\documentclass[11pt]{scrartcl}
\usepackage{evan}
\begin{document}
\title{Sylver coinage}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 95}
\maketitle
\section*{Problem}
Two players alternate naming positive integers.
The first player to name $1$ or a
sum of some previous numbers (possibly with repetition) loses.
Prove that the first player wins.
\newpage
\section*{Solution}
Call the players Alice and Bob, by tradition.
Let Alice start with some prime $p \ge 5$,
and let Bob's reply be the integer $q$ which is relatively prime
(but not necessarily itself a prime).
We now employ strategy stealing on the third turn.
Alice considers what happens if she plays $pq-p-q$.
If she still loses to a sequence of moves from Bob,
then she uses that sequence, instead of $pq-p-q$.
So in order to complete the proof we need to show:
\begin{claim}
After the first stolen move $x$ from Bob, $pq-p-q$ is invalidated anyways
by the combination of $p$, $q$, and the move $x$.
\end{claim}
\begin{proof}
Since $x$ is required to not be a linear combination of $p$ and $q$,
we have $x < pq-p-q$.
We claim that $(pq-p-q)-x$, which is nonnegative,
is in fact a sum of $p$'s and $q$'s.
That's because the Chicken McNugget theorem actually says more strongly
that for any $x \le pq-p-q$, exactly one of $x$ and $(pq-p-q)-x$ is a sum of $p$'s and $q$'s.
\end{proof}
\end{document}