PUMaC 2015 · 代数（A 组） · 第 8 题

PUMaC 2015 — Algebra (Division A) — Problem 8

[ 8 ] Let P ( x ) be a polynomial with positive integer coefficients and degree 2015. Given that there exists some ω ∈ C satisfying: 73 ω = 1 and 2 3 72 2015 2015 2015 2015 P ( ω ) + P ( ω ) + P ( ω ) + . . . + P ( ω ) = 0 , what is the minimum possible value of P (1)? 1

解析

[ 8 ] Let P ( x ) be a polynomial with positive integer coefficients and degree 2015. Given that there exists some ω ∈ C satisfying: 73 ω = 1 and 2 3 72 2015 2015 2015 2015 P ( ω ) + P ( ω ) + P ( ω ) + . . . + P ( ω ) = 0 , what is the minimum possible value of P (1)? Solution: Note that ω 6 = 1 since P (1) > 0 by the condition that the coefficients are all positive. Let 2015 0 P ( x ) = a x + · · · + a x . Now, we claim that 2015 is a primitive root mod 73. It 2015 0 24 36 suffices to show that 2015 , 2015 are not equal to 1 mod 73. The latter follows by quadratic reciprocity and the first can be calculated to be equal to − 9 mod 73. Now, the given expression is equivalent to: 72 73 ∑ ∑ k k 0 73 27 · 73 P ( ω ) = P ( ω ) − P (1) = 73 · ( a x + a x + . . . + a x ) − P (1) = 0 0 73 27 · 73 k =1 k =1 And since a ≥ 1, we have: i 0 73 27 · 73 P (1) = 73 · ( a x + a x + . . . + a x ) ≥ 73 · 28 = 2044 0 73 27 · 73 Thus, the minimum possible value of P (1) is 2044 . Author: Bill Huang 4