There are infinitely many $k$'s such that $p_{k+1}-p_{k}>2$ This question was part of my number theory assignment which could not be discussed due to pandemic.
Question: Let $p_k$ be the $k$th prime number. Show that there are infinitely many $k$'s such that $p_{k+1}-p_{k}>2.$
Assuming that there are only finitely many such $k$'s, for all other $p_{k}$'s I would have $p_{k+1}-p_{k}\leq2,$ but for such infinitely many primes to exist twin prime conjecture need to be assumed.
So, I think it cant be proved by contradiction.
So, can anyone please tell how to approach this particular problem?
 A: It would help to find more than two consecutive composite numbers. Note that $n(n+1)(n+2) + 2 $ and $n(n+1)(n+2) + 3$ are both composite (the first is divisible by $2$, the second by $3$).
For example the largest prime smaller than $2 \times 3 \times 4 + 2 = 26$ and larger than $2 \times 3 \times 4 + 3 = 27$ are $23$ and $29$ repsectively. Continue in a similar fashion, and you could find infinitely many such pairs.
A: Assume there are finitely many values of $k$ for which $p_{k+1}-p_k > 2$. Then there is a largest $k_{max}$ such that $k_i>k_{max}\Rightarrow p_{k_{i}+1}-p_{k_i} = 2$. This would imply that above a certain value, either there are no more primes, or all odd numbers are primes.
We know there are infinitely many primes, so there are infinitely many primes larger than $p_{k_{max}}$, and we know that three consecutive odd numbers cannot all be prime; at least one of them must contain a factor of $3$. Hence the conclusion presents a contradiction, so the assumption must be false.
A: If $p_k > 3$, then $p_k$ is of the form
$6m+1$ or $6m-1$.
Using this,
I will show that for any three
consecutive primes
$p_k, p_{k+1}, p_{k+2}$
at least one of
$p_{k+1}-k$ and $p_{k+2}-p_{k+1}$
is at least 4.
If $p_k = 6m+1$,
then $p_{k+1} \ge 6m+5$
so $p_{k+1}-p_k \ge 4$.
If $p_k = 6m-1$,
then $p_{k+1} \ge 6m+1$.
If $p_{k+1} = 6m+1$ then
$p_{k+2} \ge 6m+5$
so that $p_{k+2}-p_{k+1} \ge 4$;
if $p_{k+1} > 6m+1$ then
$p_{k+1} \ge 6m+5$
so $p_{k+1}-p_{k} \ge 4$.
