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)?
 A: $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.
A: $$97\equiv 1\pmod8$$
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 $\pm 1\pmod{8}.$
A: Hint:
$2^{24}+1 = 16777217 = 172961 \times 97$
A: 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 $\left(\dfrac{2}{p}\right)=(-1)^{(p^2-1)/8}.$
