, 2012

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

then $f(f(y + f(0))) = y + f(0)$, or
$$f(f(x)) = x, \quad \forall x \in \mathbb{Q}, \quad (1)$$
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
$$x + f(y + f(z)) = z + f(y + f(x)). \quad (2)$$

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

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 $f(x) = x$; $f(x) = -x + a, a \in \mathbb{Q}$.