\documentclass[11pt]{scrartcl}
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\begin{document}
\title{USA TST 2021/1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 56}
\maketitle

\section*{Problem}
Determine all integers $s \ge 4$
for which there exist positive integers $a$, $b$, $c$, $d$
such that $s = a+b+c+d$
and $s$ divides $abc+abd+acd+bcd$.

\section*{Video}
\href{https://www.youtube.com/watch?v=J3cOwLGB2ZY}{\texttt{https://youtu.be/J3cOwLGB2ZY}}

\section*{External Link}
\url{https://aops.com/community/p20672573}

\newpage

\section*{Solution}
The answer is $s$ composite.

\paragraph{Composite construction.}
Write $s = (w+x)(y+z)$,
where $w$, $x$, $y$, $z$ are positive integers.
Let $a=wy$, $b=wz$, $c=xy$, $d=xz$.
Then
\[ abc+abd+acd+bcd = wxyz(w+x)(y+z) \]
so this works.

\paragraph{Prime proof.}
Choose suitable $a$, $b$, $c$, $d$. Then
\[
  (a+b)(a+c)(a+d)
  = (abc+abd+acd+bcd) + a^2(a+b+c+d)
  \equiv 0 \pmod s.
\]
Hence $s$ divides a product of positive integers less than $s$,
so $s$ is composite.

\begin{remark*}
  Here is another proof that $s$ is composite.

  Suppose that $s$ is prime.
  Then the polynomial $(x-a)(x-b)(x-c)(x-d) \in \FF_s[x]$
  is even, so the roots come in two opposite pairs in $\FF_s$.
  Thus the sum of each pair is at least $s$,
  so the sum of all four is at least $2s > s$, contradiction.
\end{remark*}

\end{document}
