Proving that there are at least $n$ primes between $n$ and $n^2$ for $n \ge 6$

Solution 1:

We have the following result¹:

(Chebyshev, 1850) For any $X>30$, we have: $$\pi(X)\cdot\frac{\log X}{X} \in (A,B),$$ where $A=\log\left(\frac{2^{\frac{1}{2}}3^{\frac{1}{3}}4^{\frac{1}{4}}}{30^{\frac{1}{30}}}\right)\approx 0.946$ and $B=\frac{6}{5}A\approx 1.135$.

It follows that for any $X>30$ we have: $$ \pi(X^2)-\pi(X)\geq A\frac{X^2}{2\log X}-B\frac{X}{\log X}\geq\left(\frac{A}{2}-\frac{B}{30}\right)\frac{X^2}{\log X}\geq\color{red}{\frac{2}{5}\frac{X^2}{\log X}} $$ that is way greater than $X$. So we just need to check that the our claim holds in the range $[6,30]$.

^{1) it is a weaker version of the PNT, but not so weaker. Its proof only relies on the properties of the central binomial coefficients, and it is rather short and ingenious.}