Problem 2. Determine the numbers $\overline{a b}$ such that the number $\overline{a a a}+37 \cdot(a+b)$ is a perfect square.
Solution. Let $n=\overline{a a a}+37 \cdot(a+b)$. We have $n=111 \cdot a+37 \cdot(a+b)=37 \cdot(4 \cdot a+b)$. Since $n$ must be a perfect square, it follows that $4 \cdot a+b=37 \cdot k^{2}, k$ a non-zero natural number. However, since $a$ and $b$ are digits, it implies $4 \cdot a+b \leq 36+9$. Therefore, $k=1$ and thus $4 \cdot a+b=37$ from which $(a, b) \in\{(9,1),(8,5),(7,9)\}$. The numbers sought are $91,85,79$.