Problem with congruence relations Show that $97|2^{48}-1$
So far I managed to use Fermat's Little Theorem where I got
$2^{96}≡1\pmod {97}$
Which I then reconstructed as
$2^{48}*2^{48}≡1\pmod {97}$
And I got stuck here. I'm pretty sure I need to get
$2^{48}-1≡0 \pmod {97}$
as the end result, but I have no idea how to get there. Any help would be greatly appreciated.
 A: One can compute, or use a little theory. Since $97$ is of the shape $8k+1$, it follows that $2$ is quadratic residue of $97$. But then $2^{(97-1)/2}\equiv 1 \pmod{97}$.  
A: One naive way:
$$2^{48}-1=\left(2^{24}-1\right)\left(2^{24}+1\right)$$
But
$$2^{24}+1=\left(2^8\right)^3+1=\left(2^8+1\right)\left(2^{16}-2^8+1\right)$$
$$2^7=128=31\pmod{97}\Longrightarrow 2^8=62\pmod{97}\Longrightarrow $$
$$2^{16}=(2^8)^2=62^2=61\pmod{97}\Longrightarrow$$
$$2^{16}-2^8+1=61-62+1=0\pmod{97}$$
A: There are a number of results on primitive roots modulo primes, but the easiest way to do this is to just find the order of $2$ mod $97.$ The only possibilities are divisors of $96.$ First check $2^{12} \equiv 22 \mod 97$ (easy calculation) and then $2^{24} \equiv -1 \mod 97,$ so $2^{48} \equiv 1 \mod 97,$ as desired.
A: $\rm  mod\,\ {97}\!:\ 2\equiv 196\equiv 14^2.\ $ Now, taking $48$'th powers,
therefore $\rm\: 2^{48}\equiv\, (14^2)^{48}\equiv 14^{\color{}{96}}\equiv 1\,\ (mod\ 97) \ $ by little Fermat.
A: You're half right that you want to get to 
$$
2^{48} \equiv 1 \pmod {97}; 
$$
what you really want is
$$
2^{48} \equiv \pm 1 \pmod {97}
$$
To show this, we need a little more, and a contradiction-style proof works well. Suppose $2^{48} \equiv k \pmod {97}$ but 
$2^{96} \equiv 1 \pmod {97}$. This implies that $k^2 \equiv 1 \pmod {97}$ from which you can infer that $k \equiv \pm 1 \pmod {97}$ by checking the 97 cases or showing that since $k^2 \equiv 1 \pmod {97}$ then $k^2-1 = 97n$ for some $n$, but since $k^2-1 = (k+1)(k-1)$ then either $k+1$ or $k-1$ must be divisible by 97 (since 97 is prime).
