55rd Ukrainian National Mathematical Olympiad - Fourth Round

From the statement of the problem we conclude that $\{x\}+\{y\}$ is an integer, but $0\le \{x\}<1$, so: $\{x\}+\{y\}=0$ or $\{x\}+\{y\}=1$.

Case 1: $\{x\}+\{y\}=0$. So $\{x\}=\{y\}=0$, hence $x,y\in \mathbb{Z}$. But then $[x+y]=x+y=0$, so the solutions are $(n;-n)$, $n\in \mathbb{Z}$.

Case 2: $\{x\}+\{y\}=1$. Obviously $x,y\notin \mathbb{Z}$. Assume $x\notin \mathbb{Z}$, then $$x+y=[x]+\{x\}+[y]+\{y\}=[x]+[y]+1,$$ so $x+y$ is an integer. Hence $x+y=[x+y]=1$. In this case the solutions are $(x;1-x)$, $x\notin \mathbb{Z}$.