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

Question

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$\pmod{n}$?

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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.

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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?"

Answer 1

If $x\geq 4$ and $n$ is a prime divisor of $x-2$ then $x\not = n$ and $x\equiv 2 \pmod n.$

Answer 2

I've found the first 2 questions boil down to "are there infinitely many numbers 2 more than a prime?" and "4 more than a prime?" Since there are infinitely many primes, the answer to the first 2 questions is a proven yes. The answer to the 3rd is just another avenue for exploring the twin prime conjecture, but thoughts to this end would be welcome!

Answer 3

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 $\not\equiv 2\pmod n$.