<s>[INST] 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)?
 [/INST] $$97\equiv 1\pmod8$$
Thus, $2$ is a quadratic residue $\bmod 97$. So, there exists $a$ such that $a^2\equiv 2 \pmod {97}$. Thus, $2^{48}\equiv a^{96}\equiv 1\pmod{97}$, as desired.
This solution will work for any prime number $p$ that is $\pm 1\pmod{8}.$
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