There are infinitely many k such that $p_{k+1} - p_{k} >2$

Question

The following question is from my number theory assignment and I have been following number theory by Zukermann.

Let $p_k $ be the k-th prime number. Show that there are infinitely many k such that $p_{k+1}- p_{k} >2$.

I tried by assuming that the converse is true ie there are only finitely many k such that $p_{k+1} -p_k >2 $and then trying to get some contradiction. If$p_{k+1}- p_{k} \leq 2$ . then $p_{k+1}- p_{k} =1$ is clearly a contradiction. But I am unable to get any contradiction if $p_{k+1} -p_k =2$. Can you please help with that.

Any other method is also welcome.

See https://math.stackexchange.com/questions/1095948/confusion-on-the-proof-that-there-are-arbitrarily-large-gaps-between-successive — lhf, Sep 03 '21 at 16:45
Hint: show that if $p_{k+1}-p_k=2$ then $p_{k+1}+2$ can't be prime, by looking modulo $3$. — Greg Martin, Sep 03 '21 at 16:48

score 3 · Answer 1 · answered Sep 03 '21 at 17:03

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Suppose not, then only for finitely many $k$, $p_{k+1}-p_k>2$. Then eventually $p_{k+1}-p_k\leq 2$ for all $k$. Now it must be true, as you say that $p_{k+1}-p_k=2$. Furthermore $p_{k+2}=p_k+4$ and $p_{k+4}=p_k+8$. Now either $p_k$, $p_k+4$ or $p_k+8$ must be divisible by $3$.

answered Sep 03 '21 at 17:03

spinosarus123

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Or $p_k,p_{k+1},p_{k+2}$ – Roddy MacPhee Sep 03 '21 at 21:28

score 2 · Accepted Answer · answered Sep 03 '21 at 16:56

Suppose there are only finitely many such primes. Then there is a largest $n$ with $p_n-p_{n-1}>2$.

So $p_{n+m+1}-p_{n+m}=2$ for every $m\geq 0$. So $p_{n+m}=p_n+2m$

Now $p_{n+1}p_n$ is an odd number, as product of odd numbers.

Then $p_{n+1}p_n=p_n+2k$ for some positive integer $k$, which is clearly absurd, as this can not be a prime number.

There are infinitely many k such that $p_{k+1} - p_{k} >2$

2 Answers2