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?"