Junior Macedonian Mathematical Olympiad

We introduce the substitution $z=\gcd (x,y)$ and we get the equation $x+y^2+z^2=xyz$. There exist natural numbers $a$ and $b$ such that $x=az$ and $y=bz$. Then the equation gets the form $az+b^2{z}^2+z^2=ab{z}^3$, i.e. $a+b^{2}z+z=ab{z}^2$. Since the right-hand side is divisible by $z$ and two of the summands on the left-hand side are divisible by $z$, it follows that $a$ is divisible by $z$. Therefore, there exists a natural number $c$ such that $a=cz$. By substituting in the equation, the equation gets the form $cz+b^{2}z+z=cb{z}^3$, or $c+b^2+1=cb{z}^2$. Hence we get $b^2+1=c(b{z}^2-1)$. It is clear that $b{z}^2\ne 1$, since if that was not the case we would get $b^2+1=0$, which is impossible. Then we have $c=\frac{b^2+1}{b{z}^2-1}$.

By multiplying the equation by $z^2$, we get $c{z}^2=\frac{b^2{z}^2+z^2}{b{z}^2-1}=b+\frac{b+z^2}{b{z}^2-1}$. Since $c{z}^2$ is a natural number, $\frac{b+z^2}{b{z}^2-1}$ is also a natural number. Therefore, $b{z}^2-1\le b+z^2$, i.e. $$(z^2-1)(b-1)\le 2\dots \dots \dots \dots (1)$$ If $b=1$, then $c=\frac{2}{z^2-1}$ and hence $z^2=2$ or $z^2=3$, which is impossible. If $b=2$, then $c=\frac{5}{2z^2-1}$. If $2z^2-1=1$, then $z=1$. It follows that $c=5$, $a=5$, i.e. $x=5$ and $y=2$. If $2z^2-1=5$, then $z^2=3$, which is impossible. If $b=3$, then $c=\frac{10}{3z^2-1}$. The cases $3z^2-1=1$, $3z^2-1=5$ and $3z^2-1=10$ are impossible. If $3z^2-1=2$, then $z=1$. It follows that $c=5$, $a=5$, i.e. $x=5$ and $y=3$. If $b>3$ then from (1) it follows that $z=1$. Then $c=\frac{b^2+1}{b-1}=b+1+\frac{2}{b-1}$, from where we have $b=2$ or $b=3$, which contradicts the assumption that $b>3$. Therefore the solutions to the equation are $(x,y)=(5,2),(5,3)$.