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What are the necessary conditions for two of the terms in the Pythagorean triplet $a^2 = b^2 + c^2$ to be prime numbers?

flag	asked Oct 5 2011 at 15:17 Vassilis Parassidis 3●1●4

quid · Accepted Answer · 2011-10-05 15:35:43Z UTC

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There is a well-known parametrization of Phythagorean triples as $k(m^2 - n^2)$, $2kmn$ , $k(m^2 + n^2)$ with positive integers $k,m,n$ and $m$ greater $n$.

Now, if two are prime we get $k=1$. And also the middle term is never prime. So the question is when are $m^2 - n^2$ and $m^2 + n^2$ both prime. The former factors as $(m-n) (m+n)$. For this to be prime we need $m-n = 1$.

So we get two of three are prime if and only if $2n+1$ and $2n^2 + 2n + 1$ are prime for some positive integer $n$.

answered Oct 5 2011 at 15:35

quid
12.9k●31●51

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...which, in turn, probably happens infinitely often, but this is unknown. – Cam McLeman Oct 5 2011 at 15:41

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@Cam: thank for adding this information. A very general conjecture from which this would follow is Schinzel's Hypothesis H en.wikipedia.org/wiki/Schinzel's_hypothesis_H – quid Oct 5 2011 at 15:46

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Also, letting $p=2n+1$ we find that $2n^2+2n+1=\frac{p^2+1}{2}$. So you're basically looking for odd primes $p$ such that $\frac{p^2+1}{2}$ is also prime. – Faisal Oct 5 2011 at 15:52

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See oeis.org/A048161 – Barry Cipra Oct 5 2011 at 16:03

prime numbers and Pythagorean triplets

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