\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{India TST 2019/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 12}
\maketitle

\section*{Problem}
Show that there do not exist natural numbers
$a_1, a_2, \dots, a_{2018}$ such that the numbers
\[ a_1^{2018}+a_2, \; a_2^{2018}+a_3, \; \dots, \; a_{2018}^{2018}+a_1 \]
are all powers of $5$.



\section*{External Link}
\url{https://aops.com/community/p21566876}

\newpage

\section*{Solution}
Assume the contrary.
\begin{claim*}
  None of the numbers are divisible by $5$.
\end{claim*}
\begin{proof}
  If $\nu_5(a_1)$ is maximal and positive,
  then $\nu_5(a_1^{2018}) > \nu_5(a_2)$,
  so $a_1^{2018} + a_2$ cannot be a power of $5$.
\end{proof}

Let $5^e$ denote the smallest of the powers of $5$,
so all elements are divisible by $5^e$.
Evidently, $e \ge 2$, and for every index $i$,
\[ a_{i+2} \equiv - a_i^{2018^2} \pmod{5^e} \]
\begin{claim*}
  We have $a_i \equiv 4 \pmod 5$ for all $i$.
\end{claim*}
\begin{proof}
  This follows immediately form the previous displayed equation.
\end{proof}

Now let $x = -a_1 \pmod{5^e}$.
But if we iterate this $1009$ times we obtain
\[ x \equiv x^{2018^{2018}} \pmod{5^e} \]
Thus $x \bmod{5^e}$ has order dividing $N = 2018^{2018}-1$.
But since $N \equiv 3 \pmod 5$, we find $\gcd(N, \varphi(5^e)) = 1$,
So this forces $x \equiv 1$.

In other words, $a_1 \equiv -1 \pmod{5^e}$,
and the same is true for any other term.
But $(5^e-1)^{2018} > 5^e$ and we have a size contradiction.

\end{document}
