7

"Obviously" it is thrue that $p_{n+1}<2p_n$. Testing for $n<10$ shows it is true for small $n$ and no mathematician or wannabe has ever doubt that it is true for big $n$. But there is no real simple arithmetic proof, so far, not using the prime number theorem or other results that isn't simple to prove.

So I wonder, are there any (non trivial) prime gap theorems at all with simple proofs? Or prime recursion inequalities of the form $p_{n+1}<f(p_1,p_2,\dots,p_n)$?

Can you prove: $p_{n+1}<p_n^2$ without using PNT, Bertrand's postulate,..?

Can you find a refinement of $\displaystyle p_{n+1}\le\prod^n_{i=1}p_i+1$, about just as simple to prove?

Lehs
  • 14,252
  • 4
  • 28
  • 82
  • 1
    The original argument of Chebyshev is a little complicated, but "elementary" and does not use PNT. A much simplified proof was given by Erdos. Please see the Wikipedia article on Bertrand's postulate. – André Nicolas Jan 21 '15 at 07:28
  • See Vinogradov's Elements of Number Theory for a proof that for any $\varepsilon > 0$, there is an $n$ for which $p_{n+1} < (1 + \varepsilon)p_n$. https://archive.org/details/ElementsOfNumberTheory – user208259 Jan 21 '15 at 07:40
  • That doesn't answer the question though – Stella Biderman Jan 21 '15 at 07:48
  • Prime gap theorems with simple proofs - I'm pretty sure that there is a simple proof for "a prime between $n$ and $2n$ for every $n\in\mathbb{N}$". – barak manos Jan 21 '15 at 07:49
  • 3
    In any case, your terminology of "non trivial" and "simple" is primarily opinion-based. In my opinion, for example, a theorem is trivial if and only if it has a simple proof. So the answer to your question is No by definition. – barak manos Jan 21 '15 at 07:59
  • I don't understand how quotation marks make the initial claim less ... well, what it is. – Mariano Suárez-Álvarez Jan 21 '15 at 08:14
  • @MarianoSuárez-Alvarez: it marks a kind of false induction. – Lehs Jan 21 '15 at 08:15
  • It just signals that the premise of your question is simply untenable, really. II would love to see any basis for the claim that no mathematician has ever doubted the truth of Bertrand's postulate (and the opinion of whatever it is you call wannabes is surely irrelevant? Is there any need to even use that derogatory term at all?) – Mariano Suárez-Álvarez Jan 21 '15 at 08:18
  • @MarianoSuárez-Alvarez: I could remove the rhetoric, but I want to communicate a feeling... – Lehs Jan 21 '15 at 08:25
  • 2
    Can we count Euclid's argument that $p_{n+1} \le\prod_{i=1}^n p_i +1$ as a prime-gap-theorem? – Neil W Jan 21 '15 at 08:33
  • @Neil: yes, it's a good example. Perhaps I should have wrote recursive inequality or something, but are there any refinements then?. – Lehs Jan 21 '15 at 08:38
  • @barakmanos But the identity between "trivial" and "has simple proof" may not extend to "obvious". For example, everybody$^{\text{TM}}$ considers Jordan's curve theorem obvious until they attempt to find a simple proof. – Hagen von Eitzen Jan 21 '15 at 10:17
  • @Lehs You can easily create a condition dependent solely on the previous prime itself if you like: $p_{n+1} \le\prod_{i=1}^n p_i +1 \le p_n^n +1 \le p_n^{p_n} +1$ – Neil W Jan 23 '15 at 11:17
  • @Neil: yes, few arguments is preferable, but "good" estimates with elementary proofs are what I desire. – Lehs Jan 23 '15 at 11:22

2 Answers2

3

For the question about $p_{k+1} < p_k^2$, this should be fairly easy to deduce from the estimate $\sum_{\text{$p$ prime, }p \leq n} \frac{\ln p}{p} = \ln n + O(1)$, so long as you're careful to make the $O(1)$ bound explicit. The proof of the estimate is elementary and short. (See the exercises to Chapter 2 of Vinogradov's Elements of Number Theory.)

Namely, if the $O(1)$ term is between $A$ and $B$, then as long as $n > e^{B-A}$, there must be a prime between $n$ and $n^2$. Then you can check small values of $n$ manually.

I haven't worked out the details of what $A$ and $B$ are, but I don't think they're large.

Edit: Here is a quick proof of Mertens' estimate, which is what we used above. On Wikipedia it is stated that $O(1)$ is bounded by $2$ in absolute value, so this would establish the theorem for all $n > e^4$.

This proof can be immediately generalized to prove that if $n$ is large enough, then there is necessarily a prime number between $n$ and $n^{1 + \varepsilon}$.

user208259
  • 1,515
0

There is a refinement of $\displaystyle p_{n+1}\le\prod^n_{i=1}p_i+1$, just about as simple to prove:

$ \displaystyle p_n\leq 2\prod_{q\in\Bbb P'}^{q<p_n}q+1$, where $\Bbb P'=\{p\in\Bbb P|\exists m\in\Bbb N:p=4m+3\}$and $n>3$.

Proof: Let $ \displaystyle A_n=2\prod_{q\in\Bbb P'}^{q<p_n}q+1$ and suppose $p_n>A_n$. Then $A_n$ only has prime factors $p_k<p_n$ such that $p_k=4m+1,\; m\in\Bbb N$. Thus $A_n=4m'+1,\; m\in\Bbb N$ which is obviously not true.
Hence $p_n\leq A_n$.

A simple proof for the corresponding conjecture with $\Bbb P''=\{p\in\Bbb P|\exists m\in\Bbb N:p=4m+1\}$ is probably also possible and it might be possible to show for any similar class of primes for $n$ big enough.

Lehs
  • 14,252
  • 4
  • 28
  • 82