$2^{1008} \mod 2017$ 
Find $2^{1008} \mod 2017$   

From Fermat's we can find that $2^{2016} \equiv 1 \mod 2017$.
But now we cant say that $2^{1008} \equiv 1 \mod 2017$. It can be $-1$ too. But how to prove that it is equal to $1$? 
 A: One way would be to find $2^{1008} \pmod{2017}$ using the binary expansion of $1008 = 1111110000_{(2)}$.
A different method uses quadratic residues: Since 2017 is a prime $\equiv 1\pmod{8}$, 2 is a quadratic residue mod 2017. (see https://en.wikipedia.org/wiki/Quadratic_residue )
Then there exist values $r$ (more precisely, 986 and 1031) such that $r^2 \equiv 2 \pmod{2017}$, which implies that $2^{1008} \equiv r^{2016} \equiv 1\pmod{2017}$.
A: If you have Quadratic Reciprocity at your disposal, then Catalin Zara's answer is the way to go.  If not, a little experimentation (and a bit of luck) leads to the following:
$$\left(2\over2017\right)=\left(-2\over2017\right)=\left(2015\over2017\right)=\left(5\over2017\right)\left(403\over2017\right)=\left(5\over2017\right)\left(2420\over2017\right)\\=\left(5\over2017\right)\left(5\over2017\right)\left(4\over2017\right)\left(121\over2017\right)=1$$
The first step is based on the theorem that $\left(-1\over p\right)=1$ for $p\equiv1$ mod $4$.  As in Zara's answer, the final conclusion for $2^{1008}$ follows from the theorem that $r^{(p-1)/2}\equiv1$ mod $p$ when $\left(r\over p\right)=1$.
Added later:  A bit more experimentation gives a simpler evaluation, using the factorizations $8=2\cdot4$ and $2025=45\cdot45$:
$$\left(2\over2017\right)=\left(8\over2017\right)=\left(2025\over2017\right)=\left(45\over2017\right)^2=1$$
A: $986^2-2=2017\cdot482$.
Thus, $$2^{1008}-1\equiv986^{2016}-1\equiv0$$
