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

I read that the following can be proved using Bertrand's postulate (there's always a prime between $n$ and $2n$): $\mathrm{\forall }N\in \mathbb{N}$, 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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Johnathan Morse
You don' need Bertrand's Postulate to prove it, it is enough with Euler's theorem on the divergence of $\sum \frac{1}{p}$. The argument is as follows:

For some integer $n$, consider the $n-2$ differences ${p}_{i+1}-{p}_{i}$ for $i=2,3,\dots ,n-1$, if they take at most
${T}_{n}=⌊\frac{n-2}{N}⌋$
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 $\left\{{p}_{i+1}-{p}_{i}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}};i=2,3,\dots ,n\right\}$ then
${p}_{n}-{p}_{2}=\sum _{i=2}^{n-1}{p}_{i+1}-{p}_{i}\ge 2+4+6+\cdots +2{T}_{n}+2\left({T}_{n}+1\right)=\left({T}_{n}+1\right)\left({T}_{n}+2\right)$
but
$\left({T}_{n}+1\right)\left({T}_{n}+2\right)\ge \frac{\left(n-2{\right)}^{2}}{{N}^{2}}$
So
$\frac{1}{{p}_{n}}\le \frac{{N}^{2}}{\left(n-2{\right)}^{2}+3{N}^{2}}$
if we fix $N$ and sum for $n=N,N+1,\dots ,M$ we get
$\sum _{i\le M}\frac{1}{{p}_{i}}\le -\sum _{i\le N}\frac{1}{{p}_{i}}+{N}^{2}\sum _{N\le n\le N}\frac{1}{\left(n-2{\right)}^{2}+3{N}^{2}}$
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 $k$.

Crystal Wheeler
It's not a direct consequence of the postulate. Indeed, consider the set $P=\left\{{2}^{k}:k\ge 0\right\}$. This set also satisfies Bertrand's postulate, but the differences ${2}^{a}-{2}^{b}$ are unique.