Probability property that the longest side of primitive Pythagorean triples is prime If we consider the set of the first $n$ primitive Pythagorean triples, then the probability that the triple's longest side is prime is approximately $\dfrac{1}{\log_{11.475}n}$ based on Mathematica’s $\tt{FindFit}$ and some code of mine.
Is this result known? If so, is the base of this logarithm in closed form known?
(More information about how this approximation was generated is here.)
 A: I don't have an answer, but here's some relevant information:
All pythagorean triples can be represented as, given $\alpha,\beta\in\mathbb{N}$, $a=\alpha^2-\beta^2$, $b=2\alpha\beta$, and $c=\alpha^2+\beta^2$. I assume you are talking about irreducible triples, so we can set $GCD(\alpha,\beta)=1$.  So, your question is really the probability of a sum of two coprime squares being prime.
Notes on sums of squares and number theory:
Sums of squares are actually closed under multiplication: $$(i^2+j^2)(k^2+l^2)=(ik-jl)^2+(il+jk)^2$$ and you can check that the two resultant squares will also be coprime.  Moreover, a sum of two squares can never be a prime that is $3\pmod{4}$ (or any number $3\pmod{4}$ for that matter), while every prime $p\equiv 1\pmod{4}$ can be expressed as a sum of $2$ squares.  This means that the problem further reduces to finding the distribution of primes in numbers not divisible by odd exponents of numbers that are equivalent to $3\pmod{4}$ (sums of squares can be divisible by the square of primes that are $3\pmod{4}$).
