Clearly, a prime fits the criteria if the result of $\sqrt{24x+1}$ is an integer. By trial and error, I have found that seemingly the only primes to fit this criteria are 2, 5 and 7. How would I go about proving that they are the only ones (or, alternatively, that $x$ must be below a certain value and the only primes below this value that fit the criteria are 2, 5 and 7)?

I've gotten as far as stating that for some integer $a$:

$24x + 1 = a^2$

Then, I rearranged this to give:

$ 24x = a^2 - 1\\ 24x = (a+1)(a-1) $

I'm not quite sure where to go from here in order to complete the proof that $x$ cannot be above a certain value. Any help would be much appreciated! I'd prefer hints on where to go next rather than full solutions since I'd much rather reach the full solution myself.

  • $\begingroup$ $x$ is a prime and divides $(a+1)(a-1)$ by your last equality. Therefore ... HTH, AB, $\endgroup$ – martini Mar 7 '12 at 14:38
  • 1
    $\begingroup$ @martin HTH, AB? $\endgroup$ – Graphth Mar 7 '12 at 14:39
  • $\begingroup$ @Graphth: For HTH see en.wikipedia.org/wiki/HTH, AB is short for "Allzeit bereit" which is a German variant of "Be prepared" (en.wikipedia.org/wiki/Scout_Motto), AB, martini. $\endgroup$ – martini Mar 7 '12 at 14:44
  • $\begingroup$ You are very much on the right track. You might want to suppose that $x >7$ and try to derive a contradiction. You might also use the fact that $x$ can't divide both $a+1$ and $a-1$ if $x >7.$ $\endgroup$ – Geoff Robinson Mar 7 '12 at 14:45
  • $\begingroup$ For an approach with lots of writing and little thinking, try just writing out all the possible factorizations of $24x$ into two integers (remember $x$ is prime, so there aren't many) and see what happens if you let one factor be $a+1$ and the other be $a-1$. $\endgroup$ – Chris Eagle Mar 7 '12 at 16:25

Your approach leads to a solution. As a continuing reminder that $x$ is supposed to be prime, let's call it $p$. We have $$24p=(a-1)(a+1).$$ Since $p$ divides the left-hand side, $p$ divides the right-hand side. It follows that (i) $p$ divides $a-1$ or (ii) $p$ divides $a+1$. (We are not excluding the possibility that $p$ divides both.)

Case (i): Suppose that $p$ divides $a-1$. Then $a-1=pk$ for some integer $k$. It follows that $a+1=pk+2$, and therefore $$24p=(pk)(pk+2).$$ By cancellation, we conclude that $$24=k(pk+2).$$ Now case (i) is in principle finished. We must have $pk+2 \le 24$, so $pk \le 22$. In particular, $p \le 19$, so it is a question of checking a small number of possibilities. The checking can be done efficiently, or not so efficiently.

Case (ii): It's your turn!

  • $\begingroup$ Thank you very much, I see it now! $\endgroup$ – EdoDodo Mar 8 '12 at 15:57

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.