Z1) Determine all pairs of strictly positive integers $(a, b)$ such that

$$
a b+2=a^{3}+2 b
$$
## Solution (Arnaud):

By moving all the $b$ to the left, we get

$$
b(a-2)=a^{3}-2
$$

This means that $a-2$ is a divisor of $a^{3}-2$ and $b=\left(a^{3}-2\right) /(a-2)$. When does $a-2$ divide $a^{3}-2$? The idea is to get rid of the $a^{3}$ to see more clearly. One obtains

$$
\begin{aligned}
& a-2 \mid a^{3}-2 \\
\Rightarrow & a-2 \mid a^{3}-2-a^{2}(a-2)=2 a^{2}-2 \\
\Rightarrow & a-2 \mid 2 a^{2}-2-2 a(a-2)=4 a-2 \\
\Rightarrow & a-2 \mid 4 a-2-4(a-2)=6
\end{aligned}
$$

Thus, we must have $a-2 \mid 6$. Since $a \geq 1$, then $a-2 \geq-1$. You also have the following possibilities $a-2 \in\{-1,1,2,3,6\}$, i.e. $a \in\{1,3,4,5,8\}$. This gives us five pairs of solutions $(1,1),(3,25),(4,31),(5,41)$ and $(8,85)$.