## Task A-4.4.

Determine all pairs of natural numbers $(m, n), m, n>1$, for which $n^{3}-1$ is divisible by $m n-1$.
## Solution.

Let $m, n>1$ be such that $m n-1 \mid n^{3}-1$.

We have $\left(n^{3}-1\right) m-n^{2}(m n-1)=n^{2}-m$, so $m n-1 \mid n^{2}-m$.

Also, $m\left(n^{2}-m\right)-(m n-1) n=n-m^{2}$, so $m n-1 \mid n-m^{2}$.

If $n>m^{2}$, then $m n-1 \leq n-m^{2} \leq n-1$, which implies $m n \leq n$, which is impossible.

If $n=m^{2}$, then clearly $m^{3}-1 \mid m^{6}-1$, so all pairs $\left(m, m^{2}\right), m>1$ are solutions.

If $n<m^{2}$, then $m n-1 \leq n^{2}-m < n^{2}-1$, which implies $m n \leq n^{2}$, and since $m>1$, we have $n \geq m$. If $n=m$, then $m n-1 \mid n^{3}-1$ becomes $m^2-1 \mid m^3-1$, which is true, so all pairs $(m, m), m>1$ are also solutions.

The condition is satisfied by all pairs of the form $\left(k, k^{2}\right)$ and $\left(k^{2}, k\right)$, where $k \in \mathbb{N}, k>1$.