IMO 1994- Problem 4

Problem: Determine all ordered pairs $(m,n)$ of positive integers such that $\frac{n^3+1}{mn-1}$ is an integer.
Solution: Note that $mn-1$ and $m^3$ are relative prime. That $mn-1$ dividing $n^3+1$ is therefore equivalent to $mn-1$ dividing $m^3(n^3+1)=m^3n^3-1+m^3+1$, which is in turn equivalent to $mn-1$ dividing $m^3+1$. If $m=n$, we have $\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$. This is an integer if and only if $n=2$. We now consider the case $m>n$. If $n=1,\frac{2}{m-1}$ is an integer. This is so if and only if $m=2,3$. Suppose $n\ge 2$. Note that $n^3+1\equiv 1(\text{ mod }n)$ while $mn-1\equiv -1(\text{ mod }n)$. Hence $\frac{n^3+1}{mn-1}=kn-1$ for some positive integer $k$. Now $kn-1<\frac{n^3+1}{n^2-1}=n+\frac{1}{n-1}$ or $(k-1)n<1+\frac{1}{n-1}$. Hence $k=1$, so that $n^3+1=(mn-1)(n-1)$. This yields $m=\frac{n^2+1}{n+1}=n+1+\frac{2}{n-1}$, which is an integer if and only if $n=2,3$. In each case, we have $m=5$. In summary, there are $9$ solutions, namely $(2,2),(2,1),(3,1),(5,2),(5,3),(1,2),(1,3),(2,5),(3,5)$
the last $4$ obtained by symmetry.
Source: http://sms.math.nus.edu.sg/simo/IMO_Problems/94.pdf