I read that the following can be proved using Bertrand's postulate (there's always a prime between n and 2 n): ∀ N ∈ N , there exists an even integer k &gt; 0 for which there are at least N prime pairs p, p + k.But I have no idea how to prove it. Any help would be much appreciated.Many thanks.

Question

I read that the following can be proved using Bertrand&#039;s postulate (there&#039;s always a prime between   n and   2  n):   ∀  N  ∈      N  , there exists an even integer   k  &amp;gt;  0 for which there are at least   N prime pairs   p,   p  +  k.But I have no idea how to prove it. Any help would be much appreciated.Many thanks.

Johnathan Morse · Accepted Answer

You don&#039; need Bertrand&#039;s Postulate to prove it, it is enough with Euler&#039;s theorem on the divergence of   ∑      1    p  . The argument is as follows:For some integer   n, consider the   n  −  2 differences       p          i      +      1        −      p    i   for   i  =  2  ,  3  ,  …  ,  n  −  1, if they take at most      T    n    =      ⌊                  n        −        2            N        ⌋  different values then there is at least one taken   N times and we are done.Suppose otherwise that for every   n there are more than       T    n   different values in the set   {      p          i      +      1        −      p    i        ;  i  =  2  ,  3  ,  …  ,  n  } then      p    n    −      p    2    =      ∑          i      =      2              n      −      1            p          i      +      1        −      p    i    ≥  2  +  4  +  6  +  ⋯  +  2      T    n    +  2  (      T    n    +  1  )  =  (      T    n    +  1  )  (      T    n    +  2  )but  (      T    n    +  1  )  (      T    n    +  2  )  ≥            (      n      −      2              )        2                    N      2      So      1          p      n        ≤            N      2              (      n      −      2              )        2            +      3              N        2            if we fix   N and sum for   n  =  N  ,  N  +  1  ,  …  ,  M we get      ∑          i      ≤      M            1          p      i        ≤  −      ∑          i      ≤      N            1          p      i        +      N    2        ∑          N      ≤      n      ≤      N            1          (      n      −      2              )        2            +      3              N        2            But the right hand side is bounded and by Euler&#039;s theorem the right hand isn&#039;t so for   n large enough we get a contradiction.Note that this proves something stronger: for every   N there is an integer   k such that there are more than   N consecutive primes with difference   k.

Crystal Wheeler · Accepted Answer

It&#039;s not a direct consequence of the postulate. Indeed, consider the set   P  =  {      2    k    :  k  ≥  0  }. This set also satisfies Bertrand&#039;s postulate, but the differences       2    a    −      2    b   are unique.

I read that the following can be proved using Bertrand's postulate (there's always a prime between

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