Problem 1072

Analysis First, construct the polynomial $f(x)=\left(x-x_{1}\right)(x-x_{2})\left(x-x_{3}\right)$ with roots $x_{k}=\left(2 \sin \frac{k \pi}{7}\right)^{2}(k=1,2,3)$. Let $S_{n}=x_{1}^{n}+x_{2}^{n}+x_{3}^{n}$, and use Newton's formulas to establish a recurrence relation for $T_{n}$. Then, complete the proof using mathematical induction.

Proof Let $x_{k}=\left(2 \sin \frac{k \pi}{7}\right)^{2}, k=1,2,3$. Construct the polynomial
$$f(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)=x^{3}-\sigma_{1} x^{2}+\sigma_{2} x-\sigma_{3},$$

where $\sigma_{1}=x_{1}+x_{2}+x_{3}=\sum_{k=1}^{3}\left(2 \sin \frac{k \pi}{7}\right)^{2}=\sum_{k=1}^{3} 2\left(1-\cos \frac{2 k \pi}{7}\right)$
$$\begin{array}{l} =6-2 \sum_{k=1}^{3} \cos \frac{2 k \pi}{7}=6-\frac{1}{\sin \frac{\pi}{7}} \sum_{k=1}^{3} 2 \cos \frac{k \pi}{7} \sin \frac{\pi}{7} \\ =6-\frac{1}{\sin \frac{\pi}{7}} \sum_{k=1}^{3}\left[\sin \frac{(2 k+1) \pi}{7}-\sin \frac{(2 k-1) \pi}{7}\right] \\ =6-\frac{1}{\sin \frac{\pi}{7}}\left(\sin \pi-\sin \frac{\pi}{7}\right)=7, \end{array}$$

Similarly, we can obtain $\sigma_{2}=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=14, \sigma_{3}=x_{1} x_{2} x_{3}=7$.
Let $S_{n}=x_{1}^{n}+x_{2}^{n}+x_{3}^{n}, n \in \mathbf{N}$. We need to prove that $\left.7^{\left[\frac{n}{3}\right]} \right\rvert\, S_{n}$, and according to Newton's formulas, we have
$$S_{n}=\sigma_{1} S_{n-1}-\sigma_{2} S_{n-2}+\sigma_{3} S_{n-3}=7 S_{n-1}-14 S_{n-2}+7 S_{n-3} \text {, and } S_{0}=3, S_{1}=7, S_{2}=21 \text {. }$$

We will use mathematical induction to prove that $S_{n}$ is divisible by $7^{\left[\frac{n}{3}\right]}$.
When $n=0,1,2$, the conclusion is obviously true. Assume that for $n \leqslant k$, $S_{n}$ is divisible by $7^{\left[\frac{n}{3}\right]}$. When $n=k+1$, by $S_{k+1}=7\left(S_{k}-2 S_{k-1}+S_{k-2}\right)$, we know that $S_{k+1}$ is divisible by $7^{1+\left[\frac{k-2}{3}\right]}$. Since $1+\left[\frac{k-2}{3}\right]=\left[\frac{k+1}{3}\right]$, $S_{k+1}$ is divisible by $7^{\left[\frac{k+1}{3}\right]}$. Therefore, $S_{n}$ is divisible by $7^{\left[\frac{n}{3}\right]}$. Proof completed.