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\begin{document}
\title{Iberoamerican 2020/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 49}
\maketitle
\section*{Problem}
Let $T_n$ denotes 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}