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

\begin{document}
\title{Shortlist 2007 N1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 77}
\maketitle

\section*{Problem}
Find all pairs of positive integers $(a, b)$
such that $7^a - 3^b$ divides $a^4 + b^2$.

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

\section*{External Link}
\url{https://aops.com/community/p1106459}

\newpage

\section*{Solution}
The only answer is $(a,b) = (2,4)$,
which works as $7^2 - 3^4 = -32$ divides $2^4 + 4^2 = 32$.

To show this is the only solution, we note the following:
\begin{itemize}
  \ii The left-hand side is always even, which implies $a$ and $b$
  have the same parity.
  \ii This in turn implies the left-hand side is $0 \bmod 4$.
  \ii Hence $a^4+b^2 \equiv 0 \pmod 4$ which means $a$ and $b$ are even.
  \ii This means that $7^a -3^b \equiv 0 \pmod 8$,
  which further means $b \equiv 0 \pmod 4$.
\end{itemize}

This means we may let $a = 2x$ and $b = 2y$
with $y$ even to obtain
\[ \left( 7^x - 3^y \right)\left( 7^x + 3^y \right) \mid (2x)^4 + (2y)^2. \]

To reduce to finite casework we contend:
\begin{claim*}
  $2 \cdot 3^a > 16a^4 + 4a^2$ once $a \ge 10$.
\end{claim*}
\begin{proof}
  $3^{10} = 243^2 = 2 \cdot59049 = 177147 > 160400$.
\end{proof}
Hence we are done if $\max(x,y) \ge 10$. Next
\begin{claim*}
  $7^x > 16x^4 + 4 \cdot 10^2$ for $x \ge 6$
\end{claim*}
\begin{proof}
  $7^6 = 343^2 > 100000 > 16 \cdot 6^4 + 400$.
\end{proof}
So we may assume $x \le 5$ as well.

This gives us the following cases.
In all situations, note that $(2 \cdot 5)^4 + (2 \cdot 10)^2 = 10400$
and consequently if $(7^x-3^y)(7^x+3^y) > 10400$
there is nothing to verify.
These cases are marked as ``big''  in the table.
\[
\begin{array}{rr|ccccc}
  && 7 & 49 & 343 & 2401 & 16807 \\
  && x=1 & x=2 & x=3 & x=4 & x=5 \\ \hline
  9 &y=2 & \text{OK} & 40 \cdot 58 \nmid 272 & \text{big} & \text{big} & \text{big} \\
  81 &y=4 & 74 \cdot 88 \nmid 48 & 32 \cdot 130 \nmid 320 & \text{big} & \text{big} & \text{big} \\
  729 &y=6 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\
  6561 &y=8 &  \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\
\end{array}
\]

\end{document}
