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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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    $\begingroup$ Can anyone give a direct combinatorial proof of this? $\endgroup$
    – quanta
    May 9 '11 at 23:03
  • $\begingroup$ Would $\pi(x) \ge \log x / \log 2$ be enough? $\endgroup$
    – quanta
    May 9 '11 at 23:33
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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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    $\begingroup$ A much harder, indeed, still wide open question is whether for every $n$ there's a prime between $n^2$ and $(n+1)^2$. $\endgroup$ May 9 '11 at 23:48
  • $\begingroup$ Yes, but if you put 3 instead 2, the result it's true. $\endgroup$
    – leo
    May 10 '11 at 4:36
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    $\begingroup$ Is anything known about the number of primes in such intervals? Does it diverge to infinity for instance? $\endgroup$ May 10 '11 at 8:07
  • $\begingroup$ @Tobias There's approximation of number of primes less than given x: en.wikipedia.org/wiki/Prime_counting_function $\endgroup$ May 13 '11 at 14:55

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