HMMT 二月 2024 · 冲刺赛 · 第 26 题

HMMT February 2024 — Guts Round — Problem 26

[14] It can be shown that there exists a unique polynomial P in two variables such that for all positive integers m and n , m n ∑ ∑ 7 P ( m, n ) = ( i + j ) . i =1 j =1 Compute P (3 , − 3) .

解析

[14] It can be shown that there exists a unique polynomial P in two variables such that for all positive integers m and n , m n ∑ ∑ 7 P ( m, n ) = ( i + j ) . i =1 j =1 Compute P (3 , − 3) . Proposed by: Pitchayut Saengrungkongka Answer: − 2445 Solution: Note that for integers m > 0 , n > 1 , m ∑ 7 P ( m, n ) − P ( m, n − 1) = ( i + n ) . i =1 For any given positive integer m , both sides are a polynomial in n , so they must be equal as polynomials. In particular, 3 ∑ 7 7 7 7 P (3 , x ) − P (3 , x − 1) = ( i + x ) = ( x + 1) + ( x + 2) + ( x + 3) i =1 for all real x . Moreover, P (3 , 1) − P (3 , 0) = P (3 , 1) = ⇒ P (3 , 0) = 0 . Then 7 7 7 7 7 7 7 7 7 P (3 , − 3) = P (3 , 0) − (1 + 2 + 3 ) − (0 + 1 + 2 ) − (( − 1) + 0 + 1 ) 7 7 = − 3 − 2 · 2 − 2 = − 2445 .