Ukrainian National Mathematical Olympiad - Third Round

Add up the equations and get
$$x^2 + 2xy + y^2 = 4y + 4x \text{ or } (x+y)^2 = 4(x+y).$$

If $x = -y$, the second equation implies
$$2y^2 = -2y, \text{ hence } y = 0 \text{ or } y = -1.$$
We get two solutions, (0; 0) and (1; -1).

If $x = 4 - y$, the first equation implies
$$y^2 - y(4 - y) = 3(4 - y) + y, \text{ hence, } y^2 - y - 6 = 0.$$
The final equation has two roots, $y = 3$ and $y = -2$. They give us solutions (1; 3) and (6; -2). It can be easily verified that they satisfy both equations.