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)

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

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 a conversation at meta http://tea.mathoverflow.net/discussion/916/delete-and-reopen/#Item_0 and after Emerton's propose i decided to make a complete version of my first question according to the new facts and delete all the others that are the same.

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

As one can see at MO Scribe's question Chen's Theorem with congruence conditions. it is the same like asking if there infinitely many k such that both $ak+1$ and $ak-1$ do not have any 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? 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.

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 a conversation at meta http://tea.mathoverflow.net/discussion/916/delete-and-reopen/#Item_0 and after Emerton's propose i decided to make a complete version of my first question according to the new facts and delete all the others that are the same.

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

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

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 a conversation at meta http://tea.mathoverflow.net/discussion/916/delete-and-reopen/#Item_0 and after Emerton's propose i decided to make a complete version of my first question according to the new facts and delete all the others that are the same.

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)

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

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 a conversation at meta http://tea.mathoverflow.net/discussion/916/delete-and-reopen/#Item_0 and after Emerton's propose i decided to make a complete version of my first question according to the new facts and delete all the others that are the same.

If someone want to add something http://tea.mathoverflow.net/discussion/921/reedited/#Item_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 natural numbers?

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)

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

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 a conversation at meta http://tea.mathoverflow.net/discussion/916/delete-and-reopen/#Item_0 and after Emerton's propose i decided to make a complete version of my first question according to the new facts and delete all the others that are the same.