# Problem 1. (2 points)

The polynomial $P(x)$ is such that $P\left(x^{2}\right)$ has $2 n+1$ roots. What is the smallest number of roots that the derivative of the polynomial $P(x)$ can have?

(In both cases, distinct roots are meant, without regard to multiplicity).

#
# Answer: $n$.

## Solution:

If the polynomial $P\left(x^{2}\right)$ has a root $x_{0}$, then it also has a root $-x_{0}$, so the number of roots can only be odd if one of the roots is the number 0. For each pair of roots $x_{0},-x_{0}$ of the polynomial $P\left(x^{2}\right)$, the number $x_{0}^{2}$ is a root of the polynomial $P(x)$; the number 0 is also a root, so the polynomial $P(x)$ has at least $n+1$ roots (there may be other negative roots, about which we know nothing).

Between any two roots of the polynomial $P(x)$, there must be a root of the derivative of this polynomial, so the derivative has at least $n$ roots.

It is easy to verify that this value is achieved, for example, for the polynomial $x(x-1) \ldots(x-n)$.