Legendre's Conjecture limit version

Question

Legendre conjectured that there will always exist at least one prime between consecutive squares.

$\pi((n+1)^2)-\pi(n^2) \geq 1$ where $\pi(x)$ is the prime counting function.

As a research project i have been trying to prove this conjecture (I know my chances of correctly proving it are tiny but it is a good learning experience nonetheless).

Has the $\lim_{n\to\infty} \left(\pi((n+1)^2)-\pi(n^2)\right)=\infty$ been a proven result, or is that still an open question?

Thanks guys, I appreciate all of y'all taking the chance at looking at this question.

Answer 1

My answer assumes that the question is whether

$$\lim_{n\to\infty} \pi((n+1)^2)-\pi(n^2)=\infty. $$

Since in principle (because Legendre's conjecture remains open) we could have an infinite sequence of square intervals which are prime-free, the answer is no. That is, for any N however large there may be some interval with $n>N $ on which the cardinality of primes drops to zero.

In contrast, on intervals such as $(2^n,2^{n+1}),$ for which the PNT error is tractable, we can say that for n sufficiently large the prime content exceeds any pre-set number, and so the limit on such intervals exists and is $\infty.$

Answer 2

I'm not certain what your question is. If you're asking about $$ \lim_{n\to\infty}(n+1)^2-n^2 $$ then yes, this is $+\infty$ (as Jose mentions in a comment). If you're asking about the smallest interval which is known to have primes, that would be the Baker-Harman-Pintz result $$ \lim_{x\to\infty}\pi(x+x^{0.525})-\pi(x)=+\infty. $$

This is close to what you want in the sense that if 0.525 could be replaced with 0.5 you get Legendre's conjecture (for large enough $x$).

(Actually they proved a little more: $\pi(x+x^{0.525})-\pi(x)\sim x^{0.525}/\log x.$) Also worth mentioning is the Zhang-Maynard-polymath result that there are infinitely many $n$ such that $\pi(n+246)-\pi(n)\ge1,$ but this (of course) doesn't hold for all large enough $n$, unlike Baker-Harman-Pintz above.