The value of $x^2+y^2+z^2+w^2$ Let$x,y,z,w$ satisfy
$$\frac{x^2}{2^2 - 1^2} +\frac{y^2}{2^2 - 3^2} +\frac{z^2}{2^2 - 5^2} +\frac{w^2}{2^2 - 7^2} =1$$
$$\frac{x^2}{4^2 - 1^2} +\frac{y^2}{4^2 - 3^2} +\frac{z^2}{4^2 - 5^2} +\frac{w^2}{4^2 - 7^2} =1$$
$$\frac{x^2}{6^2 - 1^2} +\frac{y^2}{6^2 - 3^2} +\frac{z^2}{6^2 - 5^2} +\frac{w^2}{6^2 - 7^2} =1$$
$$\frac{x^2}{8^2 - 1^2} +\frac{y^2}{8^2 - 3^2} +\frac{z^2}{8^2 - 5^2} +\frac{w^2}{8^2 - 7^2} =1$$
My work
$$\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1$$
where $t $ satisfy $4,16,36,64$
$$f(t)=0$$
$$f(t) = (t – 1)(t – 9)(t – 25)(t – 49)–x^2(t – 9)(t – 25)(t – 49) –y^2(t – 1)(t – 25)(t – 49) – z^2(t–1)(t–9)(t–49) – w^2(t–1)(t–9)(t–25)$$
then I compared the coefficient with different value of $t$ .
I want to know that is there any easier alternative methods for this .
 A: Expanding 
\begin{equation*}
\frac{x^2}{t - 1^2} +\frac{y^2}{t - 3^2} +\frac{z^2}{t - 5^2} +\frac{w^2}{t - 7^2} =1 
\end{equation*}
we get 
\begin{equation*}
(t-1^2)(t-3^2)(t-5^2)(t-7^2) - (t-3^2)(t-5^2)(t-7^2)x^2 - \text{ similar terms } = 0 
\end{equation*}
This biquadratic in $t$ has four roots $2^2, 4^2, 6^2, 8^2$ and coefficient of $t^3$ is $-(2^2+4^2+6^2+8^2)$. The coefficient of $t^3$ is also given by
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2)
\end{equation*}
and hence
\begin{equation*}
-(1^2+3^2+5^2+7^2) - (x^2+y^2+z^2+w^2) = -(2^2+4^2+6^2+8^2)
\end{equation*}
Hence we have
\begin{equation*}
x^2+y^2+z^2+w^2 = (2^2+4^2+6^2+8^2)-(1^2+3^2+5^2+7^2)=36
\end{equation*}
A: Set up a matrix equation 
$$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{bmatrix} = \begin{bmatrix} 1\\1\\1\\1\end{bmatrix}$$
and then solve. You can essentially ignore the fact that your variables are squared, and just assume that means they must all be nonnegative.
A: I multiplied the equations out to get:
$$\begin{pmatrix} -4725 & 2835 & 675 & 315 \\
2079 & 4455 & -3465 & -945 \\
-3861 & -5005 & -12285 & 10395 \\
32175 & 36855 & 51975 & 135135 \\
\end{pmatrix} 
\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} = 
\begin{pmatrix} -14175 \\
31185 \\
-135135 \\
2027025 \\
\end{pmatrix}$$
Then  inverted the matrix and solved to get: 
$$\begin{pmatrix} x^{2}\\y^{2}\\z^{2}\\w^{2}\end{pmatrix} = 
\begin{pmatrix} 10.76660156 \\
10.15136719 \\
8.797851563 \\
6.284179688
\end{pmatrix}
$$
giving 
$$x^2+y^2+z^2+w^2=36$$
 - so there probably is an algebraic short cut to that.
