Conjecture about primes:

For all $n\notin\{4,9,30\}$ it holds that if $m\equiv p_n^2\pmod {p_{n+1}}$ where $0 \leq m < p_{n+1}$, then $m$ is a square.

Tested for $n<1,000,000$.

The only exceptions seems to be $p_4=7$, $p_9=23$ and $p_{30}=113$.

Would like help to prove, explain or contradict this.

  • 2
    $\begingroup$ What's the question? Obviously $p_n^2$ is a square. $\endgroup$ – lulu Feb 9 '18 at 20:58
  • 1
    $\begingroup$ I suspect the OP is considering "mod" as a binary operation rather than as a ternary predicate.... $\endgroup$ – Lord Shark the Unknown Feb 9 '18 at 21:01
  • $\begingroup$ @LordSharktheUnknown Sorry? Not following, $\endgroup$ – lulu Feb 9 '18 at 21:02
  • $\begingroup$ @Lehs of course it is. It is the square of $p_4$. $\endgroup$ – lulu Feb 9 '18 at 21:02
  • 1
    $\begingroup$ @Lehs Indeed. People who use "mod" as a binary operation are on my little list :-) $\endgroup$ – Lord Shark the Unknown Feb 9 '18 at 21:06

This is true if either $p_n^2<p_{n+1}$ (obviously) or if $(p_{n+1}-p_n)^2<p_{n+1}$ (because these two squares are congruent mod $p_{n+1}$, since $(-1)^2\equiv 1$). Since a number only has two square roots (or none) modulo a prime, these are the only possibilities.

So the statement is equivalent to the gaps between consecutive primes being less than the square root of the higher prime. This being true in general is basically Oppermann's conjecture (what we need is very slightly stronger, see comments), so it's a deep unsolved problem.

  • $\begingroup$ Out of interest, how does this align with the exceptions (7, 23 and 113) mentioned in the question? $\endgroup$ – Rob Arthan Feb 9 '18 at 21:15
  • $\begingroup$ Oppermann says there is a prime between $n^2$ and $n^2-n$ for integer $n$, whereas what we need is a prime between $p$ and $p-\sqrt p$. So they are not quite the same, and it's technically possible that if Oppermann's conjecture was only barely true that there would be more counterexamples than this. But it seems likely that it's true with lots to spare as you go higher up. $\endgroup$ – Especially Lime Feb 9 '18 at 21:21
  • $\begingroup$ e.g. Opperman says there is a prime between $121-\sqrt{121}$ and $121$, which there is. But what we'd need is a prime between $127-\sqrt{127}$ and $127$, which there isn't. $\endgroup$ – Especially Lime Feb 9 '18 at 21:23
  • 3
    $\begingroup$ So it's a slight strengthening of Opperman's conjecture? Maybe that's worth recording in your answer. $\endgroup$ – Rob Arthan Feb 9 '18 at 21:26
  • 1
    $\begingroup$ Note that if $k^2\equiv p_n^2$ mod $p_{n+1}$ then $(k-p_n)(k+p_n)\equiv 0$, and since $p_{n+1}$ is prime it follows that $k\equiv p_n$ or $k\equiv p_{n+1}-p_n$. $\endgroup$ – Especially Lime Feb 12 '18 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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