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If $p$ and $q$ are primes such that $p-q=2$, will $pq=36x^2-1$ be always true for some natural number $x$?

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  • $\begingroup$ I believe it's not true for $3$ and $5$? $\endgroup$ – Platehead Nov 6 '14 at 7:02
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    $\begingroup$ But it is true for everybody else. $\endgroup$ – André Nicolas Nov 6 '14 at 7:04
  • $\begingroup$ I just want to inform everybody that this is not supposed to be on the site any more. I had tried earning reputation by answering my own question (or sockpuppeting, as Alexander Gruber, the moderator who suspende my account for a week, called it). I've not done any such thing ever since, so I hope there won't be any trouble because this is still on the site. $\endgroup$ – ghosts_in_the_code Dec 22 '14 at 9:59
  • $\begingroup$ I assumed it had been deleted; but today I found that I was still earning reputation from an upvote on it, so I decided to write here. $\endgroup$ – ghosts_in_the_code Dec 22 '14 at 10:00
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Every natural no, 5 onwards exists in 1 of the following 6 forms: $6x-1,6x,6x+1,6x+2,6x+3,6x+4$ for some natural no. x

Of these, $6x-1$ and $6x+1$ are the only ones that can possibly be prime. As both the primes can not be of the same form (difference would be a multiple of 6),

$p=6x-1$ and $q=6x+1$

$pq=36x^2-1$

This rule works only from 5 onwards, so there is an exception, which is (3,5) whose product, 15 can not be written as $36x^2-1$

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