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Let us observe the following pattern

$N - p_1 = m_1, N - p_2 = m_2, \ldots , N - p_r = m_r$; take $p_1 = 3$ and $p_2 = 5,\ldots$ notice that $p_r$ is the larger prime less than or equal to square root of $N$. Here $N$ is an even number and large, where as $m_1,\ldots,m_r$ are composite numbers. What I guessed, with reference to above data all even numbers smaller than $N$ can be written by two primes and then we assume that all $m_1,\ldots,m_r$ are composite and we will face an absurdity (contradiction). Can we generalize my guess, if I am not wrong.

As per the request, I am adding more information on this post.

In my previous post I have discussed the same. But, Once again I am giving below.

Consider $m_1$...$m_r$ less than or equal to $m_1$.$a_1$.$a_2$...up to a_(r-1), where $a_1$ to a_r-1 are composite numbers smaller than $m_1$ and may be equal or larger than $m_2$ to $m_r$ by this information we conclude the following.

1) Take prime factors of K and restrict the value of K, where K belongs to [3, square root of $m_1$] will appeared in {$m_1$,$a_1$,$a_2$,...up to a_ r-1}.

2) If L to the member of K belongs to [3, square root of N] are in equation N - $p_i$ less than or equal to tK, where i belongs to [1, r] and L = r/K or < r/K, for all K belongs to [3, square root of N]

3) If M to be the number of S > square root of N are in N - $p_i$ less than or equal to $S_q$ that these numbers are odd and i belongs to [1, r].

4) also, we always have; log (N-3) ...(N-$p_r$) > r log(N-$p_r$)

I hope by the above in formation we can conclude the goldbach conjecture, if I am not wrong..

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Could you please clearly state what is what, then clearly state your guess/conjecture/question? It is impossible to tell what your question is. –  Alex B. Dec 1 '11 at 16:29
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@vidya, Does this account belong to you as well? Perhaps you should consider registering on the site so as to keep track of your questions and not lose the ability to comment on your posts. Also you could flag the moderators to merge the two accounts. –  Srivatsan Dec 1 '11 at 16:31
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closed as not a real question by t.b., Srivatsan, Quixotic, Sasha, Asaf Karagila Dec 3 '11 at 9:38

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

up vote 1 down vote accepted

Your "guess" may or may not be Goldbach's conjecture, which states that every even $n\ge4$ can be written as a sum of two primes. Everyone believes that the contradiction you propose is a certainty, but nobody knows how to prove it - and this is not for want of trying, as everything from the most elementary to the most advanced techniques have been brought to bear on the problem.

If I have misunderstood your question, I apologize, and I hope you will clarify your question.

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