I have the following set of equations:
$$ \frac{x^2}{2^2-1} + \frac{y^2}{2^2 - 3^2} + \frac{z^2}{2^2-5^2} + \frac{w^2}{2^2-7^2} = 1 $$
$$ \frac{x^2}{4^2-1} + \frac{y^2}{4^2 - 3^2} + \frac{z^2}{4^2-5^2} + \frac{w^2}{4^2-7^2} = 1 $$ $$ \frac{x^2}{6^2-1} + \frac{y^2}{6^2 - 3^2} + \frac{z^2}{6^2-5^2} + \frac{w^2}{6^2-7^2} = 1 $$ $$ \frac{x^2}{8^2-1} + \frac{y^2}{8^2 - 3^2} + \frac{z^2}{8^2-5^2} + \frac{w^2}{8^2-7^2} = 1 $$
Well the nature of numbers in the denominator itself freaked me out and now I am unable to approach the question. I tried to simplify the expressions which turned out to be of no use. Any help would be appreciated.
Thanks in advance
there are many methods but i prefer to use matrix
Then i try to solve the system $AX=T $, where
$$A=\left( \begin{array}{cccc} \frac{1}{2^2-1} & \frac{1}{2^2-3^2} & \frac{1}{2^2-5^2} & \frac{1}{2^2-7^2} \\ \frac{1}{4^2-1} & \frac{1}{4^2-3^2} & \frac{1}{4^2-5^2} & \frac{1}{4^2-7^2} \\ \frac{1}{6^2-1} & \frac{1}{6^2-3^2} & \frac{1}{6^2-5^2} & \frac{1}{6^2-7^2} \\ \frac{1}{8^2-1} & \frac{1}{8^2-3^2} & \frac{1}{8^2-5^2} & \frac{1}{8^2-7^2} \\ \end{array} \right)$$
and
$$X=\left( \begin{array}{c} x^2 \\ y^2 \\ z^2 \\ w^2 \\ \end{array} \right)$$
and
$$T=\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right)$$
to find $X$ we must calculate the inverse matrix of $A$ and then multuply by the $T$ we can get
$$X=\frac{1}{1024} \left( \begin{array}{c} 11025 \\ 10395 \\ 9009 \\ 6435 \\ \end{array} \right)$$
finally
$$\left( \begin{array}{c} x \\ y \\ z \\ w \\ \end{array} \right)=\pm\sqrt{X}=\pm\frac{1}{32}\left( \begin{array}{c} 105 \\ 3 \sqrt{1155} \\ 3 \sqrt{1001} \\ 3 \sqrt{715} \\ \end{array} \right)$$
i used a Math Software