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

Solution 1:

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.

Solution 2:

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.