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

Solution 1:

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.