## The Quadratic Character of 3

Theorem
If
$p \gt 3$
is prime then
$(3/p) = \{ \begin{array}{c} 1 \; if \; p \equiv 1 \; (mod \; 12) \; or \; p \equiv 11 \; (mod \; 12) \\ -1 \; if \; p \equiv 5 \; (mod \; 12) \; or \; p \equiv 7 \; (mod \; 12) \end{array}$

Proof
We use the Law of Quadratic Reciprocity
$(p/q)(q/p)=(-1)^{(p-1)(q-1)/4}$
with
$q=3$
.
$(p/3)(3/p)=(-1)^{(3-1)(p-1)/4}=(-1)^{(p-1)/2}$

The only quadratic residue of 3 is 1, and the only quadratic non residue is 2.
Let
$p=1 \; (mod \; 3)$
.
$(p/3)(3/p)=1 \times (3.p)=(-1)^{(p-1)/2}$

$(3/p)=(-1)^{(p-1)/2}$

Then
$(3/p)=1$
if and only if
$\frac{p-1}{2}=2k \rightarrow p=4k+1$

Then
$p \equiv 1 (mod \ 3), \; p \equiv 1 (mod \ 4) \rightarrow p \equiv 1 (mod \ 12)$

Let
$p=2 \; (mod \; 3)$
.
$(p/3)(3/p)=-1 \times (3/p)=(-1)^{(p-1)/2}$

$-(3/p)=(-1)^{(p-1)/2}$

Then
$(3/p)=1$
if and only if
$\frac{p-1}{2}=2k+1 \rightarrow p=4k+3$

Then
$p \equiv 2 (mod \ 3), \; p \equiv 3 (mod \ 4) \rightarrow p \equiv 11 (mod \ 12)$

Let
$p=1 \[ (mod \; 3)$
.
$(p/3)(3/p)=1 \times (3/p)=(-1)^{(p-1)/2}$

$(3/p)=(-1)^{(p-1)/2}$

Then
$(3/p)=-1$
if and only if
$\frac{p-1}{2}=2k+1 \rightarrow p=4k+3$

Then
$p \equiv 1 (mod \ 3), \; p \equiv 3 (mod \ 4) \rightarrow p \equiv 7 (mod \ 12)$

Let
$p=2 \; (mod \; 3)$
.
$(p/3)(3/p)=-1 \times (3/p)=(-1)^{(p-1)/2}$

$-(3/p)=(-1)^{(p-1)/2}$

Then
$(3/p)=-1$
if and only if
$\frac{p-1}{2}=2k \rightarrow p=4k+3$

Then
$p \equiv 1 (mod \ 3), \; p \equiv 1 (mod \ 4) \rightarrow p \equiv 5 (mod \ 12)$