, 2010

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

Substitute $x = 0$ and $y = 0 \Rightarrow$
$$f(f(2y)) = f(2y) + 2, \qquad (1)$$
$$f(x + f(x)) = f(2x) + 2, \qquad (2)$$
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
$$f(2z + k + f(k)) = f(4z + 2k) + f(-2) = 4z + 2k + 2 - 2z + 2 = 2z + 2k + 4, \quad (3)$$
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, $f(k) = k + 2$, $k \in \mathbb{Z}$.