# If $p^2 - q^2$ is a perfect square where $p$ and $q$ are primes $> 5000$ then is one of its prime factors always greater than $17$? [closed]

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 $$17$$?

• looking at MSE post, this appears to be a conjecture. This has to be spelled out, IMHO. – Dima Pasechnik Dec 28 '19 at 6:16
• 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. – Anthony Quas Dec 28 '19 at 7:30
• 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. – Anthony Quas Dec 28 '19 at 7:33
• The title asks whether all the prime factors exceed $17$, but the body asks whether one prime factor exceeds $17$. Please edit for consistency. – Gerry Myerson Dec 28 '19 at 14:38
• 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 – Yemon Choi Dec 28 '19 at 16:42