(Find all pairs of strictly positive integers $(a, b)$ such that $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are integers.

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There are two equivalent ways to approach the problem: looking at $a^{3} b-1$ modulo $a+1$, and seeing that $b+1 \equiv 0 \pmod{a+1}$, or by forming linear combinations, noting that $a+1$ divides $a^{3} b-1$ and also $a^{3}+1$, thus $b\left(a^{3}+1\right)-(b+1)$. In both cases, $a+1$ divides $b+1$. Proceeding similarly with the other fraction, we get that $b-1$ divides $a+1$. Combining these two results, we find that $b-1$ divides $b+1$, so $b-1$ divides 2, hence $b \leqslant 3$. Since $b-1 \neq 0$, we know that $b \neq 1$. Thus, $b=3$ or $b=2$. If $b=3$, we find the solutions $(1,3)$ and $(3,3)$. If $b=2$, we obtain the solution $(2,2)$.