HMMT 二月 2015 · 冲刺赛 · 第 30 题
HMMT February 2015 — Guts Round — Problem 30
题目详情
- [ 20 ] Find the sum of squares of all distinct complex numbers x satisfying the equation 10 9 8 7 6 5 4 3 2 0 = 4 x − 7 x + 5 x − 8 x + 12 x − 12 x + 12 x − 8 x + 5 x − 7 x + 4 .
解析
- [ 20 ] Find the sum of squares of all distinct complex numbers x satisfying the equation 10 9 8 7 6 5 4 3 2 0 = 4 x − 7 x + 5 x − 8 x + 12 x − 12 x + 12 x − 8 x + 5 x − 7 x + 4 . 7 Answer: − For convenience denote the polynomial by P ( x ). Notice 4 + 8 = 7 + 5 = 12 and 16 6 5 4 that the consecutive terms 12 x − 12 x + 12 x are the leading terms of 12Φ ( x ), which is suggestive. 14 7 10 3 9 2 Indeed, consider ω a primitive 14-th root of unity; since ω = − 1, we have 4 ω = − 4 ω , − 7 ω = 7 ω , and so on, so that 6 5 P ( ω ) = 12( ω − ω + · · · + 1) = 12Φ ( ω ) = 0 . 14 Dividing, we find 4 3 2 P ( x ) = Φ ( x )(4 x − 3 x − 2 x − 3 x + 4) . 14 This second polynomial is symmetric; since 0 is clearly not a root, we have 1 1 4 3 2 2 4 x − 3 x − 2 x − 3 x + 4 = 0 ⇐⇒ 4( x + ) − 3( x + ) − 10 = 0 . x x Setting y = x + 1 /x and solving the quadratic gives y = 2 and y = − 5 / 4 as solutions; replacing y with x + 1 /x and solving the two resulting quadratics give the double root x = 1 and the roots √ ( − 5 ± i 39) / 8 respectively. Together with the primitive fourteenth roots of unity, these are all the roots of our polynomial. Explicitly, the roots are √ πi/ 7 3 πi/ 7 5 πi/ 7 9 πi/ 7 11 πi/ 7 13 πi/ 7 e , e , e , e , e , e , 1 , ( − 5 ± i 39) / 8 . The sum of squares of the roots of unity (including 1) is just 0 by symmetry (or a number of other 2 2(5 − 39) 14 7 methods). The sum of the squares of the final conjugate pair is = − = − . 2 8 32 16