HMMT 二月 2025 · ALGNT 赛 · 第 5 题

HMMT February 2025 — ALGNT Round — Problem 5

Let S be the set of all nonconstant monic polynomials P with integer coefficients satisfying P 3 + 2 = √ √ P 3 − 2 . If Q is an element of S with minimal degree, compute the only possible value of Q (10) − Q (0). r 2025!

解析

Let S be the set of all nonconstant monic polynomials P with integer coefficients satisfying P 3 + 2 = √ √ P 3 − 2 . If Q is an element of S with minimal degree, compute the only possible value of Q (10) − Q (0). Proposed by: David Dong Answer: 890 √ √ √ √ 4 2 Solution: First, note that the polynomial x − 10 x + 1 has both 3 + 2 and 3 − 2 as roots. It 3 2 suffices to check whether a polynomial of degree at most 3 belongs in S . Suppose f ( x ) = ax + bx + cx + d ∈ S . We compute √ √ √ √ √ 3 3 ( 3 + 2) − ( 3 − 2) = 22 2 √ √ √ √ √ 2 2 ( 3 + 2) − ( 3 − 2) = 4 6 √ √ √ √ √ 1 1 ( 3 + 2) − ( 3 − 2) = 2 2 , so we get that √ √ √ √ √ √ √ f ( 3 + 2) − f ( 3 − 2) = (22 2) a + (4 6) b + (2 2) c. By resolving linear dependencies, it’s clear that b = 0 and c = − 11 a . It follows that if f is not the zero 3 polynomial, it must be cubic. It is then clear that f ( x ) = x − 11 x + d has minimal degree in S , and thus Q (10) − Q (0) = f (10) − f (0) = 890 . 2025! r