Re: Goldbach Idea?
From: David Bandel (dwb1729_at_yahoo.com)
Date: 09/21/04
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Date: 20 Sep 2004 22:03:34 -0700
davecornwell@comcast.net (david cornwell) wrote in message news:<40e21c11.0409201052.2b43b19a@posting.google.com>...
> Could someone explain why the following proof idea would be difficult
> please?
>
> 1. Let S={i: i an integer in [1,2n]}
> 2. Let P={p_j: p_j is prime and p_j <=Sqrt(2n) for j=1..J}
> 3. Sieve S using P as follows:
> delete 1 and 2n-1 from S and delete all multiples of p_j in S,
> including p_j itself to give S1.
>
> At this point S1 contains only the prime numbers in the interval
> [Sqrt(2n),2n-2]
> since all multiples of primes <=Sqrt(2n) have been sieved from this
> interval.
>
> Now the primes that are left are not all Goldbach primes i.e. not all
> of the primes P_a in S1 have a partner P_b so that P_a + P_b = 2n.
>
> We want to form a set containing only Goldbach primes. To do this, for
> each x that was sieved in step 3, delete its partner 2n-x from S1. If
> x=p_j for some j then we have sieved out a possible Goldbach pair but
> we accept this possibility. We only want Goldbach prime pairs in the
> interval [Sqrt(2n),2n]. If x is composite, then we need to sieve out
> 2n-x in S1 because the pair (x,2n-x) cannot be a Goldbach prime. Let
> the resulting set be S2.
>
> If we could show S2 was non-empty for all n>N for some large N then we
> have proven the Goldbach Conjecture. Why would this be difficult since
> we have just used sieving methods? Or perhaps S2 can be empty?
amounts to nothing but a brute force search that proves goldbach on a
case by case basis.. nothing interesting or useful here.
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