From the text: Given a natural number \(r\), the possible Pythagorean triples with inradius \(r\) coincide with the possible ways of factoring \(2r^2\) into a product of two numbers \(m\) and \(n\). Lemma. Let \(r = 2^{\alpha_0}p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}\) for distinct odd primes \(p_1, p_2, \dots, p_n\) and integers \(\alpha_0\geq 0\) and \(\alpha_1,\alpha_2, \dots, \alpha_n\geq 1\). Then there are
\[ (\alpha_0 + 1)(2\alpha_1 + 1)(2\alpha_2 + 1) \cdots (2\alpha_n + 1) \]
Pythagorean triples with inradius \(r\), and \(2^n\) of these triples are primitive. The formula also holds for \(n = 0\), i.e., when \(r = 2^{\alpha_0}\) for some \(\alpha_0\geq 0\).
The result for primitive triples is well-known [N. Robbins, Fibonacci Q. 44, No. 4, 368–369 (2006; Zbl 1132.11015)], but our proof is simpler also in this case.
The technique of the author’s proof moreover generates all Pythagorean triples.