Problem 921

【Analysis】This is an application problem of roots of unity. The key is to first express $\cos \frac{2 k \pi}{p}$ using roots of unity, and then use the properties of roots of unity for calculation.
Let $\varepsilon=\mathrm{e}^{\frac{2 \pi x}{p}}$. Then
$$\begin{array}{l} \varepsilon^{p}=1, \varepsilon^{\frac{p}{2}}=\mathrm{e}^{-\pi \mathrm{i}}=-1, \\ 2 \cos \frac{2 k \pi}{p}=\varepsilon^{k}+\varepsilon^{-k} . \\ \text { Hence } \prod_{k=1}^{p}\left(1+2 \cos \frac{2 k \pi}{p}\right) \\ =3 \prod_{k=1}^{p-1}\left(1+\varepsilon^{k}+\varepsilon^{-k}\right) \\ =\frac{3 \prod_{k=1}^{p-1}\left(1+\varepsilon^{k}+\varepsilon^{2 k}\right)}{\prod_{k=1}^{p-1} \varepsilon^{k}} \\ =\frac{3 \prod_{k=1}^{p-1}\left(1+\varepsilon^{k}+\varepsilon^{2 k}\right)}{\varepsilon^{\frac{2 x-11}{2}}} \\ =3 \prod_{k=1}^{p-1}\left(1+\varepsilon^{k}+\varepsilon^{2 k}\right) \\ =3 \prod_{k=1}^{n-1} \frac{\left(1-\varepsilon^{3 k}\right)}{\left(1-\varepsilon^{k}\right)} . \end{array}$$

Since the prime $p>3$, we have $(3, p)=1$.
Thus, $3,3 \times 2, \cdots, 3(p-1)$ exactly cover all non-zero residue classes of $p$.
Therefore, $\prod_{k=1}^{p-1}\left(1-\varepsilon^{3 k}\right)=\prod_{k=1}^{p-1}\left(1-\varepsilon^{k}\right)$.
Hence, $\prod_{k=1}^{p}\left(1+2 \cos \frac{2 k \pi}{p}\right)=3$.