\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Iberoamerican 2020/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 49}
\maketitle

\section*{Problem}
Let $T_n$ denote the least positive integer such that
\[ n \mid 1+2+3+\dots+T_n = \sum_{i=1}^{T_n} i. \]
Find all positive integers $m$ such that $m \ge T_m$.

\section*{Video}
\href{https://www.youtube.com/watch?v=l18c28zosHk&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/l18c28zosHk}}

\section*{External Link}
\url{https://aops.com/community/p18966950}

\newpage

\section*{Solution}
The answer is $m=1$ and all $m$ that are not a power of $2$.

Fix an integer $m$ and consider the equation
\[ x(x+1) \equiv 0 \pmod{2m}. \]
The integer $T_m$ is the smallest positive solution in $x$ to this equation.

The residues $x \equiv 0 \pmod{2m}$ and $x \equiv 2m-1 \pmod{2m}$
are always solutions to this equation; we call these \emph{trivial} solutions.
Note more generally that:
\begin{claim*}
  If $r$ is a solution, then so is $(2m-1)-r$.
\end{claim*}
\begin{proof}
  Clear.
\end{proof}

This immediately implies that as long as there is a nontrivial solution,
then $T_m \le \half(2m-1)$, because the nontrivial solutions come in pairs with sum $2m-1$.

Now:
\begin{itemize}
  \ii If $m$ is a power of $2$, these are indeed the only solutions.
  So actually $T_m = 2m-1$ when $m$ is a power of $2$.

  \ii In any other case, the Chinese remainder theorem implies the number of solutions
  is $2^{k}$ where $k$ is the number of primes dividing $2m$ (without multiplicity).
  As $2^k > 2$, it follows there are nontrivial solutions, so these $m$ do not work.
\end{itemize}

\end{document}
