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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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    Shouldn't it be migrated to Math instead of closed? :/2011-05-09
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    I think closing, instead of migrating, is not the best move...2011-05-09
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    Can anyone give a direct combinatorial proof of this?2011-05-09
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    Would $\pi(x) \ge \log x / \log 2$ be enough?2011-05-09

1 Answers 1

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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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    A much harder, indeed, still wide open question is whether for every $n$ there's a prime between $n^2$ and $(n+1)^2$.2011-05-09
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    Yes, but if you put 3 instead 2, the result it's true.2011-05-10
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    Is anything known about the number of primes in such intervals? Does it diverge to infinity for instance?2011-05-10
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    @Tobias There's approximation of number of primes less than given `x`: http://en.wikipedia.org/wiki/Prime_counting_function2011-05-13