Does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$? For a prime number $p \neq 2$, $3$, does $p = x^2 + 9y^2$ for some $x$, $y \in \mathbb{Z}$ if and only if $p \equiv 1 \text{ mod }12$?
A case where this is true as to suggest plausibility: $13 = 2^2 + 9 \times 1^2$.

Finish: 
For $p=12k+1$, we have $p=x^2+y^2$, and,  if $gcd(xy, 3)=1$, then $p\equiv x^2+y^2\equiv 2 \pmod 3 $, a contradiction.
Conversely, $x^2+9y^2\equiv 0$ or $1 \pmod 3$, and $x^2+9y^2\equiv 0, 1, 2\pmod 4$. If this number is prime, then  we have only one possibility:
$$x^2+9y^2\equiv 1 \pmod {12}.$$ 
