2025 USAMO Problems/Problem 2

Problem

Let $n$ and $k$ be positive integers with $k<n$ . Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0, a_1, \dots, a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$ , the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.

Solution

Assume for contradiction that all roots of $P(x)$ are real. Let the distinct non-zero real roots be $r_1, r_2, \ldots, r_n$ .

\subsection*{Case $k=2$ :} For any pair of roots $r_i, r_j$ , consider: $Q(x) = (x-r_i)(x-r_j) = x^2 - (r_i+r_j)x + r_ir_j$ The product of coefficients is: $-r_ir_j(r_i+r_j) = 0$ Since $r_i,r_j \neq 0$ , we must have $r_i + r_j = 0$ for all pairs.

But for three roots $r_1, r_2, r_3$ , this gives: $r_1 + r_2 = 0$ $r_1 + r_3 = 0$ $r_2 + r_3 = 0$ which implies $r_1 = r_2 = r_3 = 0$ , contradicting the nonzero constant term.

\subsection*{General $k$ :} For any $k$ roots $r_{i_1}, \ldots, r_{i_k}$ , the polynomial: $Q(x) = \prod_{m=1}^k (x-r_{i_m})$ must have some coefficient (other than constant term) equal to zero. For $k=3$ , this requires: $r_i + r_j + r_m = 0$ for all triples, which is impossible for distinct non-zero reals when $n \geq 4$ .

Thus, $P(x)$ must have at least one nonreal root. \hfill (by Jonathan Wang)

2025 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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2025 USAMO Problems/Problem 2

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