\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Shortlist 2004 A2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 7}
\maketitle
\section*{Problem}
Let $a_0$, $a_1$, $a_2$, \dots
be an infinite sequence of real numbers
satisfying the equation
\[ a_n = \left\lvert a_{n+1}-a_{n+2} \right\rvert \]
for all $n \ge 0$,
where $a_0$ and $a_1$ are two different positive reals.
Can this sequence $a_0$, $a_1$, $a_2$, \dots be bounded?
\section*{Video}
\href{https://www.youtube.com/watch?v=ESnvvXn0B6Q&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/ESnvvXn0B6Q}}
\newpage
\section*{Solution}
Answer: the sequence cannot be bounded.
By induction one may verify easily that:
\begin{itemize}
\ii all terms of the sequence are positive;
\ii no two adjacent terms are equal;
\end{itemize}
Call an index $n$ \emph{nice} if $a_n < a_{n+1}$.
\begin{claim*}
If $n$ is not nice, then $n+1$ must be.
\end{claim*}
\begin{proof}
If $a_n > a_{n+1}$ then $a_{n+2} = a_n + a_{n+1} > a_{n+1}$.
\end{proof}
If $n$ is a nice index,
then we define $\psi(n) = (a_n,a_{n+1}-a_n)$,
a pair of two positive real numbers.
Equivalently, by $\psi(n) = (x,y)$
we mean that $a_n = x$ and $a_{n+1} = x+y$.
\begin{claim*}
If $n$ is nice and $\psi(n) = (x,y)$
then either
\begin{itemize}
\ii $n+1$ is nice and $\psi(n) = (x+y,x)$; or
\ii $n+2$ is nice and $\psi(n) = (y,x+y)$.
\end{itemize}
\end{claim*}
\begin{proof}
This follows by computation.
Let $a_n = x$ and $a_{n+1} = x+y$.
Then $a_{n+2} = a_{n+1} \pm a_n$.
\begin{itemize}
\ii If the sign is $+$,
we get $a_{n+2} = 2x+y = (x+y)+x$.
\ii If the sign is $-$,
we get $a_{n+2} = y$,
and then $a_{n+3} = (x+y)+y$. \qedhere
\end{itemize}
\end{proof}
Thus the sequence of $\psi$-pairs obviously grows
without bound starting from the first nice index.
\end{document}