Question

设 $\omega \neq 1$ 是一个 $13$ 次单位根。求除以 $1000$ 的余数。 $\prod_{k=0}^{12} \left(2 - 2\omega^k + \omega^{2k}\right)$

Let $\omega \neq 1$ be a $13$ th root of unity. Find the remainder when $\prod_{k=0}^{12} \left(2 - 2\omega^k + \omega^{2k}\right)$ is divided by $1000.$

答案

答案：321

知识点：单位根复数多项式

难度评级：3060

解答：

因为 $2 - 2x + x^2 = (x - 1)^2 + 1$ $= (x - (1+i))(x - (1-i))$ ，乘积中的每个因式都可分解；当 $k$ 从 $0$ 到 $12$ ，变化时， $\omega^k$ 遍历所有 $13$ 次单位根。由于 $\prod_k (x - \omega^k) = x^{13} - 1$ ，对任意 $\alpha$ 有 $\prod_k (\omega^k - \alpha) = (-1)^{13}(\alpha^{13} - 1)$ $= 1 - \alpha^{13}$ 。因此该乘积等于 $\left(1 - (1+i)^{13}\right)\left(1 - (1-i)^{13}\right).$

因为 $(1+i)^2 = 2i$ ，所以 $(1+i)^{13} = (1+i)(2i)^6$ $= -64(1 + i) = -64 - 64i$ ，由共轭可得 $(1-i)^{13} = -64 + 64i$ 。所以乘积为除以 $1000$ 的余数为 $321$ 。 $\begin{gathered} (65 + 64i)(65 - 64i) \\ = 65^2 + 64^2 = 4225 + 4096 \\ = 8321, \end{gathered}$

Since $2 - 2x + x^2 = (x - 1)^2 + 1$ $= (x - (1+i))(x - (1-i)),$ each factor of the product splits, and as $k$ runs from $0$ to $12,$ $\omega^k$ runs over all $13$ th roots of unity. Because $\prod_k (x - \omega^k) = x^{13} - 1,$ for any $\alpha$ we get $\prod_k (\omega^k - \alpha) = (-1)^{13}(\alpha^{13} - 1)$ $= 1 - \alpha^{13}.$ Hence the product equals $\left(1 - (1+i)^{13}\right)\left(1 - (1-i)^{13}\right).$

Since $(1+i)^2 = 2i,$ we get $(1+i)^{13} = (1+i)(2i)^6$ $= -64(1 + i) = -64 - 64i,$ and by conjugation $(1-i)^{13} = -64 + 64i.$ So the product is $\begin{gathered} (65 + 64i)(65 - 64i) \\ = 65^2 + 64^2 = 4225 + 4096 \\ = 8321, \end{gathered}$ whose remainder upon division by $1000$ is $321.$

2024 AIME II 第 13 题

答案：321

难度评级：3060

解答：

其他年份的第 13 题