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

\begin{document}
\title{IMO 1998/4}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 102}
\maketitle

\section*{Problem}
Determine all pairs $(x,y)$ of positive integers
such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$.

\section*{Video}
\href{https://www.youtube.com/watch?v=Gv7-k9dXa6s&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/Gv7-k9dXa6s}}

\section*{External Link}
\url{https://aops.com/community/p124428}

\newpage

\section*{Solution}
The answer is $(7k^2,7k)$ for all $k \ge 1$,
as well as $(11,1)$ and $(49,1)$.

We are given $xy^2+y+7 \mid x^2y+x+y$.
Multiplying the right-hand side by $y$ gives
\[ xy^2+y+7 \mid x^2y^2+xy+y^2 \]
Then subtracting $x$ times the left-hand side gives
\[ xy^2+y+7 \mid y^2-7x. \]
We consider cases based on the sign of $y^2=7x$.
\begin{itemize}
  \ii If $y^2 > 7x$, then $0 < y^2-7x < xy^2+y+7$,
  contradiction.
  \ii If $y^2=7x$, let $y = 7k$, so $x = 7k^2$.
  Plugging this back in to the original equation reads
  \[ 343k^4 + 7k + 7 \mid 343k^5 + 7k^2 + 7k \]
  which is always valid, hence these are all solutions.
  \ii  If $y^2 < 7x$, then $|y^2-7x| \le 7x$,
  so $y \in \{1,2\}$.

  When $y=1$ we get
  \[ x+8 \mid x^2+x+1 \iff x+8 \mid 64-8+1=57.\]
  This has solutions $x=11$ and $x=49$.

  When $y=2$
  \begin{align*}
    4x+9 \mid 2x^2+x+2 \\
    \implies 4x+9 &\mid 16x^2+8x+16 \\
    \implies 4x+9 &\mid 81-18+16 = 79
  \end{align*}
  which never occurs.
\end{itemize}

\end{document}
