## For Which Primes is 6 a Quadratic Residue?

For which primes is 6 a quadratic residue?
By the properties of the The Legendre Symbol,
$(6/p)=(2/p)(3/p)$

The The Quadratic Character of 2 is fixed according to
$(2/p) = \{ \begin{array}{c} 1 \; if \; p \equiv 1 \; (mod \; 8) \; or \; p \equiv 7 \; (mod \; 8) \\ -1 \; if \; p \equiv 3 \; (mod \; 8) \; or \; p \equiv 5 \; (mod \; 8) \end{array}$

The The Quadratic Character of 3 is fixed according to
$(3/p) = \{ \begin{array}{c} 1 \; if \; p \equiv 1 \; (mod \; 12) \; or \; p \equiv 11 \; (mod \; 12) \\ -1 \; if \; p \equiv 5 \; (mod \; 12) \; or \; p \equiv 7 \; (mod \; 12) \end{array}$

For 6 to be a quadratic residue of 6 we need either
$(2/p)=(3/p)=1$
or
$(2/p)=(3/p)=-1$
.
Suppose
$(2/p)=(3/p)=1$
. Then
$p \equiv 1 \; (mod \; 8) \; or \; p \equiv 7 \; (mod \; 8)$
and
$p \equiv 1 \; (mod \; 12) \; or \; p \equiv 11 \; (mod \; 12)$
.
Hence
$p \equiv 1, \; 7, \; 9, \; 15, 17, 23 \; (mod \; 24)$
and
$p \equiv 1, \; 11, \; 13, \; 23 \; (mod \; 24)$
.
Only
$[ \equiv 1, \; 23 \; (mod \; 24)$
are in both lists.
Suppose
$(2/p)=(3/p)=-1$
. Then
$p \equiv 3 \; (mod \; 8) \; or \; p \equiv 5 \; (mod \; 8)$
and
$p \equiv 5 \; (mod \; 12) \; or \; p \equiv 7 \; (mod \; 12)$
.
Hence
$p \equiv 3, \; 5, \; 11, \; 13, 19, 21 \; (mod \; 24)$
and
$p \equiv 5, \; 7, \; 17, \; 19 \; (mod \; 24)$
.
Only
$[ \equiv 5, \; 19 \; (mod \; 24)$
are in both lists.
Then 6 is a quadratic residue of
$p \equiv 1, \; 5, \; 19, \; 23 \; (mod \; 24)$
.