# Is there always a prime number between $p_n^2$ and $p_{n+1}^2$?

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

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.

• +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.

• 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

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

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

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