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Topic: Twin prime conjecture restated without reference to primes
Replies: 6   Last Post: Aug 18, 2008 12:16 AM

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 Paul A. Tanner III Posts: 5,920 Registered: 12/6/04
Re: Twin prime conjecture restated without reference to primes
Posted: Sep 1, 2003 6:54 PM

> In the mid-nineties, I stumbled across a trivial way of restating the
> twin prime conjecture without reference to primes. I've sat on it
> since it's so trivial, but found that others thought it a fun bit of
> trivia when I shared it. Perhaps the following has been shared many
> times before, but I've not seen it elsewhere. So, thinking that some
> out there might find it an enjoyable bit of trivia on the side, here
> it is:
>
> To eliminate all positive c such that 6c-1 or 6c+1 is a composite,
> take the cross product of the set of all positive integers of the form
> (6a+-1) with itself: (6a+-1)(6b+-1) = 6(6ab+-a+-b)+-1 = 6c+-1 for all
> positive a,b and for all 4 combinations of "+" and "-"(only 3 are
> necessary, obviously).
>
> So the twin prime conjecture is true if and only if C = {c | 6ab+-a+-b
> = c for some a,b} leaves infinitely many positive integers uncovered.
>
> For economy, let C = {6ab+-a+-b), and so let R = {6ab-a-b}, S = S_1 =
> {6ab-a+b} = S_2 = {6ab+a-b}, and T = {6ab+a+b}. Although S = S_1 =
> S_2, I note both because of the matrical representations below, where
> S_1 and S_2 are the transposes of each other:
>
> C forms an infinite symmetric matrix M (that is actually just a
> modified subset of the "infinite multiplication table" on the positive
> integers), where each row or column is an ordered subset of the
> residue classes +-a modulo 6a-1 or modulo 6a+1:
>
> 4 6 9 11 14 16 . . .
>
> 6 8 13 15 20 22 . . .
>
> 9 13 20 24 31 35 . . .
>
> 11 15 24 28 37 41 . . .
>
> 14 20 31 37 48 54 . . .
>
> 16 22 35 41 54 60 . . .
>
> . . . . . .
> . . . . . .
> . . . . . .
>
> So the twin prime conjecture is true if and only if M leaves
> infinitely many positive integers uncovered.
>
>

Hindsight really is 20/20, is it not? Although the context makes it
clear for some, I should have started my original post with:

"All variables are positive integers."

Anyway, below is a little extra bit of trivia that I stumbled across
also in the mid-nineties along with the material in my original post,
since it's similar to that material. Some have found it a bit of fun
on the side, and I hope that you do, too:

Theorem.

Where a and b are positive integers with b > a:

1) Given prime p = 6a-1 and composite q = 6b-1: p | q if and only if p
| b-a
2) Given prime p = 6a-1 and composite q = 6b+1: p | q if and only if p
| b+a
3) Given prime p = 6a+1 and composite q = 6b-1: p | q if and only if p
| b+a
4) Given prime p = 6a+1 and composite q = 6b+1: p | q if and only if p
| b-a

Proof.

Where n is some positive integer:

1) [p | q] <=> [(6a-1)(6n+1) = (6b-1)] <=> [6(6an-n+a)-1 = 6b-1] <=>
[6an-n+a = b] <=> [(6a-1)n = b-a] <=> [p | b-a]
2) [p | q] <=> [(6a-1)(6n-1) = (6b+1)] <=> [6(6an-n-a)+1 = 6b+1] <=>
[6an-n-a = b] <=> [(6a-1)n = b+a] <=> [p | b+a]
3) [p | q] <=> [(6a+1)(6n-1) = (6b-1)] <=> [6(6an+n-a)-1 = 6b-1] <=>
[6an+n-a = b] <=> [(6a+1)n = b+a] <=> [p | b+a]
4) [p | q] <=> [(6a+1)(6n+1) = (6b+1)] <=> [6(6an+n+a)+1 = 6b+1] <=>
[6an+n+a = b] <=> [(6a+1)n = b-a] <=> [p | b-a]

Paul

Date Subject Author
9/1/03 Paul A. Tanner III
9/1/03 Paul A. Tanner III
8/15/08 LorenP
8/15/08 b92057@yahoo.com
8/17/08 amy666
8/18/08 b92057@yahoo.com
9/2/03 Asger Grunnet