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).