67th Romanian Mathematical Olympiad

The equality can be written $(x+y)-\sqrt{x}=\sqrt{xy}+\sqrt{y}$. Squaring yields $x^2+xy+y^2+x-y=2(2y+x)\sqrt{x}$. Since $2y+x\ne 0$, $\sqrt{x}$ must be rational, therefore $x$ is a perfect square. In the same way, $y$ is a perfect square.

Denote now $\sqrt{x}=a$, $\sqrt{y}=b$. The equality $a^2+b^2=ab+a+b$ leads now to $(a-b{)}^2+(a-1{)}^2+(b-1{)}^2=2$. This gives $(a,b)\in \{(1,2),(2,1),(2,2)\}$, therefore the answer is $(x,y)\in \{(1,4);(4,1);(4,4)\}$.