Largest Number that cannot be expressed as 6nm +- n +- m I'm looking to find out if there is a largest integer that cannot be written as $6nm \pm n \pm m$ for $n,m$ elements of the natural numbers.  For example, there are no values of $n,$m for which $6nm \pm n \pm m = 17.$  Other numbers which have no solution are $25$, $30$, and $593$.  Can all numbers above a certain point be written as $6nm \pm n \pm m$ or will there always be gaps?
Thanks.

This question is equivalent to the Twin Prime Conjecture.
Let's look at the three cases:
$$6nm+n+m =\frac{(6n+1)(6m+1)-1}{6}\\
6nm+n-m = \frac{(6n-1)(6m+1)+1}{6}\\
6nm-n-m = \frac{(6n-1)(6m-1)-1}{6}
$$
So if $N$ is of one of these forms, then:
$$6N+1=(6n+1)(6m+1)\\
6N-1=(6n-1)(6m+1)\\
6N+1=(6n-1)(6m-1)$$
If $6N-1$ and $6N+1$ are both primes, then there is no non-zero solutions.
If $6N-1$ is composite, its factorization is of the form $(6n-1)(6m+1)$.
If $6N+1$ is composite, its factorization is either of the form $(6n-1)(6m-1)$ or $(6n+1)(6m+1)$.
So the question is equivalent to the Twin Prime Conjecture. I don't think this will be solved on this site, unless it is just a reference to a resolution of the conjecture elsewhere. Currently, the state of knowledge about twin primes is thin.
For example, when $N=31$, $6N-1=5\cdot 37 = (6\cdot 1-1)(6\cdot 6+1)$. So with $m=1,n=6$, we get $31 = 6\cdot 1\cdot 6 +1 -6$. Alternative, $6N+1=11\cdot 17=(6\cdot 2-1)(6\cdot 3-1)$. So $(m,n)=(2,3)$ and $31=6\cdot2\cdot 3 -2-3$.
