Find the factor
Find a factor of $2^{48}-1$.
$2^{48} − 1$ is divisible by two whole numbers between $60$ and $70.$ Find the smaller of these.
This problem is taken from the World Mathematics Championships
Answer: $63$
Use $n^2-1=(n+1)(n-1)$ for all values of $n$
$2^{48}=\left(2^{24}\right)^2$, so $$\begin{align}
2^{48}-1&=\left(2^{24}\right)^2-1\\
&=\left(2^{24}+1\right)\left(2^{24}-1\right)\\
&=\left(2^{24}+1\right)\left(\left(2^{12}\right)^2-1\right)\\
&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{12}-1\right)\\
&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(\left(2^{6}\right)^2-1\right)\\
&=\left(2^{24}+1\right)\left(2^{12}+1\right)\left(2^{6}+1\right)\left(2^{6}-1\right)\\
\end{align}$$
$2^4=16, 2^5=32, 2^6=64$
$2^{48}-1=\left(2^{24}+1\right)\left(2^{12}+1\right)\ \underbrace{\left(2^{6}+1\right)}_{65}\ \underbrace{\left(2^{6}-1\right)}_{63}$
The smaller one is $63$