The given information tells us that the equation \[\frac{x^2}{t-1} + \frac{y^2}{t-3^2} + \frac{z^2}{t-5^2} + \frac{w^2}{t-7^2} = 1\]holds for $ t = 2^2, 4^2, 6^2, 8^2 $. Clearing fractions, we have the equation \[\begin{aligned} &\quad x^2(t-3^2)(t-5^2)(t-7^2) + y^2(t-1)(t-5^2)(t-7^2) \\ &+ z^2(t-1)(t-3^2)(t-7^2) + w^2(t-1)(t-3^2)(t-5^2) = (t-1)(t-3^2)(t-5^2)(t-7^2), \end{aligned}\]or \[\begin{aligned} &(t-1)(t-3^2)(t-5^2)(t-7^2) - x^2(t-3^2)(t-5^2)(t-7^2) - y^2(t-1)(t-5^2)(t-7^2) \\ &- z^2(t-1)(t-3^2)(t-7^2) - w^2(t-1)(t-3^2)(t-5^2) = 0.\end{aligned}\]Upon expansion, the left side becomes a fourth-degree polynomial in $ t, $ with leading coefficient $ 1 $. We know that this equation holds for $ t = 2^2,4^2,6^2,8^2, $ so by the factor theorem, the linear terms $ t-2^2, $ $ t-4^2, $ $ t-6^2, $ and $ t-8^2 $ must divide this polynomial. But the polynomial has degree $ 4, $ so it must be the case that \[\begin{aligned} &(t-1)(t-3^2)(t-5^2)(t-7^2) - x^2(t-3^2)(t-5^2)(t-7^2) - y^2(t-1)(t-5^2)(t-7^2) \\ &- z^2(t-1)(t-3^2)(t-7^2) - w^2(t-1)(t-3^2)(t-5^2) = (t-2^2)(t-4^2)(t-6^2)(t-8^2) \end{aligned}\]for all $ t $. To finish, we compare the coefficients of $ t^3 $ on both sides: \[-(1+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2) = -(2^2+4^2+6^2+8^2),\]which gives \[x^2+y^2+z^2+w^2 = \boxed{36}.\]