Of the first $9.4 \times 10^9$ primitive Pythagorean triplets (generated using $s^2 - r^2, 2rs, s^2 + s^2$), nearly $16.5\%$ are such that the sum of the two orthogonal sides is a primes i.e. $s^2 - r^2 + 2rs = p$ for some prime $p$.
Question 1: Are there infinitely many primes of this form?
Question 2: If $f(x)$ and $p(x)$ be the number of such primes and the number of primitive triplets for $s \le x$ then is it true that $\dfrac{f(x)}{p(x)} \sim \dfrac{2}{\log x}$.
This can be written as $(s+r)^2-2r^2=p.$
There are infinitely many primes $p$ with solutions $(u,v)$ to $u^2-2v^2=p,$ because $\mathbb Z[\sqrt2]$ is a principal ideal domain and $-1=u^2-2v^2$ has a solution.
In particular, $p$ can be written in this form if and only if $2$ is a square modulo $p,$ if and only if $p=2$ or $p\equiv\pm 1\pmod{8}.$
However, you also need $(r,s)=(v,u-v)$ to have $r<s,$ or $2v<u.$ Showing you can always find such a case for the above $p$ takes a little more work. In particular, unless you allow $r=s,$ you can’t get $p=2.$
Claim: If $u^2-2v^2=p$ has a solution for positive integer $p,$ then there is a solution with $u\geq 2v.$
Proof: Take $u$ the smallest positive integer with a positive integer solution $(u,v)$, and assume $u<2v.$ You also have $u>\sqrt 2v,$ or $u^2-2v^2<0.$
But if $(u,v)$ is a solution, then $(u’,v’)=(3u-4v,|3v-2u|)$ is a solution and $u’=3u-4v>(3\sqrt 2 -4)v>0$ and $u’=u+2(u-2v)<u,,$ contradicting $u$ was our least positive solution. Therefore, $u\geq 2v.$
So, any prime $p\equiv\pm 1\pmod 8$ can be represented as such a sum. $2$ can be represented as a such a sum if $r=s=1$ is allowed. No others primes can be.
I believe it is true that if there is a solution, then there is exactly one pair $(r,s)$ with $s\geq r>0.$