Understanding quadratic reciprocity I'm reading this proof and I don't think I understand the notion of quadratic reciprocity properly.
$q$ is a prime. $p = 2q + 1$.
I'm referring to section (6) which claims:

$p\equiv 2\pmod{5}$ so by quadratic reciprocity $5$ is a quadratic non-residue modulo $p$.

I know that $$\left( \frac{p}{5} \right) \cdot \left( \frac{5}{p} \right) = (-1)^{\frac{(p-1)(5-1)}{4}} = 1$$
How can deduce that $5$ is quadratic non residue from this?
Thanks
By the way,
As a side question: Is there a quick way finding out that if $q\equiv \pm 2  \pmod {5}$ then $p\equiv 0,2 \pmod{5}$?
I did this:


*

*If $q\equiv 2 \pmod{5}$ then $p = 2(5k+2) +1 = 10k + 5 \equiv 0 \pmod{5}$

*If $q \equiv -2 \pmod{5}$ then $p = 2(5k-2) + 1 = 10k - 3 \equiv 2 \pmod{5}$


As $p\equiv 2\mod 5$, $\biggl(\dfrac p5\biggr)=\biggl(\dfrac 25\biggr)=-1$, so by the quadratic reciprocity law  $\biggl(\dfrac 5p\biggr)=-1$.
Second point: if $q\equiv \pm2\mod 5$ and $p=2q+1$, then $p\equiv 2\cdot2+1\equiv 0$ or $p\equiv 2(-2)+1\equiv1+1=2$.
