Primes of the form $p=3a^2+3ab+b^2$ or $p=27a^2+27ab+7b^2$ and the number of points of $y^2=x^3+2$ modulo $p$

Question

Trying to generalize this answered question based on limited numerical evidence.

Let $E / \mathbb{F}_p : y^2=x^3+2$.

Conjecture 1 Let $p=3a^2+3ab_0+b_0^2$ be prime and $a,b_0$ positive integers.

Let $b=|p+1-\#E(\mathbb{F}_p)|$. For some integer $a$ (not necessarily positive) $p=3a^2+3ab+b^2$.

In some non-rigorous sense if $p=3a^2+3ab+b^2$ then $b=|p+1-\#E(\mathbb{F}_p)|$.

Conjecture 2 Let $p=27a^2+27ab+7b^2$ be prime and $a,b$ positive integers. Then $b=|p+1-\#E(\mathbb{F}_p)|$. If the representation of $p$ in this form is not unique the conjecture is false (or possibly proceed as in Conjecture 1).

Example sage session:

sage: b=11;a=2^100+28;p=27*a^2+27*a*b+7*b^2;Kp=GF(p);E=EllipticCurve([Kp(0),Kp(2)])
sage: b2=abs(p+1-E.order());b==b2
True

Answer 1

These conjectures are also true, and they follow similarly as in my proofs here and here.

Proof of Conjecture 1. Let us denote $b=|p+1-\#E(\mathbb{F}_p)|$. The assumption on $p$ implies that $p\equiv 1\pmod{3}$, hence by my earlier posts, $4p=b^2+27c^2$ for some integer $c$. Then, $b$ and $c$ have the same parity, so that $3c=2a+b$ for some integer $a$. Finally, $$4p=b^2+27b^2=b^2+3(2a+b)^2=4(3a^2+3ab+b^2),$$ and the conclusion $p=3a^2+3ab+b^2$ follows. Done.

Proof of Conjecture 2. The assumption on $p$ implies that $4p=b^2+27(2a+b)^2$, whence $b=|p+1-\#E(\mathbb{F}_p)|$ by my earlier posts. Done.