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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).

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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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