連續數相乘一問~高手快進!

( -1 x 0 ) x 2 = 0 x 1

因為0x任何數都係0...
咁樣得唔得?

2009-02-10 15:31:35 補充：
我想問...
其實..條式係

(a*b)*2=b*c

定係

(a*b)*2=y*z

因為如果係前者...你所要o既答案範圍會收窄左好多...

2009-02-11 08:50:08 補充：
你最先問時係要『還有無其他可能?有請舉例,無請證明。』
我已舉例...
現在先說『想要的是對所有數的分析』??太遲了..

加上下面有人已作分析..我亦無必要再打多次分析內容..
未免有人當作抄襲...!!

個人應為mr_gary_cheung 所列我式概念都係前者...只不過運算上錯誤..
而你舉例的...都屬前者...
我的例子並不是創意...而是你忽略了『0』這個重要的數字...

所以綜合各點...我覺得我第一個答案是沒有問題...
而且是合乎原來問題o既要求...

還有...你認為我會為回答採用率降低要不作補充嗎?
當然你唔鍾意都無法...我亦不介意回答採用率降低...

2009-02-11 14:33:04 補充：
『有請舉例,無請證明。』
何以見得你想要一個最全面最詳細的分析呢?!
你講到明無先要證明bor...

e.g.考試個題目叫你寫答案...你走去寫晒計算過程之後先俾答案...
唔通咁樣會高分d?
唔會lor...反而仲會俾人話你唔審清楚題目要你答咩野tim la...

如果你真係要反映問題的根本性質...就唔應該『有請舉例,無請證明。』..

我諗你如果真係想反映問題的根本性質...
下次問問題就唔好再用『有請舉例,無請證明。』..
就用『有無都要證明』吧啦...!!

只要你問題清晰...就不會有誤會吧...

2009-02-12 19:15:40 補充：
補充內容字數有限...
最後我也想賜教一下...何為『有請舉例,無請證明。』^﹏^v

2m^2 + 2m = n^2+n => 2m^2 + 2m - (n^2 +n) = 0

由於m是整數，由判別式可得 2n^2 + 2n + 1 = k^2

或 (n+1)^2 + n^2 = k^2，因(n+1,n)=1

=> n+1 = a^2-b^2 , n=2ab or n = a^2-b^2 , n+1=2ab

=> ±1 = (a-b)^2 - 2b^2

let (x,y) is all solution of Pell's equation for ±1

then b=y,a=x+y is the solution

另外還有 14x15x2 = 20x21.

2009-02-07 12:59:58 補充：
Suppose 2n(n+1) = m(m+1).
Note that n(n+1) = 2(1+...+n), so we have
4(1+...+n) = 2(1+...+m)
2(1+...+n) = 1+...+m.
Obviously, n =< m. Let m=n+k. Then we have
1+...+n = (n+1) + ... + (n+k) = kn + (1+...+k)
n(n+1)/2 = kn + k(k+1)/2
n^2 + n = 2kn + k^2 + k
n^2 - (2k-1)n - (k^2+k) = 0.
View this as a quadratic equation in n. Since n is an integer, the discriminant must be a perfect square. So
(2k-1)^2 + 4(k^2+k) = h^2 for some integer h.
h^2 - 8k^2 = 1.
This type of equation is called Pell's equation. We have one set of solution (h1,k1)=(3,1). (This pair is called the fundamental solution.) Then we have infinitely many pairs of solution given by the formula
(hi + ki √8) = (h1 + k1 √8)^i,
or equivalently,
h_(i+1) = h1hi + 8k1ki
k_(i+1) = h1ki + k1hi.
For example, we have (h2,k2)=(17,6).
Also, there is trivial solutions (h,k) = (1,0)
For each pair of solutions (h,k), we get two values of n. (One positive and one negative.) These are all the possible cases.

2009-02-07 13:07:22 補充：
For example, for the case when k=6, we get n=14 or -3.
For n=14, we have 2x14x15=20x21
For n=-3, we have 2x(-3)x(-2) = 3x4.

2009-02-07 13:08:55 補充：
In my proof, I have assumed n,m are positive. But it is easy to see that the case when n,m are negative can be reduced to the positive cases.

Yes, I saw the wrong question... i mixed up the "x" with "+"....