PUMaC 2011 · 代数（A 组） · 第 8 题
PUMaC 2011 — Algebra (Division A) — Problem 8
题目详情
- [ 8 ] Let 1 , α , α , ..., α be the roots of the polynomial x − 1. It is a fact that there ex- 1 2 10 10 9 ists a unique polynomial of the form f ( x ) = x + c x + · · · + c x such that each c is 9 1 i 2 an integer, f (0) = f (1) = 0, and for any 1 ≤ i ≤ 10 we have ( f ( α )) = − 11. Find i | c + 2 c c + 3 c c + 4 c c + 5 c c | . 1 2 9 3 8 4 7 5 6 1
解析
- The existence and uniqueness of this polynomial (up to sign) are beyond the scope of this 11 contest, and as such we will take them for granted. Since all of the roots of x − 1 are powers k 11 of each other, we note that f ( x ), reduced to a degree 10 polynomial by using α = 1 for all i k i , we see that this new polynomial f ( x ) must also satisfy every condition of f ( x ). Therefore since f is unique up to sign, this new polynomial is either − f ( x ) or f ( x ). Since the new 10 coefficient of x can be any of the c ’s, we know that each each of the c ’s is ± 1. j j Since f (1) = 1 + c + · · · + c = 0, we know that 5 of the c ’s are − 1, and the other 4 are +1. 9 1 j 2 Now, look at f ( x ) . While this is a degree 20 polynomial, since again the only inputs we care 11 2 about all have the property that α = 1, we can restrict f ( x ) to a degree 10 polynomial by 11 simply identifying x with 1. 10 Thus our reduced polynomial looks like F ( x ) = B x + · · · + B x + B . Note that since we 10 1 0 11 know the values of F ( x ) at all of the 11 roots of x − 1, by Lagrange Interpolation, F ( x ) is uniquely determined. We can now perform the same trick we performed on f ( x ), by replacing k x with x for 1 ≤ k ≤ 10, since all of the inputs we are interested in are powers of each other. 3 As before, this will shuffle the coefficients of the polynomial, and will send B to each of the 10 other B ’s that are non-constant. Therefore B = B = ... = B . We also know that f (1) = 0, i 1 2 10 so F (1) = 0 as well, and we are given that F ( α ) = − 11. From these facts, we obtain the i following equations: 2 10 10 B + B = 0 and B ( α + α + · · · + α ) + B = − 11 . 1 0 1 0 10 Since 1 + α + · · · + α = 0, the second equation becomes B − B = − 11, and the solution to 0 1 these two equations is ( B , B ) = ( − 10 , 1). However, the only way the constant term of F ( x ) 0 1 11 2 can be − 10 is if the coefficient of x in f ( x ) is − 10, and the only way this can occur is if 11 every pair of coefficients that multiplies to form an x term has opposite sign. Therefore f ( x ) is anti-symmetric, so c = − 1 , c c = − 1 , c c = − 1 , c c = − 1 , c c = − 1, and the answer is 1 2 9 3 8 4 7 5 6 15 . 4