# Is there a prime number between every prime and its square?

For each prime number $p$, is there always an other prime number between $p$ and $p^2$ ?

I tested it for prime numbers $< 500,000,000$, but I wanted to know if there is any mathematical proof of this ?

• Can anyone give a direct combinatorial proof of this? Commented May 9, 2011 at 23:03
• Would $\pi(x) \ge \log x / \log 2$ be enough? Commented May 9, 2011 at 23:33
• question is claiming about Prime number between P and $P^2$ so see when we takes P-1 = k < P so there will be possibility that prime number lies between $(P-1)^2$ and $P^2$ according to Oppermann's conjecture and this inequality : P< $(P-1)^2$ < $P^2$ always holds true So there will be always prime number between P and $P^2$ .hence proved.
– user999691
Commented Jul 10, 2022 at 5:43
• IMPORTANT NOTE : Oppermann's conjecture is not proven so if it become false then the above prove will not be considered as answer.
– user999691
Commented Jul 10, 2022 at 5:44

Yes. By Bertrand's postulate (actually a theorem), for every natural number $n$ (and thus every prime) there is a prime between $n$ and $2n$. As $p^2 \gt 2p$ for all primes $p$ greater than $2$, there is another prime in this interval, and when $p=2$, $3$ comes between $p$ and $p^2$.
• A much harder, indeed, still wide open question is whether for every $n$ there's a prime between $n^2$ and $(n+1)^2$. Commented May 9, 2011 at 23:48
• @Tobias There's approximation of number of primes less than given x: en.wikipedia.org/wiki/Prime_counting_function Commented May 13, 2011 at 14:55