\[a^2+3b^2\]
can also be written in the form \[3k+2\]
.To see why, working modulo 3, a square number must be congruent to either 0 or 1, so
\[a^2+3b^2 \equiv 0+3 \times 0 \; (mod \; 3) \equiv 0 \; (mod \; 3)\]
\[a^2+3b^2 \equiv 1+3 \times 0 \; (mod \; 3) \equiv 1 \; (mod \; 3)\]
\[a^2+3b^2 \equiv 0+3 \times 1 \; (mod \; 3) \equiv 0 \; (mod \; 3)\]
\[a^2+3b^2 \equiv 1+3 \times 1 \; (mod \; 3) \equiv 1 \; (mod \; 3)\]
But
\[3k+2 \equiv 2 \; (mod \; 3)\]
so no such prime exists.