Romanian Mathematical Olympiad

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

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.