$5$ is quadratic residue mod $p$ if and only if $ p\equiv \pm 1, \pm 9 \pmod {20}$

Question

5 is quadric residue mod p if and only if $ p\equiv +/- 1, +/-9 \pmod {20}$

$$(5/p)=(p/5)$$ $p\equiv 1 \pmod 4$ ⟹ $1,5,9,13,17 \pmod {20}$
$p\equiv 1 \pmod 5$ ⟹ $1,6,11,16 \pmod {20}$
then $p\equiv 1 \pmod {20}$
$p\equiv 1 \pmod 4$ ⟹ $1,5,9,13,17 \pmod {20}$
$p\equiv 4 \pmod 5$ ⟹ $4,9,14,19 \pmod {20}$
then $p\equiv 9 \pmod {20}$

And now I should show that $$(5/p)=-(p/5)$$ $p\equiv 3 \pmod 4$ ⟹ $3,7,11,15,19 \pmod {20}$
and p is congruent to 2 mod 5 it's p ≡ 7 mod 20
and p ≡3 mod 5 it's p ≡3 mod 20

And it doesn't +/-1 and +/-9 Why?

Please of help.

Answer 1

Because $5\equiv 1 \pmod 4$, we have $(5/p)=(p/5)$ by quadratic reciprocity, so there is no second case. The squares mod $5$ are $\pm 1$. Modulo $20$, this gives $1,4,6,9,11,14,16,19$. However, because primes are odd, this eliminates all the even numbers in the list, leaving just $1,9,11,19$, which is the same as $\pm 1, \pm 9 \pmod{20}$