Let $p$ be a prime and $P\in \mathbb{R}[x]$ be a polynomial of degree less than $p-1$ such that $\lvert P(1)\rvert=\lvert P(2)\rvert=\ldots=\lvert P(p)\rvert$. Prove that $P$ is constant.
Problem
Source: 2023 Serbia TST Problem 4
Tags: algebra, polynomial, Serbia, TST
22.05.2023 16:43
Here's my solution from contest. I think this is the cleanest way to do it.
23.05.2023 05:31
If $p=2$, then $P$ has degree less than $1$, which means $P$ is constant. Without loss of generality, let $p >2$ and assume $$\lvert P(1)\rvert=\lvert P(2)\rvert=\ldots=\lvert P(p)\rvert = 1$$By Lagrange Interpolation Theorem we have $$\sum_{i= 0}^{p - 1} (-1)^i\binom{p-1}{i}P(i + 1) = 0$$Since $$\binom{p-1}{i} = \frac{(p-1)!}{i! (p-1-i)!} \equiv (-1)^i \frac{(p-1)!}{(p-1)(p-2)\cdots (p-i)(p-1-i)!} \equiv (-1)^i \pmod{p}$$we get $$0 = \sum_{i= 0}^{p - 1} (-1)^i\binom{p-1}{i}P(i + 1) \equiv \sum_{i= 0}^{p - 1} P(i+ 1) \pmod{p}$$Using $P(i + 1) = \pm 1$, we get $$-p \leq \sum_{i= 0}^{p - 1} P(i+ 1) \leq p \qquad \Longrightarrow \qquad \sum_{i= 0}^{p - 1} P(i+ 1) \in \{-p, 0, p\}$$Since $p$ is odd, the above sum can not be equal to $0$. So, the sum should be either $p$ or $-p$, and this can be if and only if all given values of $P$ are $1$ or all are $-1$. This means $P$ gets the same value at $p$ distinct points and therefore it must be a constant polynomial.
23.05.2023 08:13