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Re: Vindication of Goldbach's conjecture
Posted:
Jun 21, 2012 1:50 PM
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El jueves, 14 de junio de 2012 10:24:12 UTC-4:30, Count Dracula escribió:
> On Jun 14, 5:28 am, mluttgens <lutt...@gmail.com> wrote:
> > In P, the number 26 = 3+23, 5+21, 7+19, 9+17, 11+15, 13+13, 15+11,
> > 17+9, 19+7, 21+5 and 23+3, all the terms coming from the successive
> > uneven numbers of S and S' from 3 to 23, where 23 = 26 - 3.
> > As S and S' contain all the prime numbers inferior or equal to 26-3 ,
> > the even number 26 is necessarily the sum of two primes.
> > Let?s note that it is not necessary to indentify those primes. Knowing
> > that they exist in S and S? is enough..
>
> The argument you give here is fallacious. The sets S and S' also
> contain odd
> numbers that are not prime. This makes it necessary to identify the
> particular
> primes and it is not enough to know that primes exist in S and S'. In
> your
> example for 26, it is not possible to use the sums 5+21, 9+17, 11+15,
> 15+11,
> 17+9, 21+5 since each contains a composite integer. My earlier
> question to you was: what if there is an integer for which _all_ the
> sums
> are inadmissable, not just some as in the case of 26? Nothing in your
> argument rules out this possibility.
>
> If a counterexample exists it will have to be larger that 10^17
> because
> computer verifications have been performed up to about that value. A
> correct
> proof is highly likely to require some highly technical innovations.
>
> Count Dracula
The true difficult with Goldbach's Conjecture is this:
Given a even number 2k to warrant that the pairing of the
two arithmetic progressions 2n + 1 and 2k - (2n+1) will produce
nor less than one coincidence of two primes.
Example.- Given the even number 98. The two progressions are:
3, 5, 7, 9, 11, 13, 15, 17,.........95
95, 93, 91, 89, 87, 85, 83, 81,......... 3
Who guarantee that there will be a column with two primes?
Ludovicus
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