Re: [中學] 高中數學一題
※ 引述《adamchi (adamchi)》之銘言：
: n=2,4,6,8 時,
: ((x^2)/(n^2-1^2)) +((y^2)/(n^2-3^2))+((z^2)/(n^2-5^2))+((w^2)/(n^2-7^2)) = 1
: ,求x^2+y^2+z^2+w^2 = ?
: 本題以計算機暴力解法得答案為36,
: 想請教有何更快的解法,謝謝~
令u=n^2，兩邊同乘(u-1)(u-9)(u-25)(u-49)可得
x^2(u-9)(u-25)(u-49)+y^2(u-1)(u-25)(u-49)
+z^2(u-1)(u-9)(u-49)+w^2(u-1)(u-9)(u-25) = (u-1)(u-9)(u-25)(u-49)
移項整理為u的4次多項式，4根為4,16,36,64
4次方項係數為-1，3次方項係數為x^2+y^2+z^2+w^2+84 (84=1+9+25+49)
由根與係數關係得x^2+y^2+z^2+w^2+84 = 4+16+36+64
=> x^2+y^2+z^2+w^2 = 36
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