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Is it true that if $p^2 - q^2$ is a perfect square where $p$ and $q$ are primes $> 5000$ then it has a prime factor greater than $17$?

Note: This question was asked in MSE but did not receive an answer. Hence posting in MO.

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    $\begingroup$ looking at MSE post, this appears to be a conjecture. This has to be spelled out, IMHO. $\endgroup$ – Dima Pasechnik Dec 28 '19 at 6:16
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    $\begingroup$ This is not an answer, but just an observation. If $p^2-q^2=a^2$, then $(a,q,p)$ form a Pythagorean triple. It follows that there are $c$ and $d$ such that $p=c^2+d^2$; $a=2cd$ and $q=c^2-d^2$. Since $q$ is supposed to be prime, $c-d$ must be 1, so $c=(q+1)/2$ and $d=(q-1)/2$. It follows that $p=(q^2+1)/2$ and $a=(q^2-1)/2$. Hence you’re asking that if $q$ is prime over 5000 and is such that $(q^2+1)/2$ is prime, must one of $q-1$ and $q+1$ have a factor greater than 17. $\endgroup$ – Anthony Quas Dec 28 '19 at 7:30
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    $\begingroup$ This seems very likely: there are something like $n^7$ numbers up to $e^n$ with all factors 17 or less, so it seems likely that there will only be finitely many pairs of such numbers differing by 2. $\endgroup$ – Anthony Quas Dec 28 '19 at 7:33
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    $\begingroup$ The title asks whether all the prime factors exceed $17$, but the body asks whether one prime factor exceeds $17$. Please edit for consistency. $\endgroup$ – Gerry Myerson Dec 28 '19 at 14:38
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    $\begingroup$ I'm voting to close this question because it is now attracting useful answers on MSE which in my view is a more fitting place for recreational number theory $\endgroup$ – Yemon Choi Dec 28 '19 at 16:42