Serbian Mathematical Olympiad

Solution:

By substituting $x=y$ we obtain $v_{2}(f(x))=v_{2}(x)$. If $v_{2}(a)=k>0$, by considering the function $g(x)=f\left(2^{k} x\right) / 2^{k}$ we reduce the claim to the case of odd $a$. Therefore, from now on we assume that $2 \nmid a$.

Note that, if $x \not \equiv y\left(\bmod 2^{k}\right)$, then $f(x) \not \equiv f(y)\left(\bmod 2^{k}\right)$. Indeed, if $z \equiv-x\left(\bmod 2^{k}\right)$, then $2^{k} \nmid z+y$, so $f(y) \not \equiv-f(z) \equiv f(x)\left(\bmod 2^{k}\right)$. From this it also follows that the function $f$ is injective.

Let $2^{k-1}<3 a<2^{k}$, where $k \in \mathbb{N}$. Since $f(1), f(3), \ldots, f\left(2^{k}-1\right)$ are mutually distinct modulo $2^{k}$, there exists an odd number $x<2^{k}$ such that $f(x) \equiv 3 a$ $\left(\bmod 2^{k}\right)$. Suppose that $f(x) \neq 3 a$. Then $f(x)>2^{k}$, so from $f(x)+$ $f\left(2^{k}-x\right) \equiv 2^{k}\left(\bmod 2^{k+1}\right)$ it follows that $f(x)+f\left(2^{k}-x\right) \geqslant 3 \cdot 2^{k}$. However, on the other hand $f(x)+f\left(2^{k}-x\right) \leqslant 3\left(x+\left(2^{k}-x\right)\right)=3 \cdot 2^{k}$, so this is possible only if $f(x)=3 x$. From this $x \equiv a\left(\bmod 2^{k}\right)$, so $x=a$, i.e. again $f(x)=3 a$.