Computing the Jacobi symbol

Question

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 $\bmod3$

=$-1$

Answer 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 $$ \left(\frac{77}{257}\right)=\left(\frac{7}{257}\right)\cdot \left(\frac{11}{257}\right) = \left(\frac{257}{7}\right)\cdot \left(\frac{257}{11}\right) = \left(\frac{5}{7}\right)\cdot \left(\frac{2^2}{11}\right)= \left(\frac{7}{5}\right)=\left(\frac{2}{5}\right)=-1. $$

Answer 2

Why not first decompose $\;77=7\cdot11\;$ ? I think it is much simpler:

$$\left(\frac{77}{257}\right)=\left(\frac7{257}\right)\left(\frac{11}{257}\right)$$

and now:

$$\left(\frac7{257}\right)=\left(\frac{257}7\right)=\left(\frac{5}7\right)=-1$$

and also

$$\left(\frac{11}{257}\right)=\left(\frac{257}{11}\right)=\left(\frac{4}{11}\right)=1$$

Answer 3

Harry in this case the question "use reciprocity to compute (77/257)" and "use Jakobi to compute (77/257)" can be answered in a single answer as shown above because in both cases factorizing 77 = 11*7 is the only way to simplify the symbol. You can't factor out some powers of 2 without factoring any further as usually shown in notes that want to differentiate Legendre from the more general Jacobi. So my guess is that the answer is valid for both cases