\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Twitch 067.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 67}
\maketitle
\section*{Problem}
Let $m$ be an odd positive integer.
Show that there exist infinitely many positive integers
$n$ where $mn+1$ divides $2^n-1$.
\section*{Video}
\href{https://www.youtube.com/watch?v=9tL7vedNX_M&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/9tL7vedNX\_M}}
\newpage
\section*{Solution}
We will restrict our attention to $n$ such that $mn+1 = p$ is prime.
In other words, we are hoping for
\[ p \mid 2^{\frac{p-1}{m}} - 1. \]
In fact, we are going to prove there are infinitely many primes $p$ such
that \[ X^m - 2 \in \FF_p[X] \] splits completely.
For this, it's sufficient to let $E = \QQ(\sqrt[m]{2})$,
let $K$ be its Galois closure, and use the following theorem.
\begin{theorem}
[Chebotarev density]
Each conjugacy class $C$ of $G = \Gal(K/\QQ)$,
is obtained as the Frobenius above $p$
for a density $|C|/|G|$ of rational primes $p$.
\end{theorem}
In particular, there are infinitely many primes $p$ such that
the Frobenius above $p$ is the identity element,
which solves the problem.
\end{document}