I'm reading [this proof](https://math.stackexchange.com/a/93110/399795) 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:

1. If $q\equiv 2 \pmod{5}$ then $p = 2(5k+2) +1 = 10k + 5 \equiv 0 \pmod{5}$
2. If $q \equiv -2 \pmod{5}$ then $p = 2(5k-2) + 1 = 10k - 3 \equiv 2 \pmod{5}$