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

\begin{document}
\title{MODS 2019/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 25}
\maketitle

\section*{Problem}
Find all positive integers $b$ such that
\[ 1 + b + b^2 + \dots + b^e \]
is a triangular number for every positive integer $e$.

\section*{Video}
\href{https://www.youtube.com/watch?v=EE3z66dcd4s&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/EE3z66dcd4s}}



\newpage

\section*{Solution}
The answer is $b=9$ only, which works since
\[ 1 + 9 + \dots + 9^e =
  \half \frac{3^{e+1}+1}{2} \frac{3^{e+1}-1}{2}. \]

To prove $b=9$ is the only one possible,
note that for $e=2$ and $e=4$
we find that
\[ 8(b+1)+1 \qquad \text{and}\qquad 8(b^3+b^2+b+1)+1 \]
are perfect squares.
Hence the product $A$ defined by
\begin{align*}
  A &\coloneqq 4096 \left[ 8(b+1)+1 \right]
    \left[ 8(b^3+b^2+b+1)+1 \right] \\
  &= 4096(8b+9)(8b^3+8b^2+8b+9) \\
  &= 4096(64b^4 + 136b^3 + 136b^2 + 144b + 81) \\
  &= 4096 \cdot \left[ \left( 8b^2 + \frac{17}{2} b + \frac{255}{64} \right)^2
  + \frac{4881b}{64} + \frac{266751}{4096} \right] \\
  &= \left( 512b^2 + 544b + 255 \right)^2
  + \left( 312384b + 266751 \right)
\end{align*}
is a perfect square divisible by $4096$.

However, $312384b+266751 < 33(1024b^2+1088b+533)$
holds for all $b \ge 10$.
So when $b \ge 10$ we have
$(512b^2+544b+255)^2 < A < (512b^2+544b+288)^2$.
Since $A$ was supposed to be a perfect square divisible by $4096$,
one would have to have $A = (512b^2+544b+256)^2$, so
\begin{align*}
  (512b^2+544b+256)^2 &= \left( 512b^2 + 544b + 255 \right)^2
    + \left( 312384b + 266751 \right) \\
  \iff 312384b + 266751 &= 1024b^2 + 1088b + 511 \\
  \iff 0 &= 1024b^2 - 311296b + 266240 \\
  \iff 0 &= b^2 - 304b + 260
\end{align*}
But the discriminant $304^2 - 4 \cdot 260$ isn't a square,
so this is a contradiction.

Hence, it suffices to exclude $b = 1, 2, \dots, 8$ exhaustively.
For $1+b$ to be a triangular number for $b \le 8$,
one needs $b \in \{2,5\}$,
but $1+2+2^2=7$ and $1+5+5^2=31$ are not triangular.
This concludes the proof.

\end{document}
