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Let $p_{k} $be the $k^{\text{th}}$ prime number. Show that there are infinitely many $k$ such that $p_{k+1} − p_{k} > 2$.

I was thinking about Dirichlet's theorem as i don't know to prove its

Please help me

Thanks in advance

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closed as off-topic by user21820, José Carlos Santos, Did, Leucippus, Cesareo Feb 6 at 7:49

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    $\begingroup$ If $p>3$ is a prime is it possible that $p+2$ and $p+4$ are both also prime? $\endgroup$ – RJM Mar 2 '18 at 11:28
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Lemma: If $p_k = 6t + 1$, then $p_{k+1}-p_k > 2$.

Proof: Suppose on contrary $p_{k+1}-p_k=2$, then we have: $$p_{k+1} = p_k+2 = (6t + 1) + 2 = 6t + 3 \Longrightarrow 3 \mid p_{k+1},$$ which is a contradiction.


Also note that by Dirichlet's theorem there are infinitely many primes of the form $6t+1$.

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  • $\begingroup$ thanks a lots accursed $\endgroup$ – user525416 Mar 2 '18 at 14:02
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Let $g_n = p_{n+1}-p_n$ denote the prime gap.

In 1931, Erik Westzynthius proved that maximal prime gaps grow more than logarithmically. That is, $$ \limsup_{{n\to \infty }}{\frac{g_{n}}{\log p_{n}}}=\infty .$$

Therefore, there exists infinitely many $n > 2$ such that $g_n > 2 \log p_n > 2$.

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    $\begingroup$ THANKS GNU supporter $\endgroup$ – user525416 Mar 2 '18 at 14:01
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Let $g_k = p_{k+1} - p_k$, be the size of the $k^\text{th}$ gap.

  • $g_k \neq 0$ because otherwise $p_k = p_{k+1}$ and no prime is equal to its successor.
  • For $k > 1$, $g_k \neq 1$ because all primes greater than $2$ are odd.

Now, for contradiction, suppose that $g_k > 2$ holds for only finitely many $k$. Then there is an integer, $K$, such that for all $k > K$, $g_k = 2$. In particular, this says all odd numbers greater that $p_K$ are prime. But $p_K^2$ is also a composite odd number greater than $p_K$, exhibiting a contradiction.

(To summarize, the sequence of primes has to skip all the odd squares, so must have infinitely many gaps of length $\geq 4$.)

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