Example 12 Solve the system of equations $\left\{\begin{array}{l}x-y+z-w=2, \\ x^{2}-y^{2}+z^{2}-w^{2}=6, \\ x^{3}-y^{3}+z^{3}-w^{3}=20, \\ x^{4}-y^{4}+z^{4}-w^{4}=60 .\end{array}\right.$
(National High School Competition, 2006)
Define $a_{n}=x^{n}+z^{n}\left(n \in \mathbf{N}^{+}\right), b_{n}=y^{n}+w^{n}\left(n \in \mathbf{N}^{+}\right)$. Thus, the original system of equations becomes $\left\{\begin{array}{l}a_{1}=b_{1}+z, \\ a_{2}=b_{2}+6, \\ a_{3}=b_{3}+20, \\ a_{4}=b_{4}+66 .\end{array}\right.$ Let $x, z$ be the roots of the polynomial $f(t)=t^{2}+A_{1} t+B_{1}$, then $A_{1}=-(x+z)=-a_{1}, B_{2}=x z=\frac{1}{2}\left(a_{1}^{2}-a_{2}\right)$.
By Newton's formulas, we have $a_{3}=-A_{1} a_{2}-B_{1} a_{1}=\frac{1}{2} a_{1}\left(3 a_{2}-a_{1}^{2}\right)$, $a_{4}=-A_{1} a_{3}-B_{1} a_{2}=\frac{1}{2}\left(2 a_{1}^{2} a_{2}-a_{1}^{4}+a_{2}^{2}\right)$.
Let $y, w$ be the roots of the polynomial $f(s)=s^{2}+A_{2} s+B_{2}$, similarly,
we have $b_{3}=\frac{1}{2} b_{1}\left(3 b_{2}-b_{1}^{2}\right), b_{4}=\frac{1}{2}\left(2 b_{1}^{2} b_{2}-b_{1}^{4}+b_{2}^{2}\right)$.
Thus, the original system of equations is further transformed into
$$
\begin{array}{l}
\left\{\begin{array}{l}
a_{1}=b_{1}+2 \\
a_{2}=b_{2}+6 \\
\frac{1}{2} a_{1}\left(3 a_{2}-a_{1}^{2}\right)=\frac{1}{2} b_{1}\left(3 b_{2}-b_{1}^{2}\right)+20 \\
\frac{1}{2}\left(2 a_{1}^{2} a_{2}-a_{1}^{4}+a_{2}^{2}\right)=\frac{1}{2}\left(2 b_{1}^{2} b_{2}-b_{1}^{4}+b_{2}^{2}\right)+66
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ a _ { 1 } = b _ { 1 } + 2 } \\
{ a _ { 2 } = b _ { 2 } + 6 } \\
{ b _ { 1 } ^ { 2 } - b _ { 1 } - b _ { 2 } + 2 = 0 } \\
{ 2 b _ { 1 } ^ { 3 } + 3 b _ { 1 } ^ { 2 } - 2 b _ { 1 } b _ { 2 } - 4 b _ { 1 } - 5 b _ { 2 } + 1 6 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a_{1}=b_{1}+2 \\
a_{2}=b_{2}+6 \\
b_{1}^{2}-b_{1}-b_{2}+2=0 \\
5 b_{1}^{2}-8 b_{1}-5 b_{2}+16=0
\end{array}\right.\right. \\
\cdot\left\{\begin{array} { l } 
{ a _ { 1 } = b _ { 1 } + 2 } \\
{ a _ { 2 } = b _ { 2 } + 6 } \\
{ b _ { 1 } = 2 } \\
{ b _ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ a _ { 1 } = 4 } \\
{ a _ { 2 } = 1 0 } \\
{ b _ { 1 } = 2 } \\
{ b _ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x + z = 4 } \\
{ x ^ { 2 } + z ^ { 2 } = 1 0 } \\
{ y + w = 2 } \\
{ y ^ { 2 } + w ^ { 2 } = 4 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+z=4, \\
x z=3, \\
y+w=2, \\
y w=0 .
\end{array}\right.\right.\right.\right.
\end{array}
$$

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