Example 4 Let $p(x)$ be a polynomial of degree $3n$, satisfying $p(0)=p(3)=\cdots=p(3n)=2, p(1)=$ $p(4)=\cdots=p(3n-2)=1, p(2)=p(5)=\cdots=p(3n-1)=0$, and $p(3n+1)=$ 730, find $n$.
(1984 USA Mathematical Olympiad)
We have $\Delta^{3 n+1} P(0)=\sum_{k=0}^{3 n+1}(-1)^{k} C_{3 n+1}^{k} P(3 n+1-k)$
$$
\begin{array}{l}
=730+\sum_{j=0}^{n}(-1)^{3 j+1} C_{3 n+1}^{3 j+1} \cdot 2+\sum_{j=1}^{n}(-1)^{3 j} C_{3 n+1}^{3 j} \\
729-2 \sum_{j=0}^{n}(-1)^{j} C_{3 n+1}^{3 j+1}+\sum_{j=1}^{n}(-1)^{j} C_{3 n+1}^{3 j}=0 .
\end{array}
$$

To find $M=\sum_{j=0}^{n}(-1)^{j} C_{3 n+1}^{3 j+1}$ and $N=\sum_{j=0}^{n}(-1)^{j} C_{3 n+1}^{3 j}$, we use the cube root of unity $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}$
and the polynomial $f(x)=(1-x)^{3 n+1}=\sum_{k=0}^{3 n+1}(-1)^{k} C_{3 n+1}^{k} x^{k}$
$$
=\sum_{j=0}^{n}(-1)^{j} C_{3 n+1}^{3 j} \cdot x^{3 j}-\sum_{j=0}^{n}(-1)^{j} C_{3 n+1}^{3 j+1} \cdot x^{3 j+1}-\sum_{j=1}^{n}(-1)^{j} C_{3 n+1}^{3 j-1} \cdot x^{3 j-1},
$$

According to this expansion and the properties of $\omega$, we have
$$
\begin{array}{l}
f(1)=N-M-L=0, \text { where } L=\sum_{j=1}^{n}(-1)^{j} C_{3 n+1}^{3 j-1}, \\
f(\omega)=N-M \omega-L \omega^{2}=(1-\omega)^{3 n+1}, \\
f\left(\omega^{2}\right)=N-M \omega^{2}-L \omega=\left(1-\omega^{2}\right)^{3 n+1} .
\end{array}
$$

Thus, $\quad N=\frac{1}{3}\left[f(1)+f(\omega)+f\left(\omega^{2}\right)\right]=2(\sqrt{3})^{3 n-1} \cdot \cos \frac{3 n+1}{6} \pi$,
$$
M=-\frac{1}{3}\left[f(1)+\omega^{2} f(\omega)+\omega f\left(\omega^{2}\right)\right]=2(\sqrt{3})^{3 n-1} \cdot \cos \frac{3 n-1}{6} \pi \text {. }
$$

Substituting these into (*) gives $729-4(\sqrt{3})^{3 n-1} \cdot \cos \frac{3 n-1}{6} \pi+2(\sqrt{3})^{3 n-1} \cdot \cos \frac{3 n+1}{6} \pi=0$. When $n=$ $2 t$, we find $t=2$, i.e., $n=4$; when $n=2 t+1$, the equation has no integer solutions. Therefore, $n=4$ is the solution.