南京大学真题 第20477题

五、（10 分）设 n 为正整数，$f_{1}(x), f_{2}(x) \ldots \ldots f_{n}(x)$ 都是多项式，并且 $x^{n}+x^{n-1}+\ldots \ldots+x^{2}+x+1 \mid f_{1}\left(x^{n+1}\right)+x f_{2}\left(x^{n+1}\right)+\ldots \ldots x^{n-1} f_{n}\left(x^{n+1}\right)$ ，证明： $(x-1)^{n} \mid f_{1}(x) f_{2}(x) \ldots \ldots f_{n}(x)$ 证 明：令 $\varepsilon_{1}, \varepsilon_{2} \ldots \ldots \varepsilon_{n}$ 为 $x^{n}+x^{n-1}+\ldots \ldots+x+1=0$ 的 解，所 以 $\varepsilon_{i}{ }^{n+1}-1=0(i=1,2, \ldots \ldots \mathrm{n})$ 因 $x^{n}+x^{n-1}+\ldots \ldots+x^{2}+x+1 \mid f_{1}\left(x^{n+1}\right)+x f_{2}\left(x^{n+1}\right)+\ldots \ldots x^{n-1} f_{n}\left(x^{n+1}\right)$ ，所以 $\varepsilon_{1}, \varepsilon_{2} \ldots \ldots \varepsilon_{n}$ 必然是 $f_{1}\left(x^{n+1}\right)+x f_{2}\left(x^{n+1}\right)+\ldots \ldots x^{n-1} f_{n}\left(x^{n+1}\right)=0$ 的解，即： $$ \left\{\begin{array}{c} f_{1}(1)+\varepsilon_{1} f_{2}(1)+\ldots \ldots+\varepsilon_{1}^{n-1} f_{n}(1)=0 \\ \ldots \ldots \ldots \ldots \ldots \\ f_{1}(1)+\varepsilon_{n} f_{2}(1)+\ldots \ldots+\varepsilon_{n}^{n-1} f_{n}(1)=0 \end{array}\right. $$ 解此方程组得到 $f_{1}(1)=f_{2}(1)=\ldots \ldots=f_{n}(1)=0$ ，所以 $(x-1) \mid f_{i}(x)(i=1,2, \ldots \ldots \mathrm{n})$即可得 $(x-1)^{n} \mid f_{1}(x) f_{2}(x) \ldots \ldots f_{n}(x)$ 。