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:

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 $p = 2(5k-2) + 1 = 10k - 3 \equiv 2 \pmod{5}$
• What is $q$ w.r.t. $p$? Feb 6, 2018 at 18:26
• So ... is $p$ a quadratic residue $\pmod 5$, knowing that $p\equiv 2\pmod 5$? In other words, can a square of an integer be $\equiv 2\pmod 5$? Answer: no, squares of integers are $0,1$ or $4\pmod 5$. Thus, $\left(\frac{p}{5}\right)=-1$.
– user491874
Feb 6, 2018 at 18:26
• @Bernard, $p = 2q + 1$ (I'll add that) Feb 6, 2018 at 18:27
• If $p\equiv2\pmod 5$ then $\left(\frac 2p\right)=-1$. Feb 6, 2018 at 18:30

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 $p\equiv 2(-2)+1\equiv1+1=2$.
• I see. so as $\left( \frac{p}{5} \right) \cdot \left( \frac{5}{p} \right) = 1$ is follows that $\biggl(\frac 5p\biggr)=-1$ Feb 6, 2018 at 18:34
• Yes. B.t.w., I supposed it $p=2q+1$, not $q=2p+1$ for your second point. Feb 6, 2018 at 18:42
$5$ is a prime of the form $4k+1$ and the quadratic residues in $\mathbb{Z}/(5\mathbb{Z})^*$ are just $\pm 1$, so for every prime $p$ of the form $5k\pm 1$ the number $5$ is a quadratic residue in $\mathbb{Z}/(p\mathbb{Z})^*$, for every prime $p$ of the form $5k\pm 2$ the number $5$ is a quadratic non-residue in $\mathbb{Z}/(p\mathbb{Z})^*$.
As a side note, $\left(\frac{5}{p}\right)=\left\{\begin{array}+1 & \text{ if }p\equiv \pm 1 \pmod{5} \\ -1 & \text{ if } p\equiv \pm 2\pmod{5}\end{array}\right.$ can also be proved without invoking quadratic reciprocity, but just studying the splitting field of $\Phi_5(x)=x^4+x^3+x^2+x+1$ over $\mathbb{F}_p$ and recalling the properties of Frobenius automorphism. Indeed the degree of the splitting field of $\Phi_5(x)$ over $\mathbb{F}_p$ is given by the least $k$ such that $5\mid (p^k-1)$ and $e^{2\pi i/5}+e^{-2\pi i/5}=\frac{\sqrt{5}-1}{2}$.