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

\begin{document}
\title{IMC 2023/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 130}
\maketitle

\section*{Problem}
Let $A$, $B$ and $C$ be $n \times n$ matrices with complex entries satisfying
\[ A^2=B^2=C^2 \text{ and } B^3 = ABC + 2\opname{id}. \]
Prove that $A^6 = \opname{id}$.

\section*{Video}
\href{https://www.youtube.com/watch?v=JKne3A1Pu8o&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/JKne3A1Pu8o}}

\section*{External Link}
\url{https://aops.com/community/p28295670}

\newpage

\section*{Solution}
We start with the following claim:
\begin{claim*}
  We have $A=-C$ and $B^2=-AC$.
\end{claim*}
\begin{proof}
  We started with
  \begin{align*}
    B^5 &= ABCB^2 + 2B^2 = ABC^3 + 2C^2 = AB^3C + 2C^2 \\
      &= A(ABC+2)C + 2C^2 = B^5 + 2AC + 2C^2 \\
    \implies 0 &= AC+C^2.
  \end{align*}
  Following the analogous argument gives
  \[ A^2=B^2=C^2=-AC. \]
  Now we have \[ -BAC = -ACB = ABC + 2 \]
  Notice that this means that $-2 = (AB+BA)C$, so $C$ is invertible.
  Combined with the claim that $AC+C^2 = 0$, this gives $A=-C$.
\end{proof}
Now squaring both sides of $B^3 = -ABA + 2$ gives
\begin{align*}
  B^6 &= ABA^2BA - 4ABA + 4 = AB^4A + 4B^3 - 4 \\
    &= A(A^2)(A^2)A + 4B^3 - 4\\
    &= A^6 + 4B^3 + 4 = B^6 + 4B^3 + 4
\end{align*}
and hence $B^3 = -\id$, so $A^6 = B^6 = \id$.

\end{document}
