# Does proving the following statement equate to proving the twin prime conjecture?

After some research, I found that it has been supposedly proven, that proving that there exists an infinite number of positive integers K such that;

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

implies that the twin prime conjecture is true. My sources include the following;

http://arxiv.org/abs/1106.6050 (I do not know if this particular proof is correct)

If this is in fact true, why have mathematicians abandoned this approach?

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A simpler statement is that for $a,b\in\Bbb Z\setminus\{0\}$ the equation is $6|ab|+a+b$; this was proved by Maria Suzuki and documented in the January 2000 edition of American Mathematical Monthly. – abiessu Mar 3 '14 at 20:14
So why have mathematicians given up on this approach or is it being used in more complicated ways? – user129967 Mar 3 '14 at 20:16
The approach may or may not have been abandoned in general, but personally I stopped following it after several years of unsuccessful attempts to use what is essentially a sieve mechanism. I believe it has been stated somewhere that a sieve mechanism has never proven an infinity; other methods prove the infinity, and the sieve mechanism is used for other purposes. – abiessu Mar 3 '14 at 20:17
At first glance, because this isn't an 'approach'; it's a nearly-trivial equivalency. Saying these things is exactly equivalent to saying that there are no $a$ or $b$ such that $6K\pm 1=(6a\pm 1)(6b\pm 1)$, which is precisely the statement that $6K\pm 1$ are prime. In short, it's unlikely that there's insight to be gained from this approach just because the 'translation' back and forth is so easy. – Steven Stadnicki Mar 3 '14 at 20:19
Have they proven that a sieve mechanism cannot prove a statement regarding infinity, or is it just a feeling? – user129967 Mar 3 '14 at 20:20

Consider that $6k-1,6k+1$ are both prime exactly when there exists no positive integer $a$ such that $6a\pm 1\mid 6k-1, 6a\pm 1\mid 6k+1$.

Since the congruence classes which are relatively prime to the modulus $6$ are $\pm 1$, the only possible numbers which could divide $6k\pm1$ are numbers of the form $6a\pm 1$, and therefore the above divisibility tests turn into:

$$(6a-1)(6b-1)= 6k+1\\ (6a-1)(6b+1)= 6k-1\\ (6a+1)(6b+1)= 6k+1$$

The $(6a+1)(6b-1)$ is skipped as a duplicate case.

These equations can then be reduced to

$$6ab-a-b=k\\6ab-a+b=k\\6ab+a+b=k$$

This set of equations is then equivalent to $6ab\pm a\pm b=k$ for $a,b\in\Bbb Z^+$ or $6|ab|+a+b$ for $a,b\in\Bbb Z\setminus\{0\}$.

All of this is simply re-expressing the divisibility statement as an equation. The study of such equations is, of course, interesting in its own right, but even proving the infinity of the set of primes is difficult (if not impossible) using the equations $k=ab+a+b$ or $k=2ab+a+b$.

It is interesting to note that the equations $k=nab+a+b$ and $k=2nab+a+b$ are directly related, but that relationship is strictly limited to the $2$ multiplier for this form of equation.

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But by proving that there cannot exist an infinite sequence of consecutive integers of this form, would that not prove the twin prime conjecture. – user129967 Mar 3 '14 at 20:36
Yes, that is what I mean by "re-expressing the divisibility statement as an equation." – abiessu Mar 3 '14 at 20:37
Ok thanks, just wanted to make sure – user129967 Mar 3 '14 at 20:38