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  
\[p \equiv 1, \; 5, \; 19, \; 23 \; (mod \; 24)\]
.