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

(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.

• quadratic reciprocity? Aug 26, 2019 at 15:01
• What do you denote$(n|p)$? Legendre's symbol? Aug 26, 2019 at 15:23
• its Quadratic reciprocity, and (n|p) is Legendre symbol.
– tt z
Aug 26, 2019 at 15:49
• Is $n$ any integer, or a prime? Aug 26, 2019 at 17:00
• in the question n is prime.
– tt z
Aug 26, 2019 at 17:02

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)