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)?