# Legendre's Conjecture: Bounded Prime Gaps

I have encountered some error in the details of what Legendre's conjecture implies about bounded prime gaps. So I am working to correct errors and to state both what is conjectured and what is implied by the conjecture concerning bounded prime gaps. I am asking about the correctness of my corrections.

First, as to what is conjectured, because 3 is the only prime of the form $n^2-1$, I state the conjecture thus: In the interval between $n^2$ and $n^2$+2$n$, there will always be at least one prime number.

Second, as to what this conjecture implies about bounded prime gaps, let $n$ be any given natural number, let $p$ be the next prime greater than $n$, and let $m$ be the limit from $n$ to $p$ such that $n$+$m$ $\geq p$. Legendre's conjecture implies that the prime gap above $n$ can be bounded by the product of two factors, one of them being a constant and the other one being related to the square root of $n$. Let $c$ be the factor which is a constant and let $f$ be the factor which is related to the square root of $n$ such that $cf$=$m$.

My question here is about the correctness of what I have stated here so far before I attempt to find the precise relationships between the things mentioned herein which I have represented by $n,p,m,c,$ and $f$.

• The error which I encountered is actually one that I made. I think I understand this matter now. – Jeffrey Young Oct 3 '14 at 22:45

Yes, Legendre's conjecture that there is a prime in $[n^2, (n+1)^2]$ ($n$ being a positive integer) is equivalent to the conjecture that there is a prime in $[n^2+1, (n+1)^2-2]$.
Legendre's conjecture doesn't imply bounded gaps, but it does imply that the gap following a prime $p$ is $O(\sqrt p)$. In particular it is at most $4\sqrt{p-1}$. Of course everyone knows in their hearts that much more is true, but that's all we get from Legendre's conjecture.
• So if $n$ + 4$\sqrt n-1$ $\geq p$, where $n$ is any given natural number and $p$ is the next prime after $n$, then Legendre's conjecture is true, right? – Jeffrey Young Oct 3 '14 at 16:51
• @JeffreyYoung: No. (Try it with $n=9$ to see why it fails.) But if you can get the gap down to $n+\sqrt n-1$ then it should work. The approximate factor of 4 between these two is a measure of the cost of the converting between the discrete world and the continuous one. – Charles Oct 3 '14 at 19:34