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

This question was asked in the entrance examination of the Indian Statistical Institute(ISI).

My approach: I tried using Bertrand's postulate, but could not get anything.

I also thought of showing its negation i.e. there are infinitely many $k$ such that $p_{k+1}-p_k \leq2$ is false but it leads to twin prime conjecture.

Help me. Thanks!

SARTHAK GUPTA
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    That is not its negation. Its negation would be "For all but finitely many $k$, $p_{k+1}-p_k\le 2$." – TonyK Apr 08 '20 at 10:25
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    Hint: consider three consecutive primes. Can the two pairs both differ by 2? – Wojowu Apr 08 '20 at 10:27

4 Answers4

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You may note that for a positive integer $n$, all the numbers $$n!+2, n!+3,n!+4,\ldots ,n!+n$$ are successive and not prime. So we can find an arbitrarily large segment of integers including no primes.

Qurultay
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Proof by contradiction: Suppose $K:=\{k \in \mathbb{N} : p_{k+1}-p_k > 2 \}$ is finite. This means it has an upper bound. Let $k_* = \max (K \cup \{2\})$. Now we have $p_{k+1} \le p_k + 2$ for all $k \ge k_*$. By definition, we also have $p_{k+1} \ge p_k + 1$ for all $k$. Note that $p_{k+1}=p_k+1$ is impossible for $k>1$, since otherwise one of $p_{k+1}$ or $p_k$ must be divisible by $2$. Thus $p_{k+1}=p_k+2$ for all $k \ge k_*$. By induction, $p_k = p_{k_*} + 2 \cdot (k-k_*)$ for all $k \ge k_*$. Now set $k=p_{k_*}+k_*$. Then $p_k = 3 \cdot p_{k_*}$, which is impossible.

Thomas
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I think this is the easiest proof:

There are infinitely many odd primes greater than or equal to 5. Divide all such primes into groups of three: {5, 7, 11}, {13, 17, 19}, {23, 29, 31}, ... There are infintely many such groups.

Now focus on an arbitrary group $p_k, p_{k+1}, p_{k+2}$. Suppose that $p_{k+1}=p_k+2$ and $p_{k+2}=p_{k+1}+2$. This is clearly impossible because in that case at least one of these numbers would be divisible by 3.

So it's either $p_{k+1}-p_k>2$ or $p_{k+2}-p_{k+1}>2$. In other words there is a pair of successive primes with difference greater than 2 for each group of three consecutive primes. We have infinitely many such groups so the number of successive primes with difference greater than 2 is infinite.

Saša
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    You are redefining what $p_k$ means: the OP defines $p_k$ as the $k$th prime. Also, for a given value of $k$, the resulting pair $(p_k,p_{k+1})$ may already have been accounted for by smaller values of $k$. You can get round this fairly easily, but then it's not the easiest proof any more. – TonyK Apr 08 '20 at 13:52
  • @TonyK I have completely reworked my proof. Looks better? – Saša Apr 09 '20 at 07:34
  • Yes! But see @Wojowu's comment to the OP. – TonyK Apr 10 '20 at 20:34
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Any time $p_{k+1} = p_k + 2$, we can't also have $p_{k+2} = p_{k+1} + 2$ unless $p_k = 3, p_{k+1} = 5, p_{k+2} = 7$. Past this, at least one of $p_k, p_k+2, p_k + 4$ would be divisible by $3$ and therefore (unless $p_k =3$) not prime.

So each prime gap of $2$ is followed by a prime gap of $>2$, and there's at least as many long gaps as there are short gaps. Since there are infinitely many primes, there must be infinitely many long gaps between primes.

Misha Lavrov
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