19. N2 (AUS) ${ }^{\mathrm{IMO}}$ Determine all pairs $(m, n)$ of positive integers such that $\frac{n^{3}+1}{m^{2}-1}$ is an integer.
19. Since $m n-1$ and $m^{3}$ are relatively prime, $m n-1$ divides $n^{3}+1$ if and only if it divides $m^{3}\left(n^{3}+1\right)=\left(m^{3} n^{3}-1\right)+m^{3}+1$. Thus
$$
\frac{n^{3}+1}{m n-1} \in \mathbb{Z} \Leftrightarrow \frac{m^{3}+1}{m n-1} \in \mathbb{Z}
$$
hence we may assume that $m \geq n$. If $m=n$, then $\frac{m^{2}+1}{n^{2}-1}=n+\frac{1}{n-1}$ is an integer, so $m=n=2$. If $n=1$, then $\frac{2}{m-1} \in \mathbb{Z}$, which happens only when $m=2$ or $m=3$. Now suppose $m>n \geq 2$. Since $m^{3}+1 \equiv 1$ and $m n-1 \equiv-1(\bmod n)$, we deduce $\frac{n^{3}+1}{m n-1}=k n-1$ for some integer $k>0$. On the other hand, $k n-1<\frac{m^{2}+1}{n^{2}-1}=n+\frac{1}{n-1} \leq 2 n-1$ gives that $k=1$, and therefore $n^{3}+1=(m n-1)(n-1)$. This yields $m=\frac{n^{2}+1}{n-1}=n+1+\frac{2}{n-1} \in \mathbb{N}$, so $n \in\{2,3\}$ and $m=5$. The solutions with $m<n$ are obtained by symmetry:
There are 9 solutions in total: $(1,2),(1,3),(2,1),(3,1),(2,2),(2,5),(3,5)$, $(5,2),(5,3)$.