\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Brazil 2017/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 26}
\maketitle

\section*{Problem}
Let $a$ be a positive integer
and $p$ a prime divisor of $a^3-3a+1$,
with $p \neq 3$.
Prove that $p$ is of the form $9k+1$ or $9k-1$,
where $k$ is integer.

\section*{Video}
\href{https://www.youtube.com/watch?v=A3AQVRvVk3g&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/A3AQVRvVk3g}}

\section*{External Link}
\url{https://aops.com/community/p9495218}

\newpage

\section*{Solution}
Write $a = x + \frac 1x$ for some $x \in \FF_{p^2}$.
\begin{claim*}
  The element $x$ has order $9$.
\end{claim*}
\begin{proof}
  Because
  \begin{align*}
    0 &= \left( x + \frac 1x \right)^3
    - 3 \left( x + \frac 1x \right) + 1 \\
    &= x^3 + x^{-3} + 1 = \frac{x^6+x^3+1}{x^3}.
  \end{align*}
  This implies $x^9=1$, so $x$ has order dividing $9$.
  However, $x^3 \neq 1$ since $p > 3$.
  Therefore, $x$ has order exactly $9$.
\end{proof}

Thus $9 \mid p^2-1$ so we're done.

\begin{remark*}
  Useful example for debugging: $a=7$ gives $p=17$.
  (Sometimes people think $9 \mid p-1$ should follow.)
\end{remark*}

\end{document}
