Oh, How Long It Takes?

$\begin{aligned} \dfrac{x^2}{2^2-1^2}+\dfrac{y^2}{2^2-3^2}+\dfrac{z^2}{2^2-5^2}+\dfrac{w^2}{2^2-7^2} \ = \ 1 \\ \dfrac{x^2}{4^2-1^2}+\dfrac{y^2}{4^2-3^2}+\dfrac{z^2}{4^2-5^2}+\dfrac{w^2}{4^2-7^2} \ = \ 1 \\ \dfrac{x^2}{6^2-1^2}+\dfrac{y^2}{6^2-3^2}+\dfrac{z^2}{6^2-5^2}+\dfrac{w^2}{6^2-7^2} \ = \ 1 \\ \dfrac{x^2}{8^2-1^2}+\dfrac{y^2}{8^2-3^2}+\dfrac{z^2}{8^2-5^2}+\dfrac{w^2}{8^2-7^2} \ = \ 1 \\ \end{aligned}$

Determine $w^2+x^2+y^2+z^2$ if they satisfy the system of equations above.