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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
Twin prime conjecture restated without reference to primes
Posted: Sep 1, 2003 12:25 PM
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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.

Notice that M can be broken up into four matrices:

R forms the symmetric matrix:

4 9 14 . . .

9 20 31 . . .

14 31 48 . . .

. . .
. . .
. . .

S_1 forms:

6 11 16 . . .

13 24 35 . . .

20 37 54 . . .

. . .
. . .
. . .

S_2 forms:

6 13 20 . . .

11 24 37 . . .

16 35 54 . . .

. . .
. . .
. . .

T forms the symmetric matrix:

8 15 22 . . .

15 28 41 . . .

22 41 60 . . .

. . .
. . .
. . .

Since we've already restated the twin prime conjecture without
reference to primes, we'll refer briefly here to a theorem on primes.
Since these sets R, S, and T represent composites of the form 6c-1 or
6c+1, by Dirichlet's Theorem (remember, 6c-1 = 6(c-1)+5), we know the
following:

R and T each leave infinitely many positive integers uncovered, and we
know that R U T, the union of R and T, also leaves infinitely many
positive integers uncovered. We know that S leaves infinitely positive
integers uncovered.

The trick now is to see whether (R U T) U S, the union of (R and S)
and T, leaves infinitely many positive integers uncovered.

Anyone care to try? Have fun. I'm going sunset watching and classical
music listening. (Some like fishing, some don't.)

Paul




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