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After some research, I have discovered that proving the statement;

There exist an infinite number of positive integers K such that;

$K \neq 6ab \pm a \pm b$ and $K \neq 6ab \mp a \pm b$

is equivalent to proving the twin prime conjecture.

I have naively attempted to think of ways in which one could use this simple fact and hence prove the twin prime conjecture. This is not an attempted proof. This is simply a few simple ideas that I have that, I would like to discuss with mathematicians, so that they may then point out the fundamental flaw in my ideas.

Assume that there exists a finite set of integers not expressible in the forms;

$ 6ab \pm a \pm b$ and $6ab \mp a \pm b$

Let K be the largest integer in this set;

Hence $K \neq 6ab \pm a \pm b$ , $K \neq 6ab \mp a \pm b$

It can be easily proven that $6k-1$ and $6k+1$ will be a twin prime pair.

Now consider all primes P, such that $P>6k+1$

It has also been proven that $6K+1$ and $6K-1$ are both primes if and only if;

$K \neq 6ab \pm a \pm b$ and $K \neq 6ab \mp a \pm b$

Hence all primes $P$ where $P>6K+1$ cannot belong to a twin prime pair, and so are of the form.

$6(6ab+a+b)-1$, $6(6ab-a-b)-1$, $6(6ab-a+b)+1$

where

$6ab \pm a \pm b \neq 6pq \pm p \mp q$

and

$6ab+a-b\neq 6pq \pm p \pm q$

Let N represent those positive integers expressible in the forms;

$ 6ab \pm a \pm b$ and $6ab \mp a \pm b$

Assume that 6N+1 is prime. We know that there are an infinite number of primes, therefore there exists some difference Q , where Q is a positive integer, until we reach the next prime.

Hence let $6N+1$ and $6N+Q$ both be prime numbers.

$Q \equiv 1,5,-1,-5mod6$

We can easily prove that there exists an infinite number of primes of the form $6M\pm Q$, so I will not include the proof.

Now we must evaluate the forms of N and M that enable $6N\pm 1$ and $6M\pm Q$ to both be prime.

$(6N+1)(6M+Q)=36NM+6NQ+6M+Q=6(6NM+NQ+M)+Q$

$(6N+1)(6M-Q)=36NM+6M-6NQ-Q=6(6NM+M-NQ)-Q$

$(6N-1)(6M+Q)=36NM-6M+6NQ-Q=6(6NM-M+NQ)-Q$

$(6N-1)(6M-Q)=36NM-6M-6NQ+Q=6(6NM-M-NQ)+Q$

Hence if

$K \neq 6NM \pm NQ \pm M$ and $K \neq 6NM \pm NQ \mp M$

Then $6K\pm 1$ and $6K\pm Q$ will both be prime.

Of course the case when $Q=-1$ is the twin prime conjecture. But we know there must exist an infite number of integer not expressible in the forms;

$K \neq 6NM \pm NQ \pm M$ and $K \neq 6NM \pm NQ \mp M$

I think this may somehow imply that there cannot exist an infinite consecutive sequence of integers expressible in the forms;

$6ab \pm a \pm b$ and $K \neq 6ab \mp a \pm b$

since amogst this sequence lies those integers not expressible in the form;

$6NM \pm NQ \pm M$ , $6NM \pm NQ \mp M$ , but I am quite sure that there is a mistake somewhere, I am just curious as to where or what it is?

Edit: I have a few ideas how to finish off the proof, but I would just like some feedback as it is 99.99999% likely to be wrong.

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4  
It's good that you discovered on your own the connection between $6ab+a+b$ and twin primes. Many others have discovered it before you, but you discovered it on your own, and that's good. No one has been able to make the slightest use of it in terms of settling the twin prime conjecture, and you won't either, as long as you limit yourself to highschool algebra and elementary manipulations of congruences --- if there were a proof like that, Euler would have published it 250 years ago. Don't let that stop you from enjoying what you're doing. –  Gerry Myerson Mar 4 at 9:59
    
Is there anything particularly wrong with the idea ? –  user129967 Mar 4 at 10:13
    
What idea? There is definitely something wrong with trying to settle the twin prime problem with what you learn in the first week of an intro Number Theory course. But finding what in particular is wrong with any given argument is the job of the person putting forward the argument. –  Gerry Myerson Mar 4 at 10:18
    
I was just wondering that, as I lack experience in this field, is there any evident flaws that are obvious, but I cannot see? I am not insulting anyone here, its just an idea that is probably not original –  user129967 Mar 4 at 10:20
6  
Life is too short to find flaws in non-proofs of notorious conjectures. –  Gerry Myerson Mar 4 at 10:22

1 Answer 1

The argument uses a fixed finite number of congruence conditions and nothing else. This is fatal to its chances of success no matter the number of additional cases and sub-cases that might be considered.

All that can be achieved by such argument is to give an explicit finite collection of arithmetic progressions whose union contains all (sufficiently large) $p$ for which $p$ and $p+2$ are both prime. Those arithmetic progressions contain a nonzero proportion of the integers and of the primes, but the twin primes have density $0$ in the primes, which have density $0$ in the integers. As a consequence, the necessary congruence conditions are almost never sufficient to force an integer to be prime, or a prime number to be a twin prime; they are close to being irrelevant to the conjecture, and completely different ideas would be needed to solve it.

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