Romanian Mathematical Olympiad

If $x \in \mathbb{Z}$, then $[kx] = k[x]$ for every $k \in \mathbb{N}^*$.

For the converse, notice that the hypothesis implies, for all $n$,
$$n([x] + [nx]) + 2[(n+1)x] = (n+1)([x] + [(n+1)x]).$$
This comes to $n[nx] = [x] + (n-1)[(n+1)x]$, $\forall n \ge 1$. Replacing $n$ with $n+1$ and subtracting the two relations, we obtain that the sequence $a_n = [nx]$ is an arithmetic progression. Since $(nx)_{n \ge 1}$ is also an arithmetic progression, the difference sequence $\{nx\}_{n \ge 1}$ is an arithmetic progression. Since $\{nx\}_{n \ge 1}$ is bounded, its ratio must be 0, whence the conclusion.