<s>[INST] 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 .
 [/INST] 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.
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