# Exact Equivalence of Legendre's Conjecture Impossible?

If the upper bound for the prime gap above $n$ is such that $n$+4$\sqrt{n-1}$$\geq p, where n is any given natural number and p is the next prime after n, then Legendre's conjecture is true. If Legendre's conjecture is true, then n+4\sqrt{n-1}$$\geq p$ is not implied by Legendre's conjecture.

Have I correctly stated these two things and are they both proven facts? If not, then the answer to my question would be a correction of these two statements. Does there exist such an inequality that would both be implied by and prove Legendre's conjecture? Is this impossible because Legendre's conjecture does not imply any bound on prime gaps?

So Legendre's conjecture can be proven if a tight enough bound were proven, but no bound can be proven by proving Legendre's conjecture. Is this right?

Is it the case that a bound on prime gaps above $n$ could prove Legendre's conjecture up to a point and then beyond that, assuming Legendre's conjecture, as $n$ goes to infinity, the bound would not always work as the gaps between consecutive perfect squares grow?

Legendre's conjecture (the assertion that there is a prime between every two perfect squares) implies that there's always a prime between $n$ and $n+4\sqrt{n-1}$. (The worst case would be if $n$ were one more than a perfect square $m^2$, and the next prime weren't until $(m+2)^2-1$. Small constants could be improved throughout.)
On the other hand, if it's true that there's always a prime between $n$ and $n+2\sqrt n$, then there would always be a prime between two perfect squares. (Take $n=m^2$.)
In fact, we believe it's true that for any $\epsilon>0$, there's always a prime between $n$ and $n+\epsilon\sqrt n$ when $n$ is sufficiently large. Indeed, we believe there's a constant $C$ such that there's always a prime between $n$ and $n+C\log^2n$. So in my opinion, it's not an important question whether one exact inequality is equivalent to Legendre's conjecture, when both of them are probably weaker than the truth.