Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
 \implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
 \implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
 \implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$$
None of the terms on LHS seem to be  divisible by 97!!
Direct calculation shows that: $97\mid 2^{24}+1$ , but how to find it mathematically (of course not using calculator)?

$2^{48} - 1 = ( (2^{24})^2 - 1) = (2^{24} - 1)(2^{24}+1) = ( (2^{12})^2 - 1)(2^{24}+1) = (2^{12} - 1)(2^{12} + 1)(2^{24} + 1). $
Now $2^6 = 64 $ thus $2^{12} = 64^2 = 4096 = 22 \mod 97$. 
Therefore $2^{24} = (2^{12})^2 = 22^2 = 484 = 96 \mod 97$. 
Hence $2^{24} + 1 = 96 +1 = 97 = 0 \mod 97$. 
Therefore $97$ divides $2^{24} + 1$ so $97$ divides $2^{48} - 1$ as well.
