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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 16, 2012 12:08 PM

On 14 juin, 16:24, Gus Gassmann <horand.gassm...@googlemail.com>
wrote:
> On Jun 14, 6:28 am, mluttgens <lutt...@gmail.com> wrote:
>
>
>
>
>

> > On 10 juin, 00:20, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:
>
> > > In article <0b8c04e9-f6c1-48c0-8f63-c115150dd...@f30g2000vbz.googlegroups.com>,
>
> > > mluttgens  <lutt...@gmail.com> wrote:
> > > >[deleted]
>
> > > You refuse to address the question; I conclude you are a troll.
>
> > > -- Richard
>
> > My response is contained in the following new version:
>
> > Vindication of Goldbach's conjecture
> > ______________________________

>
> > According to the conjecture, every even integer greater than 4 can be
> > expressed as the sum of two primes.

>
> > Let's consider two identical infinite series S and S' of uneven
> > numbers, where S and S' = 3, 5, 7, 9, 11, 13, etc...
> > In other words, S(n) = S'(n) = 2n +1, with n beginning at 1.
> > Let?s note that S and S? contain only natural numbers.

>
> > The fact that some uneven numbers contained in the series S and S' can
> > be
> > the result of mathematical operations, for instance exponentiation, is
> > thus irrelevant.

>
> > By adding successively each number of S to all the numbers of S', one
> > gets a series P of even numbers.

>
> > Example:
> > _______

>
> > n     S    S'
> > _     _    _

>
> > 1     3    3
> > 2     5    5
> > 3     7    7
> > 4     9    9
> > 5    11   11
> > 6    13   13
> > 7    15   15
> > 8    17   17
> > 9    19   19
> > 10   21   21
> > 11   23   23

>
> > By adding, one gets the following even numbers of series P:
>
> > From S = 3:            6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26
> > From S = 5:            8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28
> > From S = 7:          10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
> > From S = 9:          12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32
> > From S = 11:        14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34
> > From S = 13:        16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36
> > From S = 15:        18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38
> > From S = 17:        20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40
> > From S = 19:        22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42
> > From S = 21:        24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44
> > From S = 23:        26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46

>
> > Thus, series P contains all the even numbers from 6 to 46.
>
> > 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..
> > This applies mutatis mutandi to any even integer

>
> > Conclusion:
>
> > Every even integer greater than 4 can be expressed as the sum of two
> > primes.
> > In other words, the above procedure vindicates Goldbach's conjecture.

>
> It appears you do not even understand the problem. How do you *KNOW*
> that you will never miss an even number using this schema? It is fine
> and good to verify that every even integer less than 47 can be written
> as a sum of two primes, but what about 83749105938571693046? Your
> method says nothing about that number.

Bigger is the number, and more primes it contains.
Hence, the number of possible sums of two primes corresponding to a
number
is proportional to the magnitude of the number.

Thus, it would be ridiculous to try a number as big as
83749105938571693046

Here is an example where N is greater then 26:

Example: N = 138 and N-3 = 135
______________________________

The relevant primes p are

3 5 7 11 13 17 19 23 29
31 37 41 43 47 53 59 61 67 71
73 79 83 89 97 101 103 107 109 113
127 131

N-p =

135 133 131 127 125 121 119 115 109
107 101 97 95 91 85 79 77 71 67
65 59 55 49 41 37 35 31 29 25
11 7

Thus N = 5+133, 7+131, 11+127, 29+109, 31+107, etc...

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.