推 Desperato : 推原PO也推樓上 我在耍白痴qw q 02/26 23:14
推 a016258 : 推 02/27 00:07
※ 引述《skyfan2008 (la..la..)》之銘言：
: 若x,y皆是正整數，滿足x^2=y^2(x+y^4+2y^2)的數對(x,y)=?
: 請教各位大大如何解
: 謝謝
亂猜的 各為高手看一下對不對
let t == y^2 then we have
x^2 - t*x - (t^3 + 2t^2) = 0
So the positive root of this equation in terms of x is
x = 0.5*(t + \sqrt{4t^3 + 9t^2})
But since x is in N, so t must be some 2*a for which a = 1,2,3,4...
so that the 0.5 can be balanced off
Then we plug in t = 2*a in the sqrt term, which must be some squared number
it follows that
4t^3 +9t^2 = a^2(32a + 36) for a = 1, 2, 3, 4...
and we need only care about 32a + 36 being some squared number
...
then we go over all possible a's to get the answer
the smallest answer is a = 2 -> t = 4 (y = 2) -> x = 12
--
有錯請指正^^"
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 163.18.23.206
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1519640983.A.904.html
※ 編輯: LiamIssac (163.18.23.206), 02/26/2018 18:29:59
※ 編輯: LiamIssac (163.18.23.206), 02/26/2018 18:30:22