$$\forall k\in\mathbb{N},k\ge1,\exists p:k^3\lt p\lt (k+1)^3$$ with $p$ prime number. In other words is it possible to prove that for every $k\gt1$, with $k$ integer number it exists a prime number between $k^3$ and $(k+1)^3$? Legendre conjecture states there is a prime number between $k^2$ and $(k+1)^2$, but it's unsolved. Thanks.
$\begingroup$
$\endgroup$
asked Jun 13 '14 at 13:49
Add a comment
|
$\begingroup$
$\endgroup$
4
Yes, it has been recently shown that there is at least one prime between every two consecutive cubes $x^{3}$ and $(x+1)^{3}$ if $\log (\log (x))≥15$, see the paper of Cheng of $2013$. Dudek has shown the result for all $x\ge \exp(\exp(33.217)$ and discusses some issues in the paper of Cheng. For the paper of Dudek see here.
Edit: As pointed out in the comments, this solves the question for almost all $x$, but not for all.
answered Jun 13 '14 at 13:57
-
$\begingroup$ $10^{10^{15}}$ is an awful lot of numbers to check. Has this been done? $\endgroup$ – RghtHndSd Jun 13 '14 at 14:02
-
$\begingroup$ @rghthndsd: That is way more than is possible to check. I find this result very remarkable from Dudeks paper: "Furthermore, we show that there is a prime betwee n any two consecutive mth powers for m≥4.971×$10^9$" $\endgroup$ – Michael Stocker Jun 13 '14 at 14:04
-
$\begingroup$ @rghthndsd: no, this is impossible. One should mention this, you are right. $\endgroup$ – Dietrich Burde Jun 13 '14 at 14:04
-
$\begingroup$ Just to defend my question as not being completely naive: when I say "check", I do not mean one-by-one. Of course we can't do that. But if you could prove a lemma like "the statement is true for any number with a prime factor greater than 101" or something, then it becomes possible to do. $\endgroup$ – RghtHndSd Jun 13 '14 at 14:07
