HMMT 二月 2015 · 冲刺赛 · 第 30 题

英文解析

30. [ 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 .
    Answer: − For convenience denote the polynomial by P ( x ). Notice 4 + 8 = 7 + 5 = 12 and 7
    6 5 416
    that the consecutive terms 12 x − 12 x + 12 x are the leading terms of 12Φ ( x ), which is suggestive.
    7 10 3 9 214
    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 .
    Dividing, we find 14
    4 3 2
    P ( x ) = Φ ( x )(4 x − 3 x − 2 x − 3 x + 4) .
    This second polynomial is symmetric; since 0 is clearly not a root, we have 14
    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; replacingy 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 theroots 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(5 − 39)2
    14 7
    methods). The sum of the squares of the final conjugate pair is = − = − .
    8 32 162