HMMT 二月 2005 · TEAM1 赛 · 第 11 题

英文解析

11. [25] Let P ( x ) = a x + a x + · · · + a be a polynomial with real coefficients,
    n n − 1 0
    a 6 = 0. Suppose every root of P is a root of unity, but P (1) 6 = 0. Show that thencoefficients of P are symmetric; that is, show that a = a , a = a , . . .
    n 0 n − 1 1
    Solution: Since the coefficients of P are real, the complex conjugates of the roots of − 1
    P are also roots of P . Now, if x is a root of unity, then x = ¯ x . But the roots ofn − 1 n n − 1
    x P ( x ) = a x + a x + · · · + a
    0 1 nare then just the complex conjugates of the roots of P , so they are the roots of P .
    n − 1
    Therefore, P ( x ) and x P ( x ) differ by a constant multiple c . Since a = ca andn 0
    a = ca , c is either 1 or − 1. But if it were − 1, then
    0 n
    P (1) = a + a + · · · + a = (( a + a ) + ( a + a ) + · · · + ( a + a )) = 0 ,1
    n n − 1 0 n 0 n − 1 1 0 na contradiction. Therefore c = 1, giving the result.2
    Early Re-tile-ment [125]
    Let S = { s , . . . , s } be a finite set of integers, and define S + k = { s + k, . . . , s + k } . We
    0 n 0 nsay that two sets S and T are equivalent , written S ∼ T , if T = S + k for some k . Given a
    (possibly infinite) set of integers A , we say that S tiles A if A can be partitioned into subsetsequivalent to S . Such a partition is called a tiling of A by S .