<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] An elementary method follows:
Let $S$ denote the set $\{1,\cdots,48\}.$ Then every integer $n$ satisfies $n\equiv\pm k\pmod{97}$ for some $k\in S$ and a suitable choice of the sign. Now we consider $2k$ for $k\in S.$
For $k=1,\cdots,24,$ it is clear that $2k\in S$ so that we can take $2k\equiv l\pmod{97}$ for $l=2k.$
For $k=25,\cdots,48,$ clearly $97-2k\in S,$ so that $2k\equiv -l\pmod{97}$ where $l=97-2k.$
Thus $$\begin{align}
2^{48}\prod\limits_{k\in S}k&\equiv\prod\limits_{k\in S}2k\\
&=\prod\limits_{1\le k\le24}2k\prod\limits_{25\le k\le48}2k\\
&\equiv\prod\limits_{1\le k\le24}l\prod\limits_{25\le k\le48}(-l)\\
&\equiv(-1)^{24}\prod\limits_{k\in S}k\pmod{97}
\end{align}$$
This shows that $2^{48}\equiv1\pmod{97}.$  
Hope this helps.  
Edit:
In fact this argument can also be generalized to prove that $\left(\dfrac{2}{p}\right)=(-1)^{(p^2-1)/8}.$
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