55rd 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.