1. In the set of integers, solve the equation

$$
a^{2}+2 a b+2 b^{2}=13
$$
Solution. We will write the equation in the form

$$
(a+b)^{2}+b^{2}=13
$$

Since $a$ and $b$ are integers, we get $-3 \leq a+b \leq 3$. Indeed, if $a+b \geq 4$ or $a+b \leq-4$, then $(a+b)^{2}+b^{2} \geq 16+b^{2} \geq 16>13$. Therefore, we need to consider the seven cases:

$$
a+b= \pm 3, a+b= \pm 2, a+b= \pm 1 \text { and } a+b=0 \text {. }
$$

If $a+b= \pm 1$ or $a+b=0$, then the equation takes the form $b^{2}=12$ or $b^{2}=13$. The last two equations have no solution in the set of integers.

We will consider the remaining four cases.
a) $a+b=-3$. Then $b^{2}=4$, i.e., $b=2$ or $b=-2$. For $b=2$, the solution to the original equation is $a=-5, b=2$. For $b=-2$, the solution to the original equation is $a=-1, b=-2$.

b) $a+b=3$. Then $b^{2}=4$, i.e., $b=2$ or $b=-2$. For $b=2$, the solution to the original equation is $a=1, b=2$. For $b=-2$, the solution to the original equation is $a=5, b=-2$.

c) $a+b=-2$. Then $b^{2}=9$, i.e., $b=-3$ or $b=3$. For $b=-3$, the solution to the original equation is $a=1, b=-3$. For $b=3$, the solution to the original equation is $a=-5, b=3$.

d) $a+b=2$. Then $b^{2}=9$, i.e., $b=-3$ or $b=3$. For $b=-3$, the solution to the original equation is $a=5, b=-3$. For $b=3$, the solution to the original equation is $a=-1, b=3$.

Finally, the solutions to the equation are

$$
(a, b) \in\{(-5,2),(-1,-2),(1,2),(5,-2),(1,-3),(-5,3),(5,-3),(-1,3)\}
$$