The primes $p$ are, of course, in one-to-one correspondence with the squares of primes $p^2$. But is there any interval $a < x < b$ possible where the primes thin out so much, that it contains more squares of primes than primes?

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    $\begingroup$ Trivially, yes. Any interval where $a = p^2-1$ and $b = p^2+1$, for some prime $p$. $\endgroup$ – LarsH Jan 17 '16 at 3:36
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    $\begingroup$ @Ross: Maybe you can make it less easy by requiring that the interval contains $n$ primes with $n>>0$. $\endgroup$ – Tito Piezas III Jan 17 '16 at 3:57
  • $\begingroup$ How embarrassing. Oh well. $\endgroup$ – Ross Presser Jan 18 '16 at 13:31
  • $\begingroup$ @TitoPiezasIII: Having been spanked on the simpleminded phrasing of the question I'm hesitant to try again. $\endgroup$ – Ross Presser Jan 18 '16 at 13:32
  • $\begingroup$ I'd be interested to know the motivation for this question. $\endgroup$ – Josh Chen Jan 20 '16 at 10:38

The interval $23 \lt x \lt 29$ is one such

  • $\begingroup$ And $47<x<53$, etc. $\endgroup$ – Tito Piezas III Jan 17 '16 at 3:52

I think the intent of your question was that the number of primes in the interval, call it $\pi(n)'$, is non-zero. If so, then the simplest case is,

$$p_n^2\leq x \leq p_{n+1}^2$$

for prime $p_i$. Your question then assumes the count as $\pi(n)'<2.$ However,

  1. A consequence of Legendre's conjecture is that the number of primes in that interval is at least 2.
  2. More strongly, if Brocard's conjecture is true then, for $n>1$, there are at least 4.

As the count $\pi(n)' =2, 5, 6, 15, 9, 22, 11, 27, 47,\dots$ (A050216) goes up fast, it is highly doubtful that Brocard's conjecture is false.

  • $\begingroup$ I believe Bertrand's postulate would also be relevant. $\endgroup$ – Tito Piezas III Jan 20 '16 at 10:07
  • $\begingroup$ No, in fact when I asked my question, I was imagining a desert containing no primes, not even one, but where at least two prime squares fall. You're saying that Legendre's conjecture makes that impossible. $\endgroup$ – Ross Presser Jan 20 '16 at 15:11
  • $\begingroup$ And (after reading it) I see Bertrand's postulate, which is proven, completely rules out my scenario. $\endgroup$ – Ross Presser Jan 20 '16 at 15:15
  • $\begingroup$ @RossPresser: Yes, Legendre's (and Brocard's) conjecture make it impossible since they postulate at least $2$ and $4$ primes, respectively, in that interval. $\endgroup$ – Tito Piezas III Jan 20 '16 at 15:46
  • $\begingroup$ Those two are conjectures, but Bertrand's postulate is proven. $\endgroup$ – Ross Presser Jan 20 '16 at 15:58

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