jmc

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 $$\begin{array}{rl}& x^2(t-3^2)(t-5^2)(t-7^2)+y^2(t-1)(t-5^2)(t-7^2)\\ & +z^2(t-1)(t-3^2)(t-7^2)+w^2(t-1)(t-3^2)(t-5^2)=(t-1)(t-3^2)(t-5^2)(t-7^2),\end{array}$$or $$\begin{array}{rl}& (t-1)(t-3^2)(t-5^2)(t-7^2)-x^2(t-3^2)(t-5^2)(t-7^2)-y^2(t-1)(t-5^2)(t-7^2)\\ & -z^2(t-1)(t-3^2)(t-7^2)-w^2(t-1)(t-3^2)(t-5^2)=0.\end{array}$$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 $$\begin{array}{rl}& (t-1)(t-3^2)(t-5^2)(t-7^2)-x^2(t-3^2)(t-5^2)(t-7^2)-y^2(t-1)(t-5^2)(t-7^2)\\ & -z^2(t-1)(t-3^2)(t-7^2)-w^2(t-1)(t-3^2)(t-5^2)=(t-2^2)(t-4^2)(t-6^2)(t-8^2)\end{array}$$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 $$x^2+y^2+z^2+w^2=\boxed{36}.$$