HMMT 二月 2023 · ALGNT 赛 · 第 1 题

HMMT February 2023 — ALGNT Round — Problem 1

Suppose P ( x ) is a cubic polynomial with integer coefficients such that P ( 5) = 5 and P ( 5) = 5 5. Compute P (5).

解析

Suppose P ( x ) is a cubic polynomial with integer coefficients such that P ( 5) = 5 and P ( 5) = 5 5. Compute P (5). Proposed by: Sean Li Answer: − 95 3 2 Solution: Write P ( x ) = ax + bx + cx + d , where a, b, c, d are integers. Then we have that √ √ P ( 5) − 5 = (5 a + c ) 5 + (5 b + d − 5) = 0 , √ √ √ √ 3 3 3 3 P ( 5) − 5 5 = (5 a + d ) + ( c − 5) 5 + b 25 = 0 . √ √ Recall that 5 is irrational. In particular, since (5 a + c ) 5 + (5 b + d − 5) = 0, we must have 5 a + c = 0 √ 3 and 5 b + d − 5 = 0. Similarly, from the condition on 5, we must have 5 a + d = c − 5 = b = 0. 3 This is enough to imply ( a, b, c, d ) = ( − 1 , 0 , 5 , 5), so P ( x ) = − x + 5 x + 5. Hence, our final answer is P (5) = − 125 + 25 + 5 = − 95.