 Primes represented by x squared plus three y squared

A prime number $p$ has an integer representation $p=x^2+3y^2$ exactly when $p=3$ or $p\equiv 1(\mathrm{mod}3)$. The case three is $0^2+3\cdot 1^2$, and two is impossible.

For necessity when $p\ne 3$, reducing the equation modulo three gives $p\equiv x^2(\mathrm{mod}3)$. The value cannot be zero, so it is one.

Conversely, for $p\equiv 1(\mathrm{mod}3)$, the quadratic character of minus three shows that $-3$ is a quadratic residue modulo $p$. Thus the discriminant criterion for prime representation by a binary quadratic form produces a positive definite form of discriminant $-12$ representing $p$. It is primitive because $p\nmid 12$.

Enumeration of reduced binary quadratic forms gives only $[1,0,3]$ and $[2,2,2]$ at discriminant $-12$: the bound is $a\le 2$, and the integrality and boundary conditions leave those two. The second is imprimitive, so proper equivalence carries the representing form to $[1,0,3]$. This proves sufficiency. Examples are $7=2^2+3\cdot 1^2$ and $13=1^2+3\cdot 2^2$.