Is $\left(\dfrac{3299}{2663}\right) = 1$?
The solution from the book was $-1$ :(.
My attempt was:
$\left(\dfrac{3299}{2663}\right) = \left(\dfrac{636}{2663}\right) = \left(\dfrac{2^2}{2663}\right) \cdot \left(\dfrac{159}{2663}\right)$ $ = 1 \cdot \left(\dfrac{2663}{159}\right) = \left(\dfrac{119}{159}\right) = \left(\dfrac{159}{119}\right) = \left(\dfrac{40}{119}\right) = \left(\dfrac{2^3}{119}\right) \cdot \left(\dfrac{5}{119}\right)$
And $\dfrac{119^2 - 1}{8} = 1770$. Hence, $ = \left(\dfrac{5}{119}\right) = \left(\dfrac{119}{5}\right) = \left(\dfrac{2^2}{5}\right) = 1$.
Any idea?