[ExI] Goldbach Conjecture

[Will Steinberg]

I think I may really have got it now.

Since we can represent every situation like this by a doublesieve, we need
only prove all doublesieves will have holes in them.  Here is how I do
this.  A sieve of length n--let's say 16--has a set of spaces A that are at
2k, 3k, 5k...mk where k>1 and mk<n.  It also has a set of spaces in the
opposite manner, which can be given by nmod2+2k, nmod3+3k...nmodm+mk where
once again k>1 and mk<n.  All we need to prove is one necessary hole between
those spaces.  It is given thus:  Choose a prime in the set that is under
any 2k (let's choose 5) that has the property n mod p =1. We cannot create 5
from our 2k, 3k,spaces, because the only one to do so would be 5k, and k
must be greater than 2.  So we must find a nmodp+pk that is equal to 5, k of
course being greater than 2 again.  But alas!  We find that the only way to
have nmodp+pk=5 be if p was less than 5.  so we have nmod2+2k which yields 4
and nmod3+3k which yields a minimum of seven!  The point is here:  Construct
the sieve for any length.  Choose a low prime.  As long as none of the set B
spaces are equal to it, we have a hole.  The nmodp+pk must be equal to our P
for the whole to be covered.   But there is always a P inexpressible by
nmodp+pk--simply choose a low p with a modulus of 1 for n.  then, if nmodp
is not 1, nmodp+pk is greater than P.  But if nmodp IS one, nmodp+pk can
never equal our P.


On Fri, Nov 27, 2009 at 9:30 PM, Will Steinberg <asyluman at gmail.com> wrote:

> 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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>
>
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