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Topic: Vindication of Goldbach's conjecture
Replies: 74   Last Post: Aug 9, 2012 6:50 PM

 Messages: [ Previous | Next ]
 mluttgens Posts: 80 Registered: 3/3/11
Re: Vindication of Goldbach's conjecture
Posted: Jun 21, 2012 6:32 PM

On 21 juin, 19:50, luir...@yahoo.com wrote:
> 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
>
>

All you need is a little software, which will give you all the
Goldbach's pairs.

Marcel Luttgens

Date Subject Author
6/6/12 mluttgens
6/6/12 Brian Q. Hutchings
6/6/12 GEIvey
6/7/12 Richard Tobin
6/8/12 mluttgens
6/8/12 Count Dracula
6/9/12 mluttgens
6/9/12 Brian Q. Hutchings
6/9/12 mluttgens
6/25/12 GEIvey
6/9/12 Richard Tobin
6/9/12 mluttgens
6/9/12 Richard Tobin
6/9/12 Brian Q. Hutchings
6/9/12 Brian Q. Hutchings
6/14/12 mluttgens
6/16/12 mluttgens
6/16/12 Frederick Williams
6/20/12 mluttgens
6/20/12 Rick Decker
6/21/12 mluttgens
6/21/12 Frederick Williams
6/21/12 mluttgens
6/22/12 mluttgens
6/22/12 mluttgens
6/22/12 Brian Q. Hutchings
6/25/12 Michael Stemper
6/26/12 mluttgens
6/26/12 Frederick Williams
6/28/12 Michael Stemper
7/19/12 mluttgens
7/19/12 Timothy Murphy
7/19/12 mluttgens
7/19/12 Gus Gassmann
7/20/12 mluttgens
8/1/12 Tim Little
8/4/12 mluttgens
8/4/12 Frederick Williams
8/6/12 mluttgens
8/6/12 gus gassmann
8/6/12 Brian Q. Hutchings
8/9/12 Pubkeybreaker
7/19/12 J. Antonio Perez M.
7/20/12 mluttgens
6/25/12 Michael Stemper
6/25/12 Thomas Nordhaus
6/17/12 mluttgens
6/17/12 quasi
6/18/12 Count Dracula
6/18/12 quasi
6/19/12 Count Dracula
6/19/12 quasi
6/20/12 mluttgens
6/22/12 Michael Stemper
6/22/12 mluttgens
6/22/12 Robin Chapman
6/22/12 Michael Stemper
6/23/12 mluttgens
6/22/12 Richard Tobin
6/22/12 Richard Tobin
6/25/12 Richard Tobin
6/25/12 Michael Stemper
6/14/12 Count Dracula
6/21/12 Luis A. Rodriguez
6/21/12 Brian Q. Hutchings
6/21/12 mluttgens
6/25/12 GEIvey
6/20/12 J. Antonio Perez M.