Example 5 Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}\right.
$$

Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
Consider the function
$$
f(x)=\prod_{i=1}^{4}\left[x-(2 i-1)^{2}\right]-\prod_{i=1}^{4}\left[x-(2 i)^{2}\right] \text {. }
$$

Then $f\left(k^{2}\right)=\prod_{i=1}^{4}\left[k^{2}-(2 i-1)^{2}\right](k=2,4,6,8)$.

By the Lagrange interpolation formula, we have
$$
f(x)=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \neq i<i<4 \\ 1 \leq j<4}} \frac{x-(2 i-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} .
$$

Comparing the coefficients of $x^{3}$ in equations (1) and (3), we get
$$
\begin{array}{l}
\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right] \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i \\
1 \leqslant i<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} .
\end{array}
$$

In equation (3), let $x=k^{2}(k=2,4,6,8)$, we get
$$
\begin{array}{l} 
f\left(k^{2}\right) \\
= \sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i<4 \\
1 \neq j \leqslant 4}} \frac{k^{2}-(2 j-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} \\
= \sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j+i<i \\
1<j \leqslant 4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} . \\
\prod_{t=1}^{4}\left[k^{2}-(2 t-1)^{2}\right] .
\end{array}
$$

From equations (2) and (4), we have
$$
\begin{array}{l}
\sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j \neq i<i \\
1 \leqslant \leqslant<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
\quad=1(k=2,4,6,8) .
\end{array}
$$

By the uniqueness of the solution of the system of equations, we have
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}+w^{2} \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{1 \neq i<\\
1<j<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
=\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right]=36 .
\end{array}
$$

**Summary**: The construction of the function is not unique. For example, let
$$
\begin{aligned}
f(m)= & \left(\frac{x^{2}}{m-1^{2}}+\frac{y^{2}}{m-3^{2}}+\frac{z^{2}}{m-5^{2}}+\frac{w^{2}}{m-7^{2}}\right) . \\
& \prod_{i=1}^{4}\left[m-(2 i-1)^{2}\right] .
\end{aligned}
$$

Then by the Lagrange interpolation formula, we have
$$
f(m)=\sum_{i=1}^{4} f\left((2 i)^{2}\right) \prod_{\substack{j \neq i \\ 1 \leqslant j \leqslant 4}} \frac{m-(2 j)^{2}}{(2 i)^{2}-(2 j)^{2}} \text {. }
$$

Comparing the coefficients of $m^{3}$, we get the value of $x^{2}+y^{2}+z^{2}+w^{2}$.