<s>[INST] $5$ is quadratic residue mod $p$ if and only if $ p\equiv \pm 1, \pm 9 \pmod {20}$ 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.
 [/INST] 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 $\pm 1, \pm 9 \pmod{20}$
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