\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{RSL 2018 N1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 45}
\maketitle
\section*{Problem}
Determine all polynomials $f$ with integer coefficients
such that $f(p)$ is a divisor of $2^p-2$ for every odd prime $p$.
\section*{Video}
\href{https://www.youtube.com/watch?v=YlhamV8TwSc&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/YlhamV8TwSc}}
\newpage
\section*{Solution}
The only answers are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 6$,
$\pm x$ and $\pm 2x$.
These can be seen to work, so we prove they are the only ones.
\begin{claim*}
For every odd prime $p$,
$f(p)$ only could have factors of $2$, $3$, or $p$.
\end{claim*}
\begin{proof}
Suppose $q \mid f(p)$ where $q \ge 3$ is a prime other than $p$.
Consider primes $\ell$ such that $\ell \equiv p \pmod q$
and $\ell \equiv q-2 \mod q-1$; such primes exist by Dirichlet.
Then
\[ q \mid f(\ell) \mid 2^\ell - 2 \equiv 2^{q-2} - 2 \equiv \half - 2\pmod q \]
and hence $q = 3$.
\end{proof}
\begin{claim*}
If $f(0) \neq 0$, then $f$ is constant and a divisor of $6$.
\end{claim*}
\begin{proof}
Let $p \equiv 5 \pmod 6$ be a large prime not dividing $f(0)$.
Then $f(p) \not\equiv 0 \pmod p$,
so $f(p) \mid 2 \cdot 3^\infty$ by the previous claim.
However, $2^p-2 \equiv 3 \pmod 9$ and $2^p-2 \equiv 2 \pmod 4$,
so $f(p) \mid 6$.
Since this relation holds for all
sufficiently large primes $p \equiv 5 \pmod 6$,
$f$ must be constant, and the conclusion follows.
\end{proof}
On the other hand if $f(x)$ is divisible by $x$,
redo problem with $f(x)/x$.
We can repeat this until $f$ is constant.
Hence the possible candidates are $f(x) = cx^e$ where $c \mid 6$.
We can deduce $e=1$ by simply letting $p=5$,
and the rule out $3x$ and $6x$ by letting $p=3$.
So the problem is solved.
\end{document}