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 ?
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4$\begingroup$ Can anyone give a direct combinatorial proof of this? $\endgroup$– quantaCommented May 9, 2011 at 23:03
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$\begingroup$ Would $\pi(x) \ge \log x / \log 2$ be enough? $\endgroup$– quantaCommented May 9, 2011 at 23:33
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$\begingroup$ 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. $\endgroup$– user999691Commented Jul 10, 2022 at 5:43
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$\begingroup$ IMPORTANT NOTE : Oppermann's conjecture is not proven so if it become false then the above prove will not be considered as answer. $\endgroup$– user999691Commented Jul 10, 2022 at 5:44
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1 Answer
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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 $p^2$.
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12$\begingroup$ A much harder, indeed, still wide open question is whether for every $n$ there's a prime between $n^2$ and $(n+1)^2$. $\endgroup$ Commented May 9, 2011 at 23:48
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$\begingroup$ Yes, but if you put 3 instead 2, the result it's true. $\endgroup$– leoCommented May 10, 2011 at 4:36
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1$\begingroup$ Is anything known about the number of primes in such intervals? Does it diverge to infinity for instance? $\endgroup$ Commented May 10, 2011 at 8:07
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$\begingroup$ @Tobias There's approximation of number of primes less than given
x: en.wikipedia.org/wiki/Prime_counting_function $\endgroup$ Commented May 13, 2011 at 14:55