Show that x is a quadratic (non-)residue (mod p) if p ≡ ... (mod y)

Question

How do I solve the following with quadratic residues. A thorough explanation for beginners is appreciated:

Let $p$ be a prime number that is not equal to $2$ or $5$. Show that $−5$ is a quadratic residue $\mod p$ whenever $p \equiv 1, 3, 7, 9 \mod {20}$, and that $−5$ is a quadratic non- residue (mod p) whenever $p \equiv 11, 13, 17, 19 \mod {20}$.

Answer 1

We use the Quadratic Reciprocity Law:

If $p,q$ different odd primes then:

• $\left( \frac{p}{q} \right) = \left( \frac{q}{p} \right)$ if $p \equiv 1 \mod{4}$ or $q \equiv 1 \mod{4}$

• $\left( \frac{p}{q} \right) = -\left( \frac{q}{p} \right)$ if $p \equiv 3 \mod{4}$ and $q \equiv 3 \mod{4}$

We also use the following special cases:

• $\left( \frac{-1}{p} \right) = 1$ if $p \equiv 1 \mod{4}$; $\left( \frac{-1}{p} \right) = -1$ if $p \equiv 3 \mod{4}$

• $\left( \frac{2}{p} \right) = 1$ if $p \equiv +/-1 \mod{8}$; $\left( \frac{2}{p} \right) = -1$ if $p \equiv +/-3 \mod{8}$

For $-5$ being a q.r.:

• $p \equiv 1 \mod{20}$ then $p \equiv 1 \mod{4}$ and $p \equiv 1 \mod{5}$

$\left( \frac{-5}{p} \right) = \left( \frac{-1}{p} \right)\left( \frac{5}{p} \right) = 1 * \left( \frac{p}{5} \right) = \left( \frac{1}{5} \right) = 1$

• $p \equiv 3 \mod{20}$ then $p \equiv 3 \mod{4}$ and $p \equiv 3 \mod{5}$

$\left( \frac{-5}{p} \right) = \left( \frac{-1}{p} \right)*\left(-\left( \frac{p}{5} \right)\right) = 1 * \left(-\left( \frac{3}{5} \right)\right) = 1$ etc...

For $-5$ being a q. non-r.:

• $p \equiv 11 \mod{20}$ then $p \equiv 3 \mod{4}$ and $p \equiv 1 \mod{5}$

$\left( \frac{-5}{p} \right) = \left( \frac{-1}{p} \right)*\left(-\left( \frac{p}{5} \right)\right) = 1 * \left(-\left( \frac{1}{5} \right)\right) = -1$ etc...