Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
Yes, See [Prime between Squares of Sequential Primes][1] for proof. [1]:… – Allyn Shell Sep 22 '15 at 19:39
up vote 11 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|cite|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
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

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|cite|improve this answer
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

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|cite|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

This question is Brocard's conjecture. See

share|cite|improve this answer

Yes. Brocard's conjecture is stronger: there are 4 prime numbers in this interval. I think to have proved in a publication available at: and for which I still wait some commentaries. You will understand that this proof being done in a 12 pages article, I will not explain it here. Regards.

share|cite|improve this answer

Well, an analytical estimate of the number of primes between the prime squares $p_{_{k+1}}^{2}$ and $p_{_{k}}^{2}$ for $k > 1$ is apparently $O(\frac{p_{_{k}}}{log_{e}p_{_{k}}})$. See the argument in Section IV on p.5 of this draft on defining prime probability analytically that I have just submitted to the journal Resonance of the Indian Academy of Sciences.

If so, a formal proof that there is always a prime between consecutive prime squares should then follow if we could show that, for $k > 1$, the difference between $\pi(n)$ and the analytic estimate $\pi_{_{L}}(n)$ of $\pi(n)$ (see Section III of the draft) is always less than $4(p_{_{k}}+1).\prod_{i = 1}^{k}(1 - \frac{1}{p_{i}}) + 1$.

share|cite|improve this answer

Your Answer


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.