quadratic residue such as "$(n|p)= -1$"(quadratic reciprocity)

Question

(Note: (n|p)=1 is legendre-symbol.)

So need to find primes where $(n|p)=1$ So we have

1- $1\pmod 4$ where we use quadratic residue of $n$ along with $\pmod n$ to find solutions.

2- Then we have $3\pmod 4$ where we use quadratic nonresidues of $n$ again with $(\!\bmod n)$ to find more solutions.

3-We combine solutions from 1 and 2 to get all primes in the form of $$p\equiv \langle \text{solutions-from-above}\rangle\pmod{4\times n} $$

But how to do we go about solving some thing with $'-1'$ like $ (n|p) = -1 $ for example $ (3|p) = -1 $ or $(2|p)=-1$.

Or some thing with more solutions as $(5|p)=-1$.

Examples from the book:.

in the book it says.

$(2|p) \equiv 1$ gives $p \equiv 1 (mod 8)$.

$(2|p) \equiv -1$ gives $p \equiv 3 (mod 8)$.

$(3|p) \equiv 1$ gives $p \equiv 1 (mod 12)$'

$(3|p) \equiv -1$ gives $p \equiv 7 (mod 12)$.

Thank you.

Answer 1

See my answer to another question where $(\frac{n}{p})$ is determined by applying Gauss' criterium: Using gauss's lemma to find $(\frac{n}{p})$ (Legendre Symbol)