China Hong Kong Mathematical Olympiad

Yes. We claim that $P(x, y) = (y - 3x)(y - 3x - 1)(y - 3x - 2)$ satisfies the conditions. Clearly, it is a non-constant polynomial. For any real number $\alpha$, let $n = \lfloor \alpha \rfloor$. Then we have $n \le \alpha < n + 1$ and hence $3n \le 3\alpha < 3n + 3$. This shows $\lfloor 3\alpha \rfloor = 3n, 3n + 1, 3n + 2$. Therefore, one of $\lfloor 3\alpha \rfloor - 3\lfloor \alpha \rfloor$, $\lfloor 3\alpha \rfloor - 3\lfloor \alpha \rfloor - 1$ and $\lfloor 3\alpha \rfloor - 3\lfloor \alpha \rfloor - 2$ must be 0. This shows $P(\lfloor \alpha \rfloor, \lfloor 3\alpha \rfloor) = 0$ for all $\alpha \in \mathbb{R}$.