Croatia_2018

The given condition implies that $$a^{2}bk=b^2+3a$$ for some positive integer $k$. The equation $a^{2}bk-3a=b^2$ implies $a\mid b^2$ and $b\mid 3a$. Therefore, the numbers $$\frac{b^2}{a},\frac{3a}{b}\text{and}\frac{b^2+3a}{a^{2}b}=\frac{b}{a^2}+\frac{3}{ab}$$ are integers. By multiplying the second and the third one, we conclude that $\frac{9a}{b^2}$ is an integer, meaning that $b^2\mid 9a$. Now we have $$b^{2}m=9a,an=b^2$$ for some positive integers $m$ and $n$. This implies $mn=9$. Hence, $n\in \{1,3,9\}$, which means that $b^2\in \{a,3a,9a\}$. The initial condition now implies $ab\mid n+3$.

We have three possible values for $n$.

If $n=1$, then $b^2=a$, so the initial condition becomes $b^5\mid 4b^2$, i.e. $b^3$ divides $4$. We obtain one solution $a=b=1$.

If $n=3$, then $b^2=3a$, so $\frac{b^5}{9}\mid 2b^2$, i.e. $b^3\mid 18$. Since $b$ is a multiple of $3$, there is no solution in this case.

If $n=9$, then $b^2=9a$, so $\frac{b^5}{81}\mid \frac{4}{3}{b}^2$, i.e. $b^3\mid 108$. Since $b$ is a multiple of $3$, we obtain a solution $(a,b)=(1,3)$.

The only solutions are $(1,1)$ and $(1,3)$.