# What's the history of the result that $p_{n+1} < p_n^2$, and how difficult is the proof?

In Edsger Dijkstra's monograph "Notes on Structured Programming", he describes a simple imperative program for generating an array of the first $n$ primes. For each prime $p_n$, it finds the next prime by checking each odd $j \gt p_n$, to see if it has a prime divisor $d$ in the range $2 \lt d \le \sqrt j$. The algorithm draws these candidates from the array of prime numbers it has been constructing, which at any given iteration contains the primes between $2$ and $p_n$ inclusive, so there is an implicit assumption is that $\sqrt j \le p_n$ (the algorithm iterates over the array of primes up to $p_n$ with the only termination conditions being $d|j$ or $d > \sqrt j$, so if $\sqrt j$ could be $\ge p_n$ we risk reading past the end of the array). Quoting Dijkstra:

In all humility I quote Don Knuth's comment on an earlier version of this program, where I took this fact for granted:

"Here you are guilty of a serious omission! Your program makes use of a deep result of number theory, namely that if $p_n$ denotes the $n$th prime number we always have $p_{n+1} < p_n^2$."

So I'm curious about the history of this "deep result", and how difficult the proof is.

• I was under the impression this was only conjectured, hm... – Olivier Bégassat Aug 14 '11 at 1:47
• @Olivier It's certainly true, as it's weaker than Bertrand's Postulate (which is true). – davidlowryduda Aug 14 '11 at 1:51
• Looks like a misprint. The proof by Chebyshev of the much stronger "Bertrand's Postulate" is relatively elementary. If we replace $2n$ by $n^2$, the estimates become very easy. – André Nicolas Aug 14 '11 at 1:55
• I’ve not seen that result; the stronger result known as Bertrand’s postulate has an elementary but rather involved proof that can be found here, along with a little history. – Brian M. Scott Aug 14 '11 at 2:01
• @mixedmath true! I think I confused it with the conjecture (?) that says there's a prime between each pair of consecutive squares. – Olivier Bégassat Aug 14 '11 at 3:02

## 2 Answers

I am uncertain of that particular proof, but I know a few proofs of Bertrand's Postulate. Bertrand's says that for all $n > 3$, there is a prime between $n$ and $2n$. In particular, this means that given a prime $p_0$, there is another prime $p_1$ s.t. $p_0 < p_1 < 2p_0$ As $2 < p_0$, we then have that there is a $p_1$ s.t.

$$p_0 < p_1 < 2p_0 < p_0 ^2$$

I think the Wikipedia proof is not so bad. Bertrand actually proved his conjecture himself in the late 1800's, and there are many proofs. Ramanujan made one, Erdos improved up it. There are also many better results and asymptotic results - I have used Pierre Dusart's improvement before, which says that for sufficiently large x (larger than 3275), there is a prime between $x$ and $x + \dfrac{x}{1 + 2\log^2 x}$. That's very intense!

But that idea is somewhat deep, and it relates to the Prime Number Theorem on the distribution of primes (which I would call very deep).

• Very cool, I'll go look at these proofs. Thanks! – Tom Crockett Aug 14 '11 at 2:10
• Bertrand had a proof? I've always seen the result attributed to Chebyshev. Nat Fine composed a little poem in honor of Erdos: Chebyshev said, And I say it again, There's always a prime Between $n$ and $2n$. – Gerry Myerson Aug 14 '11 at 6:28

Actually, the program as described requires no such assumption. Remember, it starts from the last prime found so far and checks every (odd) number in turn to see if it is composite; therefore, the first non-composite number thus found must be the next prime.

The primality check merely makes use of the following rather trivial result:

Lemma: Every composite number is divisible by some prime no greater than its square root.

Proof: We need to show that every composite number must have a (possibly non-prime) divisor no greater than its square root; this divisor must then have a prime factor no greater than itself. To prove this, assume the contrary, i.e. that there exists a composite number $n$ such that its smallest proper divisor $k$ satisfies $k > \sqrt n$. But then $n/k < \sqrt n$ is also a divisor of $n$, which is a contradiction.

Edit: Sorry, I see my mistake now. It is indeed not trivial to show that the program will never read past the end of the list of primes if that's not checked for. I'll leave my answer up for a record of the conversation, but I'll mark it as community wiki. Please don't upvote it.

• If all factors are greater than $\sqrt{n}$, then $n=pk > \sqrt{n}\sqrt{n}=n$. A contradiction. – Fredrik Meyer Aug 14 '11 at 12:27
• You are pointing out why we only need to check primes numbers $d$ up to $\sqrt j$ when looking for a divisor of $j$; if we don't find one by then we can conclude that it is prime. But that's not the hidden assumption Knuth was talking about. He is referring to the assumption that you can draw $d$ from the array of primes you have been constructing in a loop, with the only termination conditions being $d|j$ or $d > \sqrt j$, and have no fear of reading past the end of the array, which contains primes only up to $p_n$. – Tom Crockett Aug 14 '11 at 20:00
• If $j$ is prime, then the loop will only terminate because $d > \sqrt j$. This is where Dijkstra was silently making use of the fact that $\sqrt p_{n+1} < p_n$, always. We could of course modify the algorithm if we weren't sure of that fact, to include an additional termination condition of $d = p_n$, but the algorithm as stated didn't have that. – Tom Crockett Aug 14 '11 at 20:00
• @pelotom: Ah, I see. I missed the fact that the issue was not the correctness of the sketched algorithm as such, but rather the fact that no explicit check for reading past the end of the array was required. I've edited my answer to address this. – Ilmari Karonen Aug 14 '11 at 20:21
• @Ilmari Karonen - I think the fault is mine for not making this clearer in the original post... I was trying to keep the backstory to the question as brief as possible :) I've made a small edit to the question that will hopefully help. – Tom Crockett Aug 14 '11 at 20:55