N40 (35-4, Australia) Find all ordered pairs of positive integers $(m, n)$ such that $\frac{n^{3}+1}{m n-1}$ is an integer.



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N40 (35-4, Australia) Find all ordered pairs of positive integers $(m, n)$ such that $\frac{n^{3}+1}{m n-1}$ is an integer.
Solving: Since $n^{3}+1=n^{3}+m n-(m n-1)$, we have
$$
m n-1 \mid n\left(n^{2}+m\right) \text {. }
$$

Since $(m n-1, n)=1$, it follows that $m n-1 \mid n^{2}+m$. Also,
$$
n\left(m^{2}+n\right)-\left(n^{2}+m\right)=m(m n-1),
$$

so $m n-1 \mid m^{2}+n$. Therefore, $m, n$ are symmetric, and without loss of generality, assume $m \geqslant n$.
When $n=1$, $m n-1=m-1$ is a divisor of $1^{3}+1=2$, thus $m=2$ or 3.
Now assume $n \geqslant 2$.
When $n \geqslant 2$, if $m=n$, then from $n^{2}-1 \mid n^{3}+1$, we get
$$
n^{2}-1 \mid\left(n^{3}-n\right)+(n+1),
$$

so $n^{2}-1 \mid n+1, n-1 \mid 1$, hence $n=2$, thus $m=n=2$.
If $m>n$, then $m \geqslant n+1$, so
$$
\begin{aligned}
2(m n-1) & \geqslant n^{2}+m n+n-2 \\
& >n^{2}+m,
\end{aligned}
$$

thus $m n-1=n^{2}+m$, which means
$$
(m-n-1)(n-1)=2,
$$

so
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ m = 5 , } \\
{ n = 3 , }
\end{array} \text { or } \quad \left\{\begin{array}{l}
m=5, \\
n=2 .
\end{array}\right.\right. \\
\end{array}
$$

Therefore, there are nine pairs of positive integers $(m, n)$:
$$
(2,1),(3,1),(1,2),(2,2),(5,2),(1,3),(5,3),(3,5) \text {, }
$$