Consider the following question:

1) For a given natural number $a$, are there finitely or infinitely many natural numbers that are not of the form $anm \pm n\pm m$, where $m$ and $n$ range over positive integers? (For $a=1$ or $a=2$ you have all the natural numbers.)

Does this problem appear in the literature?

As one can see at MO Scribe's question Chen's Theorem with congruence conditions. it is the same like asking if there are infinitely many $k$ such that both $ak+1$ and $ak-1$ do not have any non-trivial factors of the form $\pm 1 \mod a$.

I give a proof that for $a=6$ the question is equivalent to the twin prime conjecture so it is known that we don't have any proof. But what about other values of $a$? Is the problem for $a=100$ or more of the same difficulty?

2) Is the density (Szemeredi's or Schnirelman's), of the numbers that are not of the form, zero for any value of $a$?

3) From Viggo Brun's theorem we have that the sum of the reciprocals of the twin primes converges. Does the sum of the reciprocals for any value of $a$ of the numbers that are not of this form converge?

4)For a given natural number $a$ are there infinitely many $k$ such that both $ak+1$ , $ak-1$ do not have any $prime$ factors of the form $\pm 1 \mod a$? (the same questions for these $k$ as in 2) and 3) )

5) And the most easy : for which $a$ do we have a proof that there are infinetely many $k$ such that such that both $ak+1$ , $ak-1$ are either prime or can be written as a product of two numbers both not of the form $\pm 1 \mod a$? I guess that if $φ(a)$ is big enough we can have such a proof (the same questions for these $k$ as in 2) and 3) )

NOTE: In 1) and 4) both both $ak+1$ and $ak-1$ can be either primes or the product of primes not of the form $\pm 1 \mod a$ ,but in the 1) no subproduct of them can be of this form.

As MO Scribe noticed the conjectural answer is that there should be infinetely many such pairs because we are aspecting to have infinitely many prime pairs of any reasonable congruence condition.

http://math.stackexchange.com/questions/15075/do-we-have-a-proof-of-the-infiniteness

There are infinitely many twin primes if and only if there are infinitely many natural numbers that are not of the form $6nm \pm n \pm m$.

**Proof:** Every number that is not a multiple of $2$ or $3$ is of the form $6N\pm 1$. So the only pairs that are not divisible by $2$ or $3$ are $(6N-1,6N+1)$ for any $N$. Now are there infinitely many such prime pairs (twin primes)?

If the number $6N-1$ is prime it should not be written as a product of some numbers $6n+1,6m-1$ for any $n,m > 0$. So $(6n+1)(6m-1)=6(6nm-n+m)-1$, which means that $N$ should not be of the form $6nm-n+m$ for any $n,m>0$.

Similarly, if $6N+1$ is a prime it should not be a product of some numbers $(6n-1)(6m-1) =6(6nm-n-m)+1$, or $(6n+1)(6m+1) =6(6nm+n+m)+1$. Which means that we have a prime couple of the form $(6N-1,6N+1)$ if and only if $N$ is not of the form $6nm \pm n \pm m$ for any $n,m$.

NOTE: After i have edited this observasion-question I realised that it is well known that for $a=6$ it is equivalent to twin prime conjecture (as it is written on the answer by Luis below too), a notice that S.Golomb seems to have done first, but my question is focused to the other values of $a$.

If someone want to add something http://tea.mathoverflow.net/discussion/921/reedited/#Item_1