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\title{IMO 2023/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 129}
\maketitle
\section*{Problem}
Determine all composite integers $n>1$ that satisfy the following property:
if $d_1 < d_2 < \dots < d_k$ are all the positive divisors of $n$ with
then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \leq i \leq k - 2$.
\section*{Video}
\href{https://www.youtube.com/watch?v=Z7G7bSGfrqU&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/Z7G7bSGfrqU}}
\section*{External Link}
\url{https://aops.com/community/p28097575}
\newpage
\section*{Solution}
The answer is prime powers.
\paragraph{Verification that these work}
When $n = p^e$, we get $d_i = p^{i-1}$.
The $i$\ts{th} relationship reads \[ p^{i-1} \mid p^i + p^{i+1} \]
which is obviously true.
\paragraph{Proof that these are the only answers}
Conversely, suppose $n$ has at least two distinct prime divisors.
Let $p < q$ denote the two smallest ones,
and let $p^e$ be the largest power of $p$ which both divides $n$
and is less than $q$ (so $e \ge 1$).
Then the smallest factors of $n$ are $1$, $p$, \dots, $p^e$, $q$.
So we are supposed to the relation that
\[ \frac{n}{q} \mid \frac{n}{p^e} + \frac{n}{p^{e-1}}
= \frac{(p+1)n}{p^e} \]
which means that the ratio
\[ \frac{q(p+1)}{p^e} \]
needs to be an integer, which is obviously not possible.
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