Legendre Symbol $\left(\frac{83}{127}\right)$ I'm failing to understand what I'm doing wrong here, I am trying to
calculate the Legendre Symbol $\left(\frac{83}{127}\right)$ and I'm ending up with $1$
and not $-1$.
$\left(\frac{83}{127}\right) = -\left(\frac{127}{83}\right)$ and $\left(\frac{127}{83}\right) =\left(\frac{44}{83}\right)$.
Then  $\left(\frac{44}{83}\right)= \left(\frac{4 \cdot 11}{83}\right)= \left(\frac{11}{83}\right) \left(\frac{4}{83}\right)$.
$\left(\frac{11}{83}\right)=-\left(\frac{83}{11}\right)$ and $\left(\frac{83}{11}\right) = \left(\frac{6}{11}\right)$.
$\left(\frac{6}{11}\right)=\left(\frac56\right) = \left(\frac65\right)=\left(\frac15\right) = 1$.
 A: You start having problems once you get to $\left(\frac{6}{11}\right)$ and want to turn it into $\left(\frac{11}{6}\right)$. 
First of all, if you're sticking to the Legendre symbol, then $\left(\frac{11}{6}\right)$ is not even defined, because $6$ is not prime. You should continue by factoring $\left(\frac{6}{11}\right) = \left(\frac{2}{11}\right) \left(\frac{3}{11}\right)$ and then computing each of the factors. (The first is $-1$, the second is $1$.)
If you decide you want to work with the Jacobi symbol, many steps like this are alright; for example, you could turn $\left(\frac{9}{11}\right)$ into $(-1)^{\frac{9-1}{2} \cdot \frac{11-1}{2}}\left(\frac{11}{9}\right) = \left(\frac{11}{9}\right) = \left(\frac{2}{9}\right)$ and that would be perfectly okay. However, even for the Jacobi symbol, $\left(\frac{11}{6}\right)$ is still not defined, because the lower argument must always remain odd. You must still proceed by factoring $\left(\frac{6}{11}\right) = \left(\frac{2}{11}\right) \left(\frac{3}{11}\right)$. 
(In general, working with the Jacobi symbol, you still have to factor out any powers of $2$, because $2$ is a special case of quadratic reciprocity; however, you don't have to factor the upper argument in other cases, which makes it more computationally efficient for very large numbers.)
