Problem 613

(solved by Axel Hovasse)

Let $a, b \in \mathbb{N}$ such that $a b+2=a^{3}+2 b$. Thus $a^{3}-2=b(a-2)$. Therefore, $a \neq 2$, because $2^{3}-2 \neq 0$. Thus $a-2 \mid a^{3}-2$, but $a-2 \mid (a-2)^{3}=a^{3}-6 a^{2}+12 a-8$, it follows that $a-2 \mid -6 a^{2}+12 a-6$. However, $a-2 \mid 6 a(a-2)$, so $a-2 \mid 6$. Thus $a-2 \in\{-6,-3,-2,-1,1,2,3,6\}$, which means $a \in\{-4,-1,0,1,3,4,5,8\}$, but $a \geqslant 0$, so $a \in\{0,1,3,4,5,8\}$. We then find $(a, b) \in\{(0,1),(1,1),(3,25),(4,31),(5,41),(8,85)\}$. We verify subsequently that these pairs are indeed solutions.