,2),(5,3)$.