\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Iberoamerican 2003/6}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 55}
\maketitle

\section*{Problem}
The sequences $(a_n)$, $(b_n)$ are defined by $a_0=1,b_0=4$ and for $n\ge 0$
\[ a_{n+1}=a_n^{2001}+b_n, \qquad b_{n+1}=b_n^{2001}+a_n\]
Show that $2003$ is doesn't divide any terms in either sequence.

\section*{Video}
\href{https://www.youtube.com/watch?v=d8d548yI5FA&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/d8d548yI5FA}}

\section*{External Link}
\url{https://aops.com/community/p475714}

\newpage

\section*{Solution}
We prove this the following sentence by induction:
\begin{claim*}
  For $n \ge 0$, $a_n b_n$ is a nonzero
  quadratic residue modulo $2003$.
\end{claim*}
\begin{proof}
  The base case is given.
  For the inductive step, first working modulo $2003$ write
  \begin{align*}
    a_{n+1} &= \tfrac{1}{a_n} + b_n \\
    b_{n+1} &= \tfrac{1}{b_n} + a_n
  \end{align*}
  and multiply the two equations together to get
  \[ a_{n+1} b_{n+1} = a_n b_n \left( \frac{1}{a_n b_n} + 1 \right)^2.  \]
  This proves $a_{n+1} b_{n+1}$ is a quadratic residue.

  But, since $2003$ is a $3 \bmod 4$ prime,
  $a_nb_n \not\equiv -1 \bmod{2003}$.
  Hence it follows $a_{n+1} b_{n+1} \not\equiv 0 \pmod{2003}$
  and the proof is complete.
\end{proof}

\end{document}
