Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
x-y+z-w=2, \\
x^{2}-y^{2}+z^{2}-w^{2}=6, \\
x^{3}-y^{3}+z^{3}-w^{3}=20, \\
x^{4}-y^{4}+z^{4}-w^{4}=66^{[2]} .
\end{array}\right.
$$
(2006, National High School Mathematics Competition)
Define the sequences
$$
\begin{array}{l}
\left\{a_{n}\right\}: a_{n}=x^{n}+z^{n}\left(n \in \mathbf{N}_{+}\right), \\
\left\{b_{n}\right\}: b_{n}=y^{n}+w^{n}\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Thus, the original system of equations is
$$
\left\{\begin{array}{l}
a_{1}=b_{1}+2, \\
a_{2}=b_{2}+6, \\
a_{3}=b_{3}+20, \\
a_{4}=b_{4}+66 .
\end{array}\right.
$$

For $\left\{a_{n}\right\}$, we have
$$
\left(\sigma_{1}, \sigma_{2}\right)=(x+z, x z)=\left(a_{1}, \frac{a_{1}^{2}-a_{2}}{2}\right) \text {. }
$$

Using Newton's formulas, we get
$$
\begin{array}{l}
a_{3}=\sigma_{1} a_{2}-\sigma_{2} a_{1}=a_{1} a_{2}-\frac{a_{1}^{2}-a_{2}}{2} a_{1} \\
=\frac{1}{2} a_{1}\left(3 a_{2}-a_{1}^{2}\right), \\
a_{4}=\sigma_{1} a_{3}-\sigma_{2} a_{2}=a_{1} a_{3}-\frac{a_{1}^{2}-a_{2}}{2} a_{2} \\
=\frac{1}{2} a_{1}^{2}\left(3 a_{2}-a_{1}^{2}\right)-\frac{1}{2} a_{2}\left(a_{1}^{2}-a_{2}\right) \\
=\frac{1}{2}\left(2 a_{1}^{2} a_{2}-a_{1}^{4}+a_{2}^{2}\right) .
\end{array}
$$

Similarly, for $\left\{b_{n}\right\}$, we get
$$
\begin{array}{l}
b_{3}=\frac{1}{2} b_{1}\left(3 b_{2}-b_{1}^{2}\right), \\
b_{4}=\frac{1}{2}\left(2 b_{1}^{2} b_{2}-b_{1}^{4}+b_{2}^{2}\right) .
\end{array}
$$

Thus, the original system of equations is further transformed into
$$
\begin{array}{l}
\left\{\begin{array}{l}
a_{1}=b_{1}+2, \\
a_{2}=b_{2}+6, \\
\frac{1}{2} a_{1}\left(3 a_{2}-a_{1}^{2}\right)=\frac{1}{2} b_{1}\left(3 b_{2}-b_{1}^{2}\right)+20, \\
\frac{1}{2}\left(2 a_{1}^{2} a_{2}-a_{1}^{4}+a_{2}^{2}\right)=\frac{1}{2}\left(2 b_{1}^{2} b_{2}-b_{1}^{4}+b_{2}^{2}\right)+66
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
a_{1}=b_{1}+2, \\
a_{2}=b_{2}+6, \\
b_{1}^{2}-b_{1}-b_{2}+2=0, \\
2 b_{1}^{3}+3 b_{1}^{2}-2 b_{1} b_{2}-4 b_{1}-5 b_{2}+16=0 \\
a_{1}=b_{1}+2, \\
a_{2}=b_{2}+6, \\
b_{1}^{2}-b_{1}-b_{2}+2=0, \\
5 b_{1}^{2}-8 b_{1}-5 b_{2}+16=0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } = b _ { 1 } + 2 , } \\
{ a _ { 2 } = b _ { 2 } + 6 , } \\
{ b _ { 1 } = 2 , } \\
{ b _ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=4, \\
a_{2}=10, \\
b_{1}=2, \\
b_{2}=4
\end{array}\right.\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ x + z = 4 , } \\
{ x ^ { 2 } + z ^ { 2 } = 1 0 , } \\
{ y + w = 2 , } \\
{ y ^ { 2 } + w ^ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+z=4, \\
x z=3, \\
y+w=2, \\
y w=0 .
\end{array}\right.\right.
\end{array}
$$

Therefore, the original system of equations has 4 solutions:
$$
(x, y, z, w)=(1,2,3,0),(1,0,3,2),
$$