Shortlisted Problems for the 64th NMO

Let $P(x)$ and $Q(x)$ be real polynomials such that $\lfloor P(x) \rfloor = \lfloor Q(x) \rfloor$ for all $x \in \mathbb{R}$.

Suppose $P(x) \ne Q(x)$. Then $P(x) - Q(x)$ is a nonzero polynomial, so $P(x) - Q(x)$ is not identically zero. Thus, there exists $x_0 \in \mathbb{R}$ such that $P(x_0) \ne Q(x_0)$.

Let $d = P(x_0) - Q(x_0)$. Then $d \ne 0$.

Consider the function $f(x) = P(x) - Q(x)$. Since $f(x)$ is a nonzero polynomial, it is continuous and unbounded (unless it is constant, in which case $d \ne 0$ everywhere).

Let $\varepsilon = |d|/2 > 0$. By continuity, there exists $\delta > 0$ such that for all $x$ with $|x - x_0| < \delta$, $|P(x) - Q(x) - d| < \varepsilon$, i.e., $|P(x) - Q(x)| > |d|/2$ in a neighborhood of $x_0$.

Now, since $\lfloor P(x) \rfloor = \lfloor Q(x) \rfloor$ for all $x$, it must be that $P(x)$ and $Q(x)$ are always in the same unit interval, i.e., $P(x) - Q(x) \in (-1, 1)$ for all $x$.

But we have found a neighborhood of $x_0$ where $|P(x) - Q(x)| > |d|/2$. If $|d| \ge 1$, then $|P(x) - Q(x)| > 1/2$ in a neighborhood, and for some $x$ it will exceed $1$, contradicting $P(x) - Q(x) \in (-1, 1)$ for all $x$.

If $0 < |d| < 1$, then $P(x_0)$ and $Q(x_0)$ are in the same unit interval, but since $P(x) - Q(x)$ is a nonzero polynomial, it is not constant, so for large $x$, $|P(x) - Q(x)|$ will exceed $1$, again contradicting $P(x) - Q(x) \in (-1, 1)$ for all $x$.

Therefore, $P(x) - Q(x) = 0$ for all $x$, i.e., $P = Q$.