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?"

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    $\begingroup$ $mn+1\not\equiv 2\pmod {n}$ for any $m$ if $n>1$. You don't need anything about primes to prove that. $\endgroup$ Feb 24, 2016 at 6:09
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    $\begingroup$ What does "congruent to $2 \pmod n$" mean to you? $\endgroup$
    – pjs36
    Feb 24, 2016 at 6:10
  • $\begingroup$ @pjs36 For instance, 5 is congruent to $2 \pmod{3}$, thus 5 doesn't work. $\endgroup$ Feb 24, 2016 at 6:12
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    $\begingroup$ @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. $\endgroup$
    – BrianO
    Feb 24, 2016 at 7:00
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    $\begingroup$ The OP answered their own question. $\endgroup$ Feb 24, 2016 at 8:26

3 Answers 3


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

  • $\begingroup$ 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. $\endgroup$ Feb 25, 2016 at 15:47
  • $\begingroup$ @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). $\endgroup$
    – Erick Wong
    Feb 25, 2016 at 19:28
  • $\begingroup$ 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. $\endgroup$ Feb 25, 2016 at 19:33

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!


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 $\not\equiv 2\pmod n$.

  • $\begingroup$ 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!) $\endgroup$ Feb 25, 2016 at 18:45

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