The Pythagorean triple is triple $(a,b,c)$ such that $a,b,c$ are natural numbers which satisfy the identity $a^2+b^2=c^2$.

Let us denote the set of prime numbers as $\mathbb P$.

The question is:

Are there infinitely many pairs of prime numbers $(p,q) \in \mathbb P \times \mathbb P$ such that for every pair there exist natural number $c(p,q)$ (I write $c(p,q)$ to denote the dependence of $c$ on $p$ and $q$) such that $(p,c(p,q),q)$ or $(q,c(p,q),p)$ is a Pythagorean triple?

Remark: I created this question in my mind maybe half an hour ago while I was waiting for my friend to send me a message on my mobile phone and somehow I believe that this is a known fact, but maybe I am wrong, am I?

Edit: I edited the question because Andre Nicolas clarified my thoughts as he stated in the comment that $c(p,q)$ cannot be a hypotenuse because if that is the case then there are no such triples. In the original question this part of the question "such that $(p,c(p,q),q)$ or $(q,c(p,q),p)$ is a Pythagorean triple" was "such that $(p,q,c(p,q))$ is a Pythagorean triple" (and that is the only change).


Right triangles with one leg and the hypotenuse of prime length were investigated by Dubner and Forbes.

The prime legs are listed at https://oeis.org/A048161 with the first 10000 examples at https://oeis.org/A048161/b048161.txt

The hypotenuses are listed at https://oeis.org/A067756 with the first 10001 at https://oeis.org/A067756/b067756.txt

It is conjectured that there are infinitely many of these. However, there is still no resolution of the question, are there infinitely many primes $n^2 + 1?$ I cannot imagine that any more is known about primes $(n^2 + 1)/ 2,$ where this time $n$ would be odd; evidently considered by Euler: these $n$ are listed at https://oeis.org/A002731 . Your condition actually asks about $(p^2 + 1)/ 2 = q,$ with both $p,q$ prime. No-one knows.


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