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?
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.
I haven't abandoned this approach to the problem. In fact, one of my curiosities has become applying more advanced methodologies to exploring this path towards a solution. In order to do so, we study sets of similar form to the numbers that fail to include each value of $k$ for which both $6k+1$ and $6k-1$ are prime, and for readability I will use a similar notation in my response.
One of the arguments is that all four forms of $k \neq 6ab \pm a \pm b$ must be acknowledged -- and the reason is not so simple, but still important. Note that the residue classes formed below with composite non-prime moduli are redundant in the set excluding $k$.
Letting $[a]^{**}_m = [a]_m \cap [a+m,\infty)$ be a residue class endowed with the infimum where $[a]_m = \{a + nm\mid a,n,m \in \mathbb{Z} \} = \{a \pmod{m}\}$ is a residue class and converting each representation of sets excluding $k$, it can be verified for $a,b \in \mathbb{N}_{\ge1}$ that
$k \neq 6ab + a + b \implies k \not\equiv b \pmod{6b+1} \implies k \not \in \bigcup[b]^*_{(6b+1)}$
$k \neq 6ab - a + b \implies k \not\equiv b \pmod{6b-1} \implies k \not \in \bigcup[b]^*_{(6b-1)}$
$k \neq 6ab + a - b \implies k \not\equiv -b \pmod{6b+1}\implies k \not \in \bigcup[-b]^{**}_{(6b+1)}$
$k \neq 6ab - a - b \implies k \not\equiv -b \pmod{6b-1}\implies k \not \in \bigcup[-b]^{**}_{(6b-1)}$
where $[a]^*_m = [a]^{**}_m : a \in [0,m) \cap \mathbb{Z}$,
For the purposes of this post, I will not provide you with proof of the TPC. The consensus of the moderators is that this off topic and beyond the scope of the site. What I will do is outline a series of observations that can be verified and that do indicate the existence of infinitely many twin primes.
Observe that applying SOE(Sieve of Eratostenes) to CRT(Sun Tsu's Chinese Remainder Theorem), implies that the union of a set of residue classes endowed with the infimum $\bigcup[a]^*_p$ and whose set of moduli is equivalent to the set of prime numbers can only form a covering on an interval $[z, \infty)$ when each residue is congruent to $z-2$ in the cyclic integer ring, for $z > 2$.
Observe that for the set of least residue classes endowed with the infimum $[ax+b]^*_{(cx+d)}$, that the intersection of residue classes with prime moduli produced by applying CRT in reverse produces a single unique residue for each prime number that is not a divisor of $c$, when $c$ and $d$ are relatively prime or $gcd(c,d)=1$. This set produces a covering that contains $\bigcup[ax+b]^*_{(cx+d)}$
Observe that when $gcd(c,d)\neq 1$, that the covering contained by the automorphic isomorphism formed by dividing out the greatest common denominator from both $c$ and $d$.
Observe then, that in order for a set of residue classes $[ax+b]^*_{(cx+d)}$ to form a covering on $[z, \infty)$, the residues $(ax+b)$ must be constant owed to observation (2), and therefore $a = 0$. Because observation (1) can only be applied to sets of residue classes possessing all the prime numbers as the set of moduli, $c = 1$. If $c = 0$ then the residues $(ax+b)$ must form the complete residue system, $\pmod{d}$, however contrary to the definition of the least residue class endowed with the infimum, there exists a value of $x$ for which $ax + b \not \in [0, cx+d)$ and as a consequence only $[b]^*_{(x+d)} \supseteq [b+d+1,\infty)$.
Note that when we refer to the set excluding $k$, $b$ refers to the index $x$, rather than the term in $ax+b$. I apologize for the confusion.
Observe that $\bigcup[5b+1]^*_{(6b+1)} = \bigcup[-b]^{**}_{(6b+1)} \setminus \{6\}$ and $\bigcup[5b-1]^*_{(6b-1)} = \bigcup[-b]^{**}_{(6b-1)} \setminus \{4\}$ implies the union of four systems of least residue classes endowed with the infimum can be used to establish whether or not there exists some finite number $z$ for which $\forall k < z$
Observe that in the above four systems of residue classes, the partitions of the system of residue classes describing the set excluding k, $\{[b]^*_{(6b+1)}, [-b]^{**}_{(6b-1)}\}$ and $\{[b]^*_{(6b-1)}, [-b]^{**}_{(6b+1)}\}$, are closed under intersection for the residue classes and that each residue class with a composite modulus is redundant and contained entirely by at least one residue class with a prime modulus in the same partition.
As a consequence of (6) the union of the residue systems cannot form any finite Erdos covering system, because the covering that excludes $k$ can be reduced to a system of residue classes whose set of moduli are precisely the primes greater than or equal to 5. (I need a reference, but Gerry Myerson relayed this theorem here) If any of the moduli are divisible by $p$, at least $p$ of the moduli are divisible by $p$. But there are only two residues per moduli and as a result none of the residue classes can be combined to form a covering system.
Therefore, the only way for the system to form a covering on $[z, \infty)$ is if it can be produced by splitting a system that does cover $[z,\infty)$ by partitioning the values of $x \in \mathbb{N}_{\ge1}$. In order for this to be a closed operation, only a covering system or a complete residue system can be used to partition $x$. There is no covering system with fewer than five residue classes and four systems that would be produced by the splitting, so only partitioning $x$ with complete residue systems will maintain a covering on some $[z, \infty)$.
Note that the value of $c$ in each set of residue classes is $6$ so therefore because there is no partition of $\mathbb{N}_{\ge 1}$ produced by a complete residue system with $4$ residue classes that each have a modulus of $6$ it follows that the set excluding $k$ does not contain any $[z, \infty) \cap \mathbb{Z}$ and therefore, there are infinitely many twin primes, provided that every observation is backed by evidence and logical proof. Have fun with the problem set. Let me know if there's any confusion about the terminology. I am eager to correct this if there are any mistakes or better vocabulary I could be using.
