FiNaL◎峰

求任何x, k 一定要整數

用quadratic formula 可以搵到



同埋



一眼睇落去就搵到0 同 -5 係整數既roots

但係我用電腦計計到有 0, 4, -5, -9

問題想問4同-9用人手要點計出黎? 同埋會唔會有更多整數答案?

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Actually, there exists only finite number of k.
k is a integer such that k(k+5)=N^2 for some integer N. Then, k and k+5 either have common integer factor other than -1 and 1, say, a, or k and k+5 are both square numbers.
For the first case, k and k+5 have common integer factor a other than -1 and 1:
Now, we can write k=ab and k+5=ac for some integers b and c, where bc is a square number also.
ac=k+5=ab+5 => a(c-b)=5
So, a=-5,  5.
If a=5, c-b=1. cb=b(b+1) and cb is square number, so b=0 and k=0.
If a=-5, by a similar process, we can find k=5.
For the second case, k and k+5 are both square numbers:
Now, k=m^2 and k+5=n^2 or k=-m^2 and k+5=-n^2, for some integer m and n.
For the first subcase, Then, n^2=5+m^2 => (n-m)(n+m)=5 => (m, n)=(2, 3) or (-2, -3) or (3, 2) or (-3, -2).
So, k=4 or 9(rejected).
For the second case, we can find k=-9.
I don't know if there are any gaps or mistakes in the proof, so please point out if there are mistakes.



比其他論壇既人答左-_-, 附上解答比好奇既會員

[ 本帖最後由 FiNaL◎峰 於 2013-3-23 03:20 編輯 ]

FiNaL◎峰

引用:

原帖由 阿平 於 2013-3-22 13:36 發表
唔會有更多整數答案
最小5~9999999999999999內唔會有
原因好難講...只可以話100%冇.
開方(自己X(自己+5/-5)=正數 就搵到y

點解會自己X(自己-5)????

其他網答我會有無限個可能性, 但係又無比其他例子-_-

我想最知道點搵4同-9... 求解

FiNaL◎峰

引用:

原帖由 愛忘情. 於 2013-3-22 22:50 發表
下.你問老師啦!係咪你打錯根本無可能計到

我答案都寫埋出黎，點會計唔到？
講野唔經大腦你知唔知醜架

FiNaL◎峰

引用:

原帖由 Believe. 於 2013-3-23 00:57 發表
打緊機 一陣幫你計
唔識計 總系覺得要假設某d野成立先

同埋唔識解 點解 冇無限個答案 ... 最小我計出泥系冇

k(k+5)=y^2 y要系整數


得4x9=36=6^2 成立... ...

-9, -5, 0, 4 係我搵既答案... 唔係全部都成立咩?

係, 我都覺得要返某樣野出黎先搵到, 看似有無限個答案, 但實際上我只搵到4個-_-''

我有上網做過research, 學左幾種方法都唔識搵
另外想問你4點搵出黎?...

FiNaL◎峰

引用:

原帖由 Believe. 於 2013-3-23 02:31 發表

唔係搵出泥只系試出泥

因為我搵唔到方法做落去-.-

以我的認知
-9, -5, 0, 4 係我搵既答案... 唔係全部都成立咩?


呢個系答案 我意思系有d條件 but 我地未搵到佢出泥 ...

我正係識得搵0同-5,  4同-9係用電腦搵出黎
我覺得4同-9既然電腦搵得出黎, 自然有方法計.