4. 148 Given
$$\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{\omega^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{\omega^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{\omega^{2}}{6^{2}-7^{2}}=1 \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{\omega^{2}}{8^{2}-7^{2}}=1
\end{array}\right.$$

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

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

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. By comparing the coefficients of the $t^{3}$ terms in equations (2) and (3), we get
$$-1-9-25-49-x^{2}-y^{2}-z^{2}-\omega^{2}=-4-16-36-64,$$

which simplifies to $x^{2}+y^{2}+z^{2}+\omega^{2}=36$.