50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010)

Answer: $f(k)=k+2$.

Substitute $x=0$ and $y=0\Rightarrow$ $$f(f(2y))=f(2y)+2,(1)$$ $$f(x+f(x))=f(2x)+2,(2)$$ Let us prove by induction, that $f(2n)=2n+2$. Base $n=0$ is already proved. Let us suppose that we have $f(2n)=2n+2$ and prove that $f(2n+2)=2n+4$. For $n=2m$ we have to show that $f(4m+2)=4m+4$. Substitute $x=2m$ in (2): we'll get $$f(2m+f(2m))=f(2m+2m+2)=f(4m+2)=f(4m)+2=4m+4.$$ For $n=2m-1$ $f(4m)=4m+2$. Substitute $y=2m-1$ in (1): $$f(f(4m-2))=f(4m)=f(4m-2)+2=4m+2.$$ Following the same lines, we can prove that $f(2n)=2n+2$ for negative $n$. Let $k$ be an odd number, then substitute $x=2z+k$, $y=-z$, $z\in \mathbb{Z}$, we get $$f(2z+k+f(k))=f(4z+2k)+f(-2)=4z+2k+2-2z+2=2z+2k+4,(3)$$ We consider the following two cases: $f(k)$ is even. Substitute to (3): $z=-\frac{f(k)}{2}$, then $f(k)=-f(k)+2k+4$, therefore $f(k)=k+2$ is odd and we get a contradiction. $f(k)$ is odd. Then $2z+k+f(k)$ is even and $f(2z+k+f(k))=2z+k+f(k)+2$. Recalling (3), we have $f(2z+k+f(k))=2z+k+f(k)+2=2z+2k+4$, and, finally, $f(k)=k+2$, $k\in \mathbb{Z}$.