Brocard's conjecture states that: If $p_{k}$ and $p_{k+1}$ are consecutive prime numbers greater than $2$, then between $p_{k}²$ and $p_{k+1}²$ there are at least four prime numbers. I know that is statement is not yet proved. But I am asking on a weaker version:
Show that there is infinitely many indices $k$ such that between $p_{k}²$ and $p_{k+1}²$ there are at least four prime numbers.
Theorem: For any constant $c$ there are infinitely many primes $p_k$ such that there are at least $c$ primes between $p_k^2$ and $p_{k+1}^2$.
Proof: Fix a $c$. Assume that for sufficiently large $k$ there are never more than $c$ primes between $p_k^2$ and $p_{k+1}^2$. Then for sufficiently large $n$ there are never more than $c$ primes between $n^2$ and $(n+1)^2$ (since for any $n$, we have some prime $p$ with $p \leq n$ and with the next prime greater than $n+1$). Let $N_c$ be a constant such that for $n>N_c$ there are no more than $c$ primes between $n^2$ and $(n+1)^2$. Let $x$ be a sufficiently large number. Then by counting how many perfect squares are less than $x$, $\pi(x) \leq \pi(N_c) + c(1+\sqrt{x})$. This implies that $\pi(x) = O(\sqrt{x})$ which contradicts the Prime Number Theorem.
This can be adapted with a small amount of work to get actual bounds on how many primes one should get between $n^2$ and $(n+1)^2$ for sufficiently large $n$ just from the PNT. In particular, some should have around a constant time $\sqrt{x}/\log x$ primes. But there's really no content here involving primes. Any sufficiently dense sequence would satisfy this sort of thing.
If you do want the strongest result that we can use given current literature, then it follows from this paper by James Maynard that for any constant $m$, there is a constant $c_m$ such that there are infinitely many primes $p$ such that there are at least $m$ primes between $p$ and $p+c_m$. This is a much stronger result, but it also implies the result you want. The constants for Maynard's sort of results a big though and his theorems give a bound for $c_m$ which grow roughly exponentially in $m$.
