Primes of the form : $x^2+4\cdot y^2$ How to prove following statement :

Conjecture:
Odd prime $p$ is expressible as : $p=x^2+4\cdot y^2$ , $x,y > 0$
if and only if : $p\equiv 1 \pmod {12}$ or $p\equiv 5 \pmod {12}$ .

Similar statements (without a proof) related to the Fermat's theorem on sums of two squares can be found here .

HINT $\rm\ (\Rightarrow)\ \ odd\ p\ =\ x^2+4\ y^2\ \Rightarrow\ x\ odd,\ so\ \ mod\ 4\!:\ x\ \equiv\: \pm1\ \Rightarrow\ p\ \equiv\ x^2\:\equiv\ 1  $  
Thus $\rm\ p\equiv 1\pmod 4\ \Rightarrow\ p\equiv 1,5,9\pmod{12}\:,\:$ but not $\:9\:$ since $\rm\:3\:|\:p\ \Rightarrow\ p = 3\ne x^2 + 4\ y^2$
$\rm(\Leftarrow)\ $ By Fermat $\rm\: p \equiv 1\pmod 4\: \Rightarrow\ p\: =\: m^2 + n^2\: \Rightarrow\ 2\ |\ m\ \ or\ \:n\ ($else $\rm\: 2\ |\ p = odd^2+odd^2)$ 
