HMMT 二月 2005 · TEAM1 赛 · 第 8 题

HMMT February 2005 — TEAM1 Round — Problem 8

[25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of 2 3 1 x x x 2 x + + + + . 2 3 4 1 + x 1 + x 1 + x 1 + x

解析

[25] Suppose x is a fifth root of unity. Find, in radical form, all possible values of 2 3 1 x x x 2 x + + + + . 2 3 4 1 + x 1 + x 1 + x 1 + x 3 Solution: Note that 2 6 4 4 6 x x x x x + x 1 4

= + = = x = , and 2 3 2 2 5 2 1 + x 1 + x 1 + x x + x 1 + x x 3 5 4 4 5 1 x x x x + x 1 4
= + = = x = . 4 5 1 + x 1 + x 1 + x x + x 1 + x x 2 1 Therefore, the sum is just 2 x + . If x = 1, this is 4. Otherwise, let y = x + . Then x x x satisfies ( ) ( ) 1 1 2 3 4 2 2 0 = 1 + x + x + x + x = x + + 2 + x + − 1 = y + y − 1 , 2 x x √ √ − 1 ± 5 so solving this quadratic yields y = , or 2 y = − 1 ± 5. Since each value of y can 2 correspond to only 2 possible values of x , and there are 4 possible values of x besides √ 1, both of these values for y are possible, which yields the answers, 4 and − 1 ± 5.