(JBMO 2013) Find the non-zero natural numbers $a$ and $b$ such that $\frac{a^{3} b-1}{a+1}$ and $\frac{b^{3} a+1}{b-1}$ are integers.
We want to find $a>0$ and $b>0$ integers such that $a^{3} b-1 \equiv 0(\bmod a+1)$ and $b^{3} a+1 \equiv 0(\bmod b+1)$. Since $a \equiv-1(\bmod a+1)$, we have $a^{3} b-1 \equiv(-1)^{3} b-1 \equiv-(b+1)$ $(\bmod a+1)$. Similarly, $b \equiv 1(\bmod b-1)$, so $b^{3} a+1 \equiv 1 a+1 \equiv a+1(\bmod b-1)$. In particular, $b-1$ then divides $a+1$, which itself divides $b+1$. Therefore, $b-1$ divides $b+1$ and even divides $(b+1)-(b-1)=2$. Since $b-1 \geqslant 0$, we deduce that $b-1 \in\{1,2\}$:

1. if $b-1=1$, $a+1$ divides $b+1=3$ and, since $a+1>1$, we deduce that $a+1=3$;
2. if $b-1=2$, $a+1$ is even and divides $b+1=4$ and, since $a+1>1$, we deduce that $a+1=2$ or $a+1=4$.

The possible solutions $(a, b)$ are therefore $(2,2),(1,3)$ and $(3,3)$. We then verify that these solutions give respectively $\frac{a^{3} b-1}{a+1}=\frac{15}{3}, \frac{2}{2}, \frac{80}{4}$ and $\frac{b^{3} a+1}{b-1}=\frac{17}{1}, \frac{28}{2}, \frac{82}{2}$. The couples sought are therefore indeed $(2,2),(1,3)$ and $(3,3)$.