Problem 2. Determine the pairs of integers $(a, b)$ that have the property that $a^{2}+2 b^{2}+2 a+1$ is a divisor of $2 a b$.
Solution. $(0, b)$ and $(a, 0)$ are solutions for any integers $a, b$ ....

1 point

$(a, b)$ is a solution if and only if $(a, -b)$ is a solution,

reduction to the case $a b > 0$

1 point

$a^{2} + 2 b^{2} + 2 a + 1 \leq 2 a b$

1 point

$(a - 2 b)^{2} + (a + 2)^{2} \leq 2$

1 point

$|a + 2| \leq \sqrt{2}, |a - 2 b| \leq \sqrt{2}, a \in \{-3, -2, -1\}$,

$(a, b) \in \{(-3, -2); (-3, -1); (-1, -1)\}$ (after verification)

2 points

$S_{1} = \{(-3, -2); (-3, -1); (-1, -1); (-3, 2); (-3, 1); (-1, 1)\}$,

$S_{2} = \{(a, 0) \mid a \in \mathbb{Z}\} \cup \{(0, b) \mid b \in \mathbb{Z}\}, S = S_{1} \cup S_{2}$.

1 point