Belorusija 2012

(Solution of S. Dabryneuski, A. Tanana, A. Zhuk.) Set $x=0$ and $z=0$ in $$f(x+f(y+f(z)))=y+f(x+z),(\ast )$$

then $f(f(y+f(0)))=y+f(0)$, or $$f(f(x))=x,\forall x\in \mathbb{Q},(1)$$ hence $f$ is bijective. Let $f(0)=a$, then $f(a)=0$. Further, the right hand side of () is symmetric with respect to $x$ and $z$, hence $f(x+f(y+f(z)))=f(z+f(y+f(x)))$. Applying the bijectivity of $f$ we conclude that $$x+f(y+f(z))=z+f(y+f(x)).(2)$$

$$f(y+z)=f(z)+f(y+f(x))-x.(3)$$

Set $x=a=f(0)$ in (3) then $f(y+z)=f(z)+f(y)-a$, or $f(y+z)-a=(f(y)-a)+(f(z)-a)$. We see that the function $f(x)-a$ is additive, thus $f(x)-a=\alpha x$, or $f(x)=\alpha x+a$. Substituting $f$ in () gives $f(x)=x$; $f(x)=-x+a,a\in \mathbb{Q}$.