British Maths Olympiad 2006 Round 1 Problem 1

My solution to an Olympiad problem.

By Vamshi Jandhyala in mathematics

September 14, 2019

Find four prime numbers less than $100$ which are factors of $3^{32}-2^{32}$.

Solution

Factorizing,we have

$$ \begin{align} 3^{32}-2^{32} &=&(3^{16}+2^{16})((3^{16}-2^{16})\
&=&(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2}) \label{y2006r1p1e2} \end{align} $$

From Fermat’s Little Theorem, we have $3^{17-1}-1 - (2^{17-1}-1)$ is divisible by $17$.

From $\ref{y2006r1p1e2}$, we see that $(3^{32}-2^{32})$ is divisible by $5$, $13$ and $97$.