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There is a well-known proof of the fact that the prime gaps can be arbitrarily large. Namely, for any natural number $n$, the consecutive integers $(n+1)!+2,...,(n+1)!+(n+1)$ are never prime.

But clearly, this proof does not guarantee that $(n+1)!+1$ and $(n+1)!+(n+2)$ are prime. How can we show that the prime gap can be exactly $n$ for any even $n$?

Levent
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    Up to now we are not able to show that. – Cave Johnson Dec 12 '16 at 07:21
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    It is conjectured that for every even $n$, there are infinitely many pairs of consecutive primes $p$ and $q$, such that $p-q=n$. In other words, not only that the prime gap can be exactly $n$ for every even $n$, but there are also infinitely many cases for every even $n$. Nevertheless, even your (weaker) conjecture has not been proven as of yet. – barak manos Dec 12 '16 at 07:36
  • @barakmanos In fact, until a few years ago, we didn't know whether there were any $n$ such that there are infinitely many consecutive prime pairs $p,q$ with $p-q=n$. We now know that the smallest such $n$ is below 70 million (https://youtube.com/watch?v=vkMXdShDdtY) (There might've been improvements on this result since.) – Arthur Dec 12 '16 at 08:11
  • @Arthur: Yes, if I'm not mistaken, it was a leading number theorist named Chen (or something similar) who proved it. And I think there is now a project attempting to reduce it (I recall reading it has been reduced all the way down to $268$ or something like that, but I might be wrong on that one). – barak manos Dec 12 '16 at 08:17

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