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

cooloicons62 2022-07-06 Answered
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 > 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.
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Answers (2)

Johnathan Morse
Answered 2022-07-07 Author has 18 answers
You don' need Bertrand's Postulate to prove it, it is enough with Euler'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's theorem the right hand isn'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.

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Crystal Wheeler
Answered 2022-07-08 Author has 4 answers
It's not a direct consequence of the postulate. Indeed, consider the set P = { 2 k : k 0 }. This set also satisfies Bertrand's postulate, but the differences 2 a 2 b are unique.

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