Find all strictly positive integers $a, b>1$ such that $a$ divides $b+1$ and $b$ divides $a^{3}-1$
(Solved by Yassine Hamdi)

We have $b=a k-1, a^{3}=b l+1$ with $k, l \geqslant 1$. Therefore, $a^{3}=a k l-(l-1)$, so $l=a m+1, m \geqslant 0$. Therefore, $a^{3}=a k(a m+1)-a m$ and

$$
a^{2}-k m a+(m-k)=0
$$

Since $a$ is an integer, the discriminant of the equation $X^{2}-k m X+(m-k)$ is also an integer. Therefore, there exists $d \geqslant 0$ such that

$$
4(m-k)=(k m-d)(k m+d)
$$

We distinguish two cases:

- If $k=m, a^{2}-a k^{2}=0, a=k^{2}$ and $b+1=k^{3}$. We verify that for all $k \geqslant 2$, the pairs

$\left(k^{2}, k^{3}-1\right)$ are solutions to the problem.

- If $k \neq m$, we find $k m \leqslant k m+d \leqslant 4|m-k|$. Again, two cases are to be distinguished.
- If $m>k, k m \leqslant 4 m-4 k \leqslant 4 m$ so $k \in\{1,2,3\}$. If $k=1, b=a-1$ and we verify that for $a \geqslant 2$, the pair $(a, a-1)$ works. If $k=2$, $a=(b+1) / 2$, and $b \mid a^{3}-1$ gives $8 b \mid(b+1)^{3}-8$. However, $(b+1)^{3}-8 \equiv-7 \bmod b$ so $b \mid 7$ and $b=7, a=4$, of the form $\left(k^{2}, k^{3}-1\right)$, already found in a previous case. If $k=3$, we similarly find $b \mid 26$ and the solution $(9,26)$, still of the form $\left(k^{2}, k^{3}-1\right)$.
- If $m<k, k m \leqslant 4 k-4 m \leqslant 4 k$ so $m \in\{0,1,2,3,4\}$. The case $m=0$ gives $K=a^{2}$ and $b=a^{3}-1$. For $a \geqslant 2$, the pairs $a, a^{3}-1$ are indeed solutions. The other four cases do not provide any new solutions.

Conclusion: The solutions are the pairs $(a, a-1)$ and $\left(a^{p}, a^{3}-1\right)$ with $a \geqslant 2$ and $p \in\{1,2\}$.