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\begin{document}
\title{Sweden 2018}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 72}
\maketitle
\section*{Problem}
For which positive integers $n$ is the polynomial
\[ p(x) = 1+x^n+x^{2n} \]
reducible over the integers?
\section*{Video}
\href{https://www.youtube.com/watch?v=drrHDLWR6bw&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/drrHDLWR6bw}}
\newpage
\section*{Solution}
The answer is only $n$ a power of $3$ (including $1$).
Indeed, letting $\Phi_\bullet$ denote the cyclotomic polynomial,
we have the decomposition of $p$ into irreducibles is
exactly given by
\[ p(x) = \frac{x^{3n}-1}{x^n-1}
= \frac{\prod_{d \mid 3n} \Phi_{d}(n)}{\prod_{d \mid n} \Phi_d(n)}
= \prod_{\substack{d \mid 3n \\ d \nmid n}} \Phi_{d}(n). \]
Hence the answer is those $n$ for which
there is only one $d$ satisfying $d \mid 3n$, $d \nmid n$.
This is exactly the powers of $3$.
\end{document}