Primes representable by either $x^2+36y^2$ or $4x^2+9y^2$ Is there a simple criterion for primes that are representable by either $x^2 + 36 y^2$  or $4x^2 + 9y^2$?
This is not my area of expertise, so any pointers appreciated.
I had a look in the Cox book "Primes of the form $x^2 + n y^2$", but if those specific forms are there, I missed it.
 A: Sure, for $p \equiv 1 \pmod 4,$   there is an expression $u^2 + 36 v^2 = p$    precisely when $x^4 + 3$  factors into four linear factors $\pmod p,$   that is four distinct roots. Let me check whether it is enough to have one root.  This is from Kenneth S. Williams and D. Liu, Tamkang Journal of Mathematics, volume 25, number 4, Winter 1994;  pages 321-334.
Once $p \equiv 1 \pmod {12},$   it is enough to have one root of $x^4 + 3 \pmod p$
Meanwhile, this can probably be done with biquadratic reciprocity, as in Ireland and Rosen.
More description:  we need only primes $p \equiv 1 \pmod {12}$ Now,    we can always find a number $a$  such that $$  (x^2 + a) ( x^2 - a) \equiv x^4 + 3  \pmod p$$
Indeed, we demanded $(-3|p) = 1,$  we are taking $a^2 \equiv -3 \pmod p$
If   $a$ is not a square $\pmod p$    then $p = 4 u^2 + 9 v^2 $  In this case the two quadratics displayed are irreducible $\pmod p.$
If we can take $a = b^2, $  so that
$$   (x^2 + b^2) ( x^2 - b^2) \equiv x^4 + 3  \pmod p$$
then $ p = u^2 + 36 v^2 .$  Note how we can see the condition $b^4 \equiv -3 \pmod p$
Let's see,  $( -1 | p) + 1,$ there is an $i$  with $i^2 \equiv -1 \pmod p,$   we may define $c = bi$  and write
$$   (x^2 -c^2) ( x^2 - b^2) \equiv x^4 + 3  \pmod p$$  to display the four roots
