哥德巴哈猜想 Goldbach Conjecture

2008-02-23 4:27 pm

個猜想人人都明: "每個大於10的偶數都是2個質數相加的和."
這雖然不是六大未解決難題之一, 但有著360 年歷史, 卻是數學界的一個重要難題.
當然, 就會有很多人去嘗試証明.
其中一個較短較淺的証明如下:
http://arxiv.org/vc/arxiv/papers/0712/0712.2381v10.pdf
問: 証明那裏出錯了?

更新1:

這個証明很容易看懂, 5.5頁長, 但有幾幅圖+double line spacing, 所以實際長度=3頁...... 比佢証明到, 都係個句: 數學界可以摺埋.

更新2:

你的解答未如滿意, 請補充

更新3:

Andrew: good observation, but in his definition, he didn't say "every even number m", only "for some m", i.e. first he only consider the m such that p0 and pn are both prime.

回答 (2)

Andrew

2008-02-28 2:56 am

✔ 最佳答案

看來「好怕Homework」冇仔細睇個proof.
其實個proof係Page 4, Def 1.1 就已經唔work.

The last prime in the Goldbach block pn = m - p0
根本唔係一個prime

楝一個例子一個even number 減 3唔係prime非常容易, 例如12就唔得。

2008-02-29 00:47:16 補充：
Eq 2.0 都唔掂啦，都冇define咩係p_(n+1)

2008-02-29 01:10:41 補充：
The true problem is Corollary 1.1.

The statement is for any prime p_a, r_a, r_t, p_q.
If k-(p_a + r_a + r_t) = p_q, then p_q + (k - p_q) is a goldbach partition.

2008-02-29 01:10:46 補充：
However, the proof requires p_a + r_a + r_t to be a prime number to start with, which is not stated as an assumption for the corollary.

Check the usage on the last page, the use of corollary do not check the unstated assumption as well.

參考: 從不抄襲。

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myisland8132

2008-02-29 12:55 am

我表面看上去這個proof是倒果為因。他能夠找到個模式是因為哥德巴哈猜想好可能是對的﹐而他就倒過來說可以用這個方法找到每一個偶數的2個對應質數。andrew大概講中了重點

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