Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question

migrated from stackoverflow.com May 9 '11 at 22:29

This question came from our site for professional and enthusiast programmers.

1  
Shouldn't it be migrated to Math instead of closed? :/ –  Oscar Mederos May 9 '11 at 22:00
    
I think closing, instead of migrating, is not the best move... –  Brian Stinar May 9 '11 at 22:05
3  
Can anyone give a direct combinatorial proof of this? –  quanta May 9 '11 at 23:03
    
Would $\pi(x) \ge \log x / \log 2$ be enough? –  quanta May 9 '11 at 23:33

1 Answer 1

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 $p^2$.

share|improve this answer
7  
A much harder, indeed, still wide open question is whether for every $n$ there's a prime between $n^2$ and $(n+1)^2$. –  Gerry Myerson May 9 '11 at 23:48
    
Yes, but if you put 3 instead 2, the result it's true. –  leo May 10 '11 at 4:36
1  
Is anything known about the number of primes in such intervals? Does it diverge to infinity for instance? –  Tobias Kildetoft May 10 '11 at 8:07
    
@Tobias There's approximation of number of primes less than given x: en.wikipedia.org/wiki/Prime_counting_function –  Victor Sorokin May 13 '11 at 14:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.