Example 19 (Problem from the 2nd American Mathematical Invitational Competition) Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1,
\end{array}\right.
$$

Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
First, note that: $x, y, z, w$ satisfying the given system of equations is equivalent to $t=4,16,36,64$ satisfying the equation
$$
\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}+\frac{w^{2}}{t-49}=1 \text {. }
$$

Clearing the denominators and rearranging terms, we get
$$
\begin{aligned}
(t-1)(t-9)(t-25)(t-49)-x^{2}(t-9)(t-25)(t-49)-y^{2}(t-1)(t- \\
25)(t-49)-z^{2}(t-1)(t-9)(t-49)-w^{2}(t-1)(t-9)(t-25)=0,
\end{aligned}
$$

Equation (2) is a quartic equation in $t$, and $t=4,16,36,64$ are the four roots of this equation. Therefore, equation (2) is equivalent to $(t-4)(t-16)(t-36)(t-64)=0$.

The leading coefficients of equations (2) and (3) are both 1, so the coefficients of the other terms should also be equal. Comparing the coefficients of the $t^{2}$ terms in equations (2) and (3), we get
$$
-1-9-25-49-x^{2}-y^{2}-z^{2}-w^{2}=-4-16-36-64 \text {. }
$$

Thus, $x^{2}+y^{2}+z^{2}+w^{2}=36$ is the desired result.