4.1. Let $p$ be an odd number with exactly $n$ distinct prime divisors. How many solutions does the equation $p^{2}+b^{2}=c^{2}$ have with coprime $b$ and $c$ (i.e., primitive Pythagorean triples $(p, b, c)$)?
4.1. Let's rewrite the equation as $p^{2}=(c-b)(c+b)$. If we denote $c-b=u, c+b=v$, then $c=(u+v) / 2, b=(v-u) / 2$, so $u$ and $v$ must be coprime, meaning each of the $n$ prime divisors of $p$ divides exactly one of the numbers $u, v$. The number of ways to partition $n$ prime divisors into two subsets is $2^{n-1}$, which is the answer to the problem.