HMMT 十一月 2014 · 冲刺赛 · 第 28 题

HMMT November 2014 — Guts Round — Problem 28

[ 15 ] Let x be a complex number such that x + x is a root of the polynomial p ( t ) = t + t − 2 t − 1. 7 − 7 Find all possible values of x + x .

解析

[ 15 ] Let x be a complex number such that x + x is a root of the polynomial p ( t ) = t + t − 2 t − 1. 7 − 7 Find all possible values of x + x . − 1 Answer: 2 Since x + x is a root, ( ) ( ) ( ) 3 2 − 1 − 1 − 1 0 = x + x + x + x − 2 x + x − 1 3 − 3 − 1 2 − 2 − 1 = x + x + 3 x + 3 x + x + 2 + x − 2 x − 2 x − 1 3 − 3 2 − 2 − 1 = x + x + x + x + x + x + 1 ( ) − 3 2 6 = x 1 + x + x + · · · + x . 7 Since x 6 = 0, the above equality holds only if x is a primitive seventh root of unity, i.e. x = 1 and 7 − 7 x 6 = 1. Therefore, the only possible value of x + x is 1 + 1 = 2.