- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,587

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Determine $x^2+y^2+z^2+w^2$ if

$\dfrac{x^2}{2^2-1^2}+\dfrac{y^2}{2^2-3^2}+\dfrac{z^2}{2^2-5^2}+\dfrac{w^2}{2^2-7^2}=1,\\\dfrac{x^2}{4^2-1^2}+\dfrac{y^2}{4^2-3^2}+\dfrac{z^2}{4^2-5^2}+\dfrac{w^2}{4^2-7^2}=1,\\\dfrac{x^2}{6^2-1^2}+\dfrac{y^2}{6^2-3^2}+\dfrac{z^2}{6^2-5^2}+\dfrac{w^2}{6^2-7^2}=1,\\\dfrac{x^2}{8^2-1^2}+\dfrac{y^2}{8^2-3^2}+\dfrac{z^2}{8^2-5^2}+\dfrac{w^2}{8^2-7^2}=1$

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