\documentclass[11pt]{scrartcl}
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\begin{document}
\title{Canada 2025/2}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 164}
\maketitle

\section*{Problem}
Determine all positive integers $a$, $b$, $c$, $p$
where $p$ and $p+2$ are odd primes and
\[ 2^a p^b = (p+2)^c - 1. \]



\section*{External Link}
\url{https://aops.com/community/p34216137}

\newpage

\section*{Solution}
The answer is only $(a,b,c,p) = (3,1,2,3)$ which works.

Let $q \coloneq p+2$.
Note that the left-hand side contains a factor of $(p+2)-1 = p+1$.
Since $\gcd(p, p+1) = 1$, it follows $p+1=q-1$ must be a power of $2$, say $2^n$.

However, $2^n-1$ can be prime only when $n$ is prime
while $2^n+1$ can be prime only when $n$ is a power of $2$.
So this forces $n = 2$, i.e.\ $p = 3$ and $q = 5$.

Hence, the problem reduces to solving $2^a \cdot 3^b = 5^c - 1$.
Zsigmondy's theorem now ensures there are no solutions for $c \ge 2$,
and when $c = 1$ we get $b = 0$ which is invalid.

\end{document}
