# Are there infinite numbers $\not\equiv 2 \pmod {n}$, where n is any prime?

It's been proven there are infinitely many primes. This means that there exist infinitely many $m$ such that for all other prime $n$, "$m \not\equiv 0 \pmod{n}$".

My question is then, would this imply there exist infinitely many $m$ such that for all prime $n$ other than $n=m-2$, "$m \not\equiv 2 \pmod{n}$"? How about $4 \pmod{n}$?

Finally how about equivalent to neither 2 nor 4$\pmod{n}$?

Edit: To clarify my final question, another way of saying this would be "Are there infinitely many numbers, $m$, such that $m-2$ and $m-4$ are prime?" This closely approximates the twin prime conjecture, but just states it differently.

Edit: Still trying to clear ambiguity. Pick a number $m$. Is $m\equiv 2 \pmod{n}$ for some prime $n$ other than when $n=m-2$? Then that $m$ doesn't work. For instance, $m=11$: $11\equiv 2 \pmod{3}$ so 11 doesn't work. But consider $m=13$. There are no prime $n$'s we could insert such that $13\equiv 2 \pmod{n}$ other than when $n=m-2$ -- $11=13-2$. Thus $m=13$ works. Are there infinitely many working $m$'s? My final question (see above): "are there infinitely many that work for both 2 and 4 simultaneously?"

• $mn+1\not\equiv 2\pmod {n}$ for any $m$ if $n>1$. You don't need anything about primes to prove that. – Thomas Andrews Feb 24 '16 at 6:09
• What does "congruent to $2 \pmod n$" mean to you? – pjs36 Feb 24 '16 at 6:10
• @pjs36 For instance, 5 is congruent to $2 \pmod{3}$, thus 5 doesn't work. – Elem-Teach-w-Bach-n-Math-Ed Feb 24 '16 at 6:12
• @user3363795 An aside about usage: Everywhere you've written "infinite", it should be "infinitely many". For example, there are no "infinite primes", but there are infinitely many primes. – BrianO Feb 24 '16 at 7:00
• The OP answered their own question. – Greg Martin Feb 24 '16 at 8:26

If $x\geq 4$ and $n$ is a prime divisor of $x-2$ then $x\not = n$ and $x\equiv 2 \pmod n.$
• I get this now. You use $x$ where I use $m$. This says basically if $x-2$ isn't prime, there must be some $n$ such that $x \equiv 2 \pmod{n}$. My edit at the bottom of my original question now addresses this. – Elem-Teach-w-Bach-n-Math-Ed Feb 25 '16 at 15:47
• @user3363795 Even if $x-2$ is prime then there will be a prime $n$ such that $x \equiv 2 \pmod n$, namely $n=x-2$. The only time $x-2$ does not have any prime divisor is if $x\in \{1,3\}$ (since $\pm1$ are the only numbers that aren't divisible by a prime). – Erick Wong Feb 25 '16 at 19:28
• True, but in the case where $x-2$ is prime, the only prime $n$ such that $x \equiv 2\pmod n$ is $x-2$ itself. This was accounted for in the original question. – Elem-Teach-w-Bach-n-Math-Ed Feb 25 '16 at 19:33
For any (not necessarily prime) integer $n>1$ there are infinitely many natural numbers $m$ with $m\not\equiv 2\pmod n$. Indeed, for any $k$ there are $kn$ natural numbers $\le kn$ and only $k$ of these are $\equiv 2\pmod n$. In fact, any set of $n$ consecutive integers contains $n-1$ numbers $\not\equiv 2\pmod n$.
• I think my ambiguity has confused you. See Thomas's comments under the original post. Rather than "for any $n$, $m\not\equiv 2\pmod n$ for that particular $n$," I mean "for all prime $n$, are there infinitely many $m$ such that $m\not\equiv 2\pmod n$." (Still slightly ambiguous, but I think you get what I mean, and I can't think of a way to remove the ambiguity. Any suggestions would be great!) – Elem-Teach-w-Bach-n-Math-Ed Feb 25 '16 at 18:45