6
$\begingroup$

Legendre`s conjecture states that there is always a prime between $n^2$ and $(n+1)^2$ for every natural $n$.

It is natural to create following generalization:

Is it true that for every $\varepsilon \in (0,1]$ there exists natural number $n_0(\varepsilon)$ such that for every $n \geq n_0(\varepsilon)$ there is always a prime between $n^{1+\varepsilon}$ and $(n+1)^{1+\varepsilon}$?

I do not know if some results from number theory do not allow this generalization to be true but if it is not known is it true or not what is known, if anything, about this generalization of Legendre`s conjecture?

$\endgroup$
  • 1
    $\begingroup$ I suspect it is known for epsilon 2 or bigger. I think Baker Harman Pintz result implies epsilon greater than 3/2 also works . Gerhard "And Probably Smaller Epsilon Too" Paseman, 2017.05.12. $\endgroup$ – Gerhard Paseman May 12 '17 at 17:25
6
$\begingroup$

As mentioned in the remarks, this conjecture is equivalent to the statement that if $p$ is a prime, then the next prime satisfies $p'\ll_\delta p+p^\delta$ for any $\delta>0$. The parameter $\delta$ corresponds to the OP's $\epsilon/(1+\epsilon)$. For $\delta>1/2$ (which corresponds to $\epsilon>1$) this follows from Legendre's conjecture and also from the Riemann Hypothesis, for $\delta>21/40$ (which corresponds to $\epsilon>21/19$) it follows unconditionally from the work of Baker-Harman-Pintz. For arbitrary $\delta>0$, the statement follows from Cramér's conjecture. The bottom line is: we don't know the truth yet.

$\endgroup$
0
$\begingroup$

This seems to be about as strong as Cramer's conjecture.

$\endgroup$
  • 2
    $\begingroup$ Why the downvote, dare I ask? $\endgroup$ – Igor Rivin Apr 12 '17 at 17:26
  • 1
    $\begingroup$ This must be one of these unjustified downvotes. See here meta.mathoverflow.net/questions/3122/… for a discussion. $\endgroup$ – tj_ Apr 12 '17 at 18:04
  • 1
    $\begingroup$ @tj_ Thanks. Personally, I have never downvoted anything, so I guess I am just not getting it. $\endgroup$ – Igor Rivin Apr 12 '17 at 18:06
  • 4
    $\begingroup$ The conjecture is equivalent to the statement that the prime gap after $n$ is $n^{o(1)}$. Cramer’s conjecture states that it is $O((\log n)^2)$. That’s much stronger. $\endgroup$ – Emil Jeřábek supports Monica Apr 12 '17 at 18:33
  • 2
    $\begingroup$ anyway, given the conjectures, it is odd that you never see tables of $g / \log^2 p,$ I had to produce that myself. $\endgroup$ – Will Jagy Apr 12 '17 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.