PUMaC 2022 · 代数（B 组） · 第 2 题

PUMaC 2022 — Algebra (Division B) — Problem 2

A pair ( f, g ) of degree 2 real polynomials is called foolish if f ( g ( x )) = f ( x ) · g ( x ) for all real x . How many positive integers less than 2023 can be a root of g ( x ) for some foolish pair ( f, g )?

解析

A pair ( f, g ) of degree 2 real polynomials is called foolish if f ( g ( x )) = f ( x ) · g ( x ) for all real x . How many positive integers less than 2023 can be a root of g ( x ) for some foolish pair ( f, g )? Proposed by Austen Mazenko Answer: 2021 We claim that if ( f, g ) is foolish, then there exist real numbers a, b such that f ( x ) = ax ( x + b ) 2 and g ( x ) = x + bx − b . To see this, let r be a root of g , and plug x = r into the functional equation to find f (0) = 0. This immediately implies that f ( x ) = ax ( x + b ) for some a, b . Next, plug this form of f into the functional equation to find ag ( x )( g ( x ) + b ) = ax ( x + b ) g ( x ). Since deg g = 2, g is not identically zero, hence g ( x ) + b = x ( x + b ). Rearranging yields the claim. 2 2 x Now, note that a positive integer x is a root of x + bx − b iff b = . It follows that any 1 − x x ̸ = 1 is a root of some g . Hence the valid positive integers between 1 and 2022 inclusive are all numbers except 1. This yields an answer of 2022 − 1 = 2021 integers.