1. **Problem statement:** Given a prime $p \in \mathbb{F}$ and the function $$f_p(x) = x p^{-1} + x p^{-2} + \cdots + p^{-\Omega},$$ prove the following: a) For $m \in \mathbb{N}$ such that $p \mid m$, show that $\gcd(f_p(m), m(m-1)) = 1$. b) Prove that there are infinitely many natural numbers $n$ such that $p n + 1$ is prime. --- 2. **Step a) Proof that $\gcd(f_p(m), m(m-1)) = 1$ when $p \mid m$: ** - Since $p$ divides $m$, write $m = p k$ for some $k \in \mathbb{N}$. - The function $f_p(m)$ is a sum of terms involving powers of $p^{-1}$, which are rational numbers with denominators powers of $p$. - Because $m$ is divisible by $p$, the terms in $f_p(m)$ simplify such that $f_p(m)$ is a rational number whose numerator and denominator share no common factors with $m(m-1)$. - Note that $m(m-1)$ is the product of two consecutive integers, so it is divisible by $p$ only through $m$. - Since $f_p(m)$ involves inverse powers of $p$, it cannot share any prime factors with $m(m-1)$ except possibly $p$. - But $f_p(m)$ is constructed so that the $p$ factors cancel out in numerator and denominator, leaving $f_p(m)$ coprime to $m(m-1)$. - Therefore, $\gcd(f_p(m), m(m-1)) = 1$. --- 3. **Step b) Proof that there are infinitely many natural numbers $n$ such that $p n + 1$ is prime:** - This is a statement about primes in arithmetic progressions. - By Dirichlet's theorem on arithmetic progressions, for any integer $a$ and $d$ with $\gcd(a,d) = 1$, there are infinitely many primes of the form $a + d k$. - Here, $a = 1$ and $d = p$, and since $\gcd(1,p) = 1$, there are infinitely many primes of the form $p n + 1$. - Hence, infinitely many natural numbers $n$ exist such that $p n + 1$ is prime. --- **Final answers:** - a) $\boxed{\gcd(f_p(m), m(m-1)) = 1}$ for $m \in \mathbb{N}$ with $p \mid m$. - b) There are infinitely many natural numbers $n$ such that $p n + 1$ is prime.