E2 4 ** Let $n$ be a positive integer, find the value of the product $\left(1+2 \cos \frac{2 \pi}{2 n+1}\right)\left(1+2 \cos \frac{4 \pi}{2 n+1}\right) \cdots(1+$ $\left.2 \cos \frac{4 n \pi}{2 n+1}\right)$.
By using the roots of unity, let $\varepsilon=\mathrm{e}^{\frac{i 2 x}{2 n+1}}$, then
$$
\begin{aligned}
A & =\prod_{k=1}^{2 n}\left(1+2 \cos \frac{2 k \pi}{2 n+1}\right)=\prod_{k=1}^{2 n}\left(1+\varepsilon^{k}+\bar{\varepsilon}^{k}\right) \\
& =\prod_{k=1}^{2 n}\left(\frac{1+\varepsilon^{k}+\varepsilon^{2 k}}{\varepsilon^{k}}\right)=\prod_{k=1}^{2 n} \frac{1-\varepsilon^{3 k}}{\varepsilon^{k}\left(1-\varepsilon^{k}\right)} .
\end{aligned}
$$

When $3 \mid 2 n+1$, there exists $k \in\{1,2, \cdots, 2 n\}$ such that $3 k=2 n+1$. At this time, $1-\varepsilon^{3 k}=0$, so $A=0$.

When $3 \nmid 2 n+1$, $(3,2 n+1)=1$, therefore when $k$ runs through a complete residue system modulo $2 n+1$, $3 k$ also runs through a complete residue system modulo $2 n+1$. Hence $\left\{1-\varepsilon^{3}, 1-\varepsilon^{3 \cdot 2} \cdots, 1-\varepsilon^{3 \cdot 2 n}\right\}=\left\{1-\varepsilon, 1-\varepsilon^{2}, \cdots, 1-\varepsilon^{2 n}\right\}$, thus from (1) we know
$$
A=\prod_{k=1}^{2 n} \frac{1}{\varepsilon^{k}}=\varepsilon^{-\frac{2 n(2 n+1)}{2}}=\left(\varepsilon^{2 n+1}\right)^{-n}=1 .
$$

In summary, when $3 \mid 2 n+1$, $A=0$; when $3 \nmid 2 n+1$, $A=1$.