HMMT 二月 2005 · TEAM1 赛 · 第 11 题

HMMT February 2005 — TEAM1 Round — Problem 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 the n coefficients of P are symmetric; that is, show that a = a , a = a , . . . n 0 n − 1 1 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 n say 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 subsets equivalent to S . Such a partition is called a tiling of A by S .

解析

[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 the n coefficients 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 of n − 1 n n − 1 x P ( x ) = a x + a x + · · · + a 0 1 n are 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 and n 0 a = ca , c is either 1 or − 1. But if it were − 1, then 0 n 1 P (1) = a + a + · · · + a = (( a + a ) + ( a + a ) + · · · + ( a + a )) = 0 , n n − 1 0 n 0 n − 1 1 0 n 2 a contradiction. Therefore c = 1, giving the result. 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 n say 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 subsets equivalent to S . Such a partition is called a tiling of A by S .