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.

The following table indicates that there is a prime number p between the square of two consecutive primes.

$$ \displaystyle \begin{array}{rrrr} \text{n} & p_n^2 & p_{n+1}^2 & \text{p} \\ \hline 1 & 4 & 9 & 7 \\ 2 & 9 & 25 & 23 \\ 3 & 25 & 49 & 47 \\ 4 & 49 & 121 & 113 \\ 5 & 121 & 169 & 167 \\ 6 & 169 & 289 & 283 \\ 7 & 289 & 361 & 359 \\ 8 & 361 & 529 & 523 \\ 9 & 529 & 841 & 839 \\ 10 & 841 & 961 & 953 \end{array} $$

Can anyone prove that for each natural number $n$ there is always a prime number $p$, such that $p_n^2<p<p_{n+1}^2$ ?

share|improve this question
add comment

3 Answers

up vote 9 down vote accepted

Maybe. Can we prove it? The answer has to be no. Since there may be an infinite number of primes $p_{n+1}-p_n = 2,$ and since we cannot now prove that there is a prime between $n^2$ and $(n+2)^2$ for all n, the answer seems clear-cut.

share|improve this answer
    
+1 Of course, we wouldn't need to prove the existence of a prime between $n^2$ and $(n+2)^2$ for all $n$, but yes. –  Thomas Andrews Jun 6 '13 at 18:44
    
@ThomasAndrews: I could go back and edit but I think your comment covers it,thanks. –  daniel Jun 6 '13 at 18:52
2  
There is a conjecture called Brocard's conjecture that there always exist 4 primes between $p_n^2$ and $p_{n+1}^2$, which is still unresolved. This seems easier, so there might be some hope of existing a proof for this conjecture. –  MathJJ Jun 6 '13 at 19:13
add comment

This is a famous unsolved problem called Legendre's conjecture. It's 'obviously' true but very hard to prove. In some sense it is stronger than the Riemann Hypothesis (which only 'gets you' $\sqrt x\log x$ instead of $2\sqrt x$), so I wouldn't expect it to be proved soon.

share|improve this answer
2  
Legendre would imply the OP, but is the converse true? –  daniel Jun 6 '13 at 22:24
    
@daniel: Oh, you're right -- I missed (somehow) the subtlety that the question applies only to prime squares. –  Charles Jun 7 '13 at 1:05
add comment

Bertrand's Postulate (now a theorem) states that, for $n > 3$, there is always at least one prime p such that $n < p < 2n - 2$ (or for all $n > 1$, $n < p \leq 2n$). Since for all $n > 1$, $2n \leq n^2$, it seems that this is very likely to be true. This conjecture admittedly underestimates the actual number of primes by a long shot. This conjecture is not yet proven because it is not always the case that $2p_n^2 < p_{n+1}^2$ and for other reasons. In short, progress is being made towards this result, which is almost certainly true, and it is my belief that it will not remain an open question for much longer.

See the Wikipedia article below: Proof of Bertrand's Postulate

or the article from Wolfram Mathworld: Bertrand's Postulate (Chebyshev's Theorem)

share|improve this answer
    
No this argument doesn't work. Suppose that $p_{n+1}=p_n+2$ which conjecturally could happen infinitely often (Twin prime conjecture). Then $p_{n+1}^2 = p_n^2 + 4p_n + 4 < 2p_n^2-2$ for $p_n$ large. –  Michalis Jun 23 '13 at 16:40
    
Sorry; made the correction. –  Adam Jun 23 '13 at 16:42
    
Actually, if you've been following recent progress in Number Theory, a mathematician called Zhang has proven that there are infinitely many primes with $p_{n+1}-p_n<C$ (He had $C=70000000$ but this has been improved significantly in the past month). So even with our knowledge today your argument doesn't work as for those $p_{n+1}$ we have $p_{n+1}^2<p_n^2+2Cp_n+C^2<2p_n^2-2$ for $n$ large enough. –  Michalis Jun 23 '13 at 16:45
    
Again, thanks for the correction. I will edit my answer. –  Adam Jun 23 '13 at 17:06
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.