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Yet another Goldbach proof

  1. Aug 31, 2010 #1
    My name is Todd and I have a posted a proof of Goldbach's Conjecture.

    I have been looking around for a while to find a place to post it and this seems like a reasonable place to start. I have stopped by here before but never registered.

    The proof is http://www.just-got-lucky.com/gc-083110-draft1.pdf" [Broken].

    The proof is somewhat geometric, wolfram NKS-like, and does not use much advanced math per se so the concepts are much different than you usually see posted here.

    Perhaps someone will look - I understand that I am both new here and probably a crackpot but we all have to do what we feel we need to do.

    So if there is a horrific glaring error please be at least somewhat kind.

    The basic proof is fairly simple in concept:

    I create a diagonal "table" where the diagonal is odd primes > 3 and under the diagonal all even numbers appear at the intersection of rows and columns.

    I then derive a "counter diagonal" that shows, for a given even number, all the unique odd number pairs that can generate it.

    I then eliminate all the non-prime odd pairs from this table and show that, with what's left, you basically have to violate Betrand's Postulate in order to contradict the conjecture.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 31, 2010 #2
    Here's the gap in your logic:

    "As we have said, the presence of the column component of strikes are insufficient alone for eliminating a counter-diagonal. However, it may be possible
    that strikes for consecutive odd non primes could perform this function, i.e. given some odd prime or non prime p with a diagonal of length k we find m
    consecutive odd non primes along the odd diagonal such that m > k."

    All you need is to find enough non-primes between p and 2p to eliminate horizontally all points along the diagonal that were not already eliminated vertically. You don't need ALL of those numbers to be non-prime.
     
  4. Aug 31, 2010 #3

    Office_Shredder

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    The prime that you found with Bertrand's postulate must intersect one of the non-struck columns, but that doesn't mean that where they intersect is going to be on the counter-diagonal that you are trying to prove still exists
     
  5. Aug 31, 2010 #4
    I believe it must because vertical columns under 3, for example, would not be covered for an intersection.
     
  6. Aug 31, 2010 #5

    Office_Shredder

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    I don't understand what you mean by this

    Let's walk through an example of your argument:

    The counter-diagonal for 28 for example. We have left the numbers 13,11,7,5 and 3. We use Bertrand's Postulate to find a prime between 28 and 40. We find the number 19 (there are others, but this just happens to be the one that we pick, since all we can do is pick one). But 19+9=28 and 9 is not a prime. The row for 19 hits the columns for 13, 11, 7, 5 and 3 but not on the counter-diagonal we were interested in
     
  7. Sep 1, 2010 #6
    I had to run out yesterday but I gave your comment some thought.

    Yes - I thought about this and you are correct - its wrong in this regard.

    EDIT - I guess I am a little further because now I know there is a prime - I just can't prove it lines up...

    So the bottom line is this then -

    1) there are only certain spots along the counter-diagonal where a prime can reside - this is easy to show.

    2) If there is a single prime then the GC says it must be at a row position where the intersection is also prime.

    3) If there are multiple primes then GC says at least one must be at a row position where the intersection is also prime.
     
    Last edited: Sep 1, 2010
  8. Sep 2, 2010 #7
    To Office_Shredder -

    What my mistake does say is this:

    If the Goldbach Conjecture is false then it would be because the Bertrand prime would intersect on a column strike.

    The column strike always represents an odd multiple of a prime.

    So for primes p and q such that p + q = 2n we know that

    q >= n
    p = some odd x (let's call it Osubx) times p' where p' is a prime
    p' <= n

    So for GC to be true Osubx = 1 for all n, otherwise there exists at least one prime where Osubx > 1.
     
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