[ExI] Goldbach Conjecture

Will Steinberg asyluman at gmail.com
Sat Nov 28 02:30:09 UTC 2009

I think I may have ended up merely restating something, but I think
approaching the problem in a new paradigm is important--namely, instead of
having each even number expressed as the sum of two primes, have each number
n have 2 distinct primes an equal distance from it (because if Px + Py = 2n,
then n - m = Px and n + m = Py.  So I wrote out the numbers from a to a
number 2n-1 (say 9) then wrote the numbers 9 to 1 below them, so each column
sums to 2n (10).

e.g. 1 2 3 4 5 6 7 8 9
      9 8 7 6 5 4 3 2 1

  Then I ran a sieve of eratosthenes each way.  The "holes" after a forwards
and backwards run-through reveal the sets of primes which are equidistant
from n and thus sum to 2n, or every even number.

Now, if we know there is one prime left unscathed by our doublesieve, it
must have a mirror partner (also prime because of our sieving).  And we KNOW
that there is at least one prime p between n and 2n given Bertrand's
postulate.  Since this is a mirror, we can consider only one half of the
line, from 1 to n.  Now, we have already marked off the composites given by
the sieve.  But now lots are recast and even some primes must be lost.
These are determined by taking whatever composite numbers are between n and
2n and subtracting them from 2n.  Now we have a set that looks like this:
{composites between 1 and n, 2n-composites between n and 2n}.

Visually, what we will have is a symmetrical pattern: two sieves laid on top
of each other oppositeways.  Each hole will have its conjugate and so the
same for each filled in space.  In essence, if we can prove that that
pattern will always contain holes, then there will always be 2 primes able
to add up to 2n.

I think the answer may lie with some sort of modulus thing.  If you look at
the center n, adding up prime numbers (to produce composites) will often lay
a stretch of number over n.  For example. the stretch of 9-12 lies over 10,
with 1 on the right and 2 on the left.  Consequently, all subsequent
additions of 3 will give 10+2+3k.  This means that the spaces with numbers
that equal 10-2-3k will be filled in on the right half, as well as 10-4-7k,
et cetera.  I think the actual answer lies near and I would love for someone
to give me insight (or tell me this has all already been done)

On Fri, Nov 27, 2009 at 6:59 PM, spike <spike66 at att.net> wrote:

> On Behalf Of Will Steinberg
>        Subject: [ExI] Goldbach Conjecture
>         I think I found something really good about the Goldbach
> Conjecture; does anyone have a background in number theory or know somebody
> who does?  It is important
>        -Will
> Hi Will!  I am a Goldbach conjecture fan.  I know there are other
> Goldbachers here, Lee Corbin is one of our math superbrains, and several
> others too.  What did you find?
> spike
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