10
$\begingroup$

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 $p_n^2<p<p_{n+1}^2$ ?

$\endgroup$
10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ +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. $\endgroup$ – Thomas Andrews Jun 6 '13 at 18:44
  • $\begingroup$ @ThomasAndrews: I could go back and edit but I think your comment covers it,thanks. $\endgroup$ – daniel Jun 6 '13 at 18:52
  • 2
    $\begingroup$ 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. $\endgroup$ – MathJJ Jun 6 '13 at 19:13
6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Legendre would imply the OP, but is the converse true? $\endgroup$ – daniel Jun 6 '13 at 22:24
  • $\begingroup$ @daniel: Oh, you're right -- I missed (somehow) the subtlety that the question applies only to prime squares. $\endgroup$ – Charles Jun 7 '13 at 1:05
0
$\begingroup$

This question is Brocard's conjecture. See https://en.wikipedia.org/wiki/Brocard%27s_conjecture.

$\endgroup$

protected by Alex M. Apr 5 '17 at 16:42

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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