1. Solve the equation in the domain of integers

$$
x^{2}+y^{2}+x+y=4
$$
1. By multiplying both sides of the given equation by four, we get

$$
4 x^{2}+4 y^{2}+4 x+4 y=16 \text {. }
$$

The expression on the left side of the thus modified equation is completed to the sum of the squares of two binomials. We obtain

$$
\left(4 x^{2}+4 x+1\right)+\left(4 y^{2}+4 y+1\right)=(2 x+1)^{2}+(2 y+1)^{2}=18 .
$$

It is therefore sufficient to investigate all decompositions of the number 18 into the sum of two positive odd numbers, since the numbers $2 x+1$ and $2 y+1$ are not divisible by two (and thus do not equal zero) for any integers $x$ and $y$.

Consider, therefore, the following decompositions:

$$
18=1+17=3+15=5+13=7+11=9+9 \text {. }
$$

Among the given sums, only one $(18=9+9)$ is the sum of the squares of two integers. Therefore, the following four cases can occur:

$$
\begin{aligned}
& 2 x+1=3, \quad 2 y+1=3, \quad \text { i.e. } x=1, \quad y=1 \\
& 2 x+1=3, \quad 2 y+1=-3, \quad \text { i.e. } x=1, \quad y=-2, \\
& 2 x+1=-3,2 y+1=3, \quad \text { i.e. } x=-2, y=1, \\
& 2 x+1=-3,2 y+1=-3, \quad \text { i.e. } x=-2, y=-2 .
\end{aligned}
$$

Conclusion. The given equation is satisfied by exactly four pairs of integers $(x, y)$, namely $(1,1)$, $(1,-2)$, $(-2,1)$, and $(-2,-2)$.

Alternative solution. The given equation can be rearranged to the form $x(x+1)+y(y+1)=4$, from which it is clear that the number 4 must be decomposed into the sum of two integers, each of which is the product of two consecutive integers. Since the smallest values of the expression $t(t+1)$ for positive and negative integers $t$ are $0,2,6,12, \ldots$, only the decomposition $4=2+2$ is feasible, so each of the unknowns $x, y$ must equal one of the numbers 1 or -2, the only integers $t$ for which $t(t+1)=2$. Moreover, it is clear that each of the four pairs $(x, y)$ composed of the numbers $1,-2$ is a solution to the given problem.

For a systematic and complete solution, award 6 points. For guessing the solution $(1,1)$, do not award any points; for each additional correctly guessed solution, award 1 point; for all correctly guessed solutions, award 2 points. For a correct approach, deduct 1 point for each missing solution. The uniqueness of the decomposition of the number 18 into the sum of two squares is so obvious that it does not need to be justified (as stated in the original solution).