Computing the Jacobi symbol I have been asked to compute $\left(\frac{77}{257}\right)$ specifically using Jacobi symbols, showing all working. I have done the following:
$\left(\frac{77}{257}\right) =\left(\frac{257}{77}\right)$
=$\left(\frac{26}{77}\right)$ reducing $257 \bmod77$
=$\left(\frac{2}{77}\right)$$\left(\frac{13}{77}\right)$
Then here is where I'm not sure if what I have done is correct:
$(-1)\left(\frac{13}{77}\right)$ since $2$ is a quadratic non residue $\bmod77$? (Can I do this or is there a different way?
=$-\left(\frac{77}{13}\right)$ by flipping
=$-\left(\frac{12}{13}\right)$
=$-\left(\frac{2}{13}\right)^2$$\left(\frac{3}{13}\right)$
=$-\left(\frac{3}{13}\right)$
=$-\left(\frac{13}{3}\right)$
=$\left(\frac{1}{3}\right)$
=$-(1)$ since $1$ is a quadratic residue $\bmod3$
=$-1$

We can use multiplicativity, and the Legendre symbol, since $257$ is prime.
Note that $257\equiv 1 \bmod 4$, and $7\equiv 11\equiv 3\bmod 4$. So we have
$$
\left(\frac{77}{257}\right)=\left(\frac{7}{257}\right)\cdot \left(\frac{11}{257}\right) = \left(\frac{257}{7}\right)\cdot \left(\frac{257}{11}\right)
= \left(\frac{5}{7}\right)\cdot \left(\frac{2^2}{11}\right)= \left(\frac{7}{5}\right)=\left(\frac{2}{5}\right)=-1.
$$
