IMC 2000 · Day 2 · P9

The statement is not true if $p$ is a constant polynomial. We prove it only in the case if $n$ is positive.

For an arbitrary polynomial $q(z)$ and complex number $c$, denote by $\mu(q, c)$ the largest exponent $\alpha$ for which $q(z)$ is divisible by $(z - c)^\alpha$. (With other words, if $c$ is a root of $q$, then $\mu(q, c)$ is the root's multiplicity. Otherwise 0.)

Denote by $S_0$ and $S_1$ the sets of complex numbers $z$ for which $p(z)$ is 0 or 1, respectively. These sets contain all roots of the polynomials $p(z)$ and $p(z) - 1$, thus $$\tag{1} \sum_{c \in S_0} \mu(p, c) = \sum_{c \in S_1} \mu(p - 1, c) = n.$$ The polynomial $p'$ has at most $n - 1$ roots ($n > 0$ is used here). This implies that $$\tag{2} \sum_{c \in S_0 \cup S_1} \mu(p', c) \le n - 1.$$ If $p(c) = 0$ or $p(c) - 1 = 0$, then $$\tag{3} \mu(p, c) - \mu(p', c) = 1 \quad \text{or} \quad \mu(p - 1, c) - \mu(p', c) = 1,$$ respectively. Putting (1), (2) and (3) together we obtain $$\begin{align*} |S_0| + |S_1| &= \sum_{c \in S_0} \bigl( \mu(p, c) - \mu(p', c) \bigr) + \sum_{c \in S_1} \bigl( \mu(p - 1, c) - \mu(p', c) \bigr) = \\ &= \sum_{c \in S_0} \mu(p, c) + \sum_{c \in S_1} \mu(p - 1, c) - \sum_{c \in S_0 \cup S_1} \mu(p', c) \ge n + n - (n - 1) = n + 1. \end{align*}$$