, 2017

Solution:

By the condition, we know that $|P(x)-Q(x)| \leq 1$ for all $x$. This can only hold if $P(x)-Q(x)$ is a constant polynomial. Now take a constant $c$ such that $P(x)=Q(x)+c$. Without loss of generality, we can assume that $c \geq 0$. Assume that $c>0$. By continuity, if $\operatorname{deg} P=\operatorname{deg} Q>0$, we can select an integer $r$ and a real number $x_{0}$ such that $Q\left(x_{0}\right)+c=r$. Then $\left\lfloor P\left(x_{0}\right)\right\rfloor=\left\lfloor Q\left(x_{0}\right)+c\right\rfloor=r$. On the other hand, $\left\lfloor Q\left(x_{0}\right)\right\rfloor=\lfloor r-c\rfloor<r$ as $r$ was an integer. This is a contradiction. Therefore, $c=0$ as desired.