Determine all pairs $(a, b) \in \mathbb{N}$ with $b \neq 1$ such that

$$
a+1 \mid a^{3} b-1 \quad \text { and } \quad b-1 \mid b^{3} a+1
$$
Let $(a, b)$ be a solution pair. We have $a^{3} b-1 \equiv(-1)^{3} b-1[a+1]$, so $b+1 \equiv 0[a+1]$ and $a+1 \mid b+1$. On the other hand, we have $b^{3} a+1 \equiv 1^{3} a+1[b-1]$, and thus $b-1 \mid a+1$. Therefore, $b-1 \mid b+1$, so $b-1 \mid 2$, and thus $b$ is 0, 2, or 3.

If $b=0, a+1 \mid 1$ and thus $a=0$. If $b=2, a+1$ is a multiple of $b-1=1$ and a divisor of $b+1=3$, so it is 1 or 3, and thus $a$ is 0 or 2. If $b=3, a+1$ is a multiple of $b-1=2$ and a divisor of $b+1=4$, so it is 2 or 4, and $a$ is 1 or 3.

We verify that if $a$ equals $b$ and is 0, 2, or 3, we have the desired divisibilities. We verify that $(0,2)$ is a solution and that $(1,3)$ is a solution.

The solutions are therefore: $(0,0),(1,1),(3,3),(1,3),(0,2)$.