# If two primes differ by $n$, then infinitely many primes differ by $n$

A proof I'm writing rests on something I can't prove, probably beyond my knowledge, but it seems right:

For any two primes $p_k, p_l$ (not necessarily consecutive) such that the distance between them $|p_l - p_k| = n$, there exist infinitely many other primes such that the distance between them is also $n$.

I can't figure out a way to show this; I'm guessing it's probably a known result and referring to it would be enough.

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This is false if $p = 2$ :-) For odd primes, see en.wikipedia.org/wiki/Polignac%27s_conjecture –  JavaMan Nov 6 '11 at 2:22

Nobody knows. The twin primes conjecture is still a conjecture. Same for your $n=4,$ or $n=6,$ and so on. Nobody knows. As pointed out, you do need to take your $n$ even.
There seem to be doubts. Let me point out that the Prime Number Theorem says that the next larger prime above some prime $p$ is approximately $p + \log p,$ where the logarithm is base $e = 2.718281828459...$ At the same time, conjectures of, for example, Shanks, are consistent with the suggestion that the next larger prime is never larger than $p \; + \; 3 \; (\log p)^2.$ What is missing is small prime gaps, maybe there is some slowly increasing function ( monotone increasing and unbounded) $f(p)$ such that the next prime is larger than $p + f(p).$ If so, you are out of luck. Nothing is known for certain except the Prime Number Theorem.
@fakaff Maybe I am missing something, but for $n=2$ how can two primes with difference 2 not be consecutive? More exactly, you think your claim holds for all primes. But then if $p_l=3, p_k=5$ you claim is exactly the Twin prime conjecture... –  N. S. Nov 6 '11 at 2:40
fakaff, maybe it would help if you wrote some computer programs to find pairs with your $n=2,$ say up to 1,000,000, then pairs with $n=4,$ then $n=6,$ see if you still think consecutive matters. A list of twin primes to a smaller bound is on the OEIS, sequence A077800 –  Will Jagy Nov 6 '11 at 3:02