I have an example in my lecture notes that says:
Check
$$\left( \frac{31}{1019} \right) = -1$$
where $\left( \frac{a}{b} \right)$ is the Legendre symbol.
I said, because $1019$ and $31%$ are both primes, we can use quadratic reciprocity and get:
$$ \left( \frac{31}{1019} \right)\left( \frac{1019}{31} \right) = (-1)^{509 \times 15} = -1$$
and so we get
$$\left( \frac{31}{1019} \right) = - \left( \frac{1019}{31} \right) = - \left( \frac{27}{31} \right).$$
Now I'm a little stuck as $27 = 3^3$ and so this isn't $+1$ and so I end up with $-(-1) = +1$ which isn't the answer. Then I thought of doing quadratic reciprocity on
$$\left( \frac{3}{31} \right)^3$$
which gives me
$$\left( \frac{3}{31} \right) \left( \frac{31}{3} \right) = (-1)^{15 \times 1} = -1$$
and so
$$\left( \frac{3}{31} \right) = - \left( \frac{31}{3} \right) = - \left( \frac{1}{3} \right) = -(+1) = -1$$
so now I have
$$- \left(-1 \right)^3 = -(-1) = +1$$
which, again, is the wrong answer.
Where have I gone wrong?
The Question seems to be wrong
$$\left(\frac{27}{31}\right)=\left(\frac{-4}{31}\right)=\left(\frac{2^2}{31}\right)\left(\frac{-1}{31}\right)=\left(\frac{-1}{31}\right)\text{ as } (\pm2)^2\equiv 4\pmod {31}$$
Again, we know, $\left(\frac{-1}p\right)=1\iff $ prime $p\equiv1\pmod 4$