49th Mathematical Olympiad in Ukraine

Put $n=1$: $f(|m|)+f(|m|+2)=2f(|m|+1)$. If we denote $l=|m|\ge 0$, then we can easily get: $f(l+2)-f(l+1)=f(l+1)-f(l)$, so our sequence is an arithmetic progression, hence $f(l)=al+b$, where $a,b\in \mathbb{Z}$, $l\in {\mathbb{Z}}^{+}$.

In the same way, substituting $n=-1$ we obtain: $f(-|m|)+f(-(|m|+2))=2f(-(|m|+1))$. If we denote $k=-|x|\le 0$, then $\forall k\in {\mathbb{Z}}^{-}$, $f(k)+f(k+2)=2f(k+1)$, and we get: $f(k+2)-f(k+1)=f(k+1)-f(k)$, which implies $f(k)=ck+d$, where $c,d\in \mathbb{Z}$, $k\in {\mathbb{Z}}^{-}$.

Substituting $|m|=0$ we get on the one hand that $f(0)=2f(1)-f(2)$, and on the other hand that $f(0)=2f(-1)-f(-2)$, and so $l=k=0$ $al+b=ck+d$ or $b=d$.

Thus, we get the following answers: $f(n)=$, $a,b,c\in \mathbb{Z}$. An easy check shows that all such functions satisfy the condition of the problem.