PUMaC 2017 · 代数（B 组） · 第 8 题

PUMaC 2017 — Algebra (Division B) — Problem 8

Together, Kenneth and Ellen pick a real number a . Kenneth subtracts a from every thousandth 1000 root of unity (that is, the thousand complex numbers ω for which ω = 1) then inverts each, then sums the results. Ellen inverts every thousandth root of unity, then subtracts a from each, and then sums the results. They are surprised to find that they actually got the same answer! How many possible values of a are there? 1

解析

Let ξ be a primitive thousandth root of unity, meaning 1000 is the least positive integer n for n k which ξ = 1, so that { ξ | 0 ≤ k < 1000 } are the thousand thousandth roots of unity. Then, Kenneth’s answer is 999 999 ∑ 1 − 1000 a = k 1000 ξ − a a − 1 k =0 k 1000 because { ξ − a | 0 ≤ k < 1000 } is the set of roots of the polynomial ( x + a ) − 1. Similarly, Ellen’s answer is ( ) 999 ∑ 1 − a = − 1000 a, k ξ k =0 k 1000 because { ξ | 0 ≤ k < 1000 } is the set of roots of the polynomial x − 1. These answers are equal iff 1001 999 a − a − a = 0 . By Descartes’ rule of signs, this equation has one positive solution and one negative solution. Since 0 is also a solution, there are 3 possible real values of a . Note: The use of the word “surprised” in the problem statement led to the interpretation that a = 0 would not be a possible value, as the outcome would certainly not be surprising in that instance. Therefore, 2 was also accepted for this problem. Problem written by Matt Tyler If you believe that any of these answers is incorrect, or that a problem had multiple reasonable interpretations or was incorrectly stated, you may appeal at http://tinyurl.com/PUMaCappeal2017. All appeals must be in by 1 PM to be considered. 2