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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
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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.

http://en.wikipedia.org/wiki/Twin_prime

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.

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The two primes don't have to be consecutive though. Twin prime is for, n=6 and so on, is for the case of consecutive primes. In this case, it's the difference between any two primes and whether any other two primes are separate by the same distance. –  iDontKnowBetter Nov 6 '11 at 2:36
    
@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
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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
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It is unknown whether there are infinitely many pairs of primes differing by, say, 6. It doesn't matter whether you restrict to consecutive primes, or not; either way, unknown. –  Gerry Myerson Nov 6 '11 at 3:36
    
@GerryMyerson damn, you're right, consecutive or not doesn't matter now that I think it through. — now I'm completely stuck with my proof :( –  iDontKnowBetter Nov 6 '11 at 4:52
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