Property of some composite Mersenne numbers

Question

I noticed this property of some composite Mersenne numbers:

If $p$ is prime, $p=1 \bmod 4$ and $(1+6 \cdot p)$ is prime

for a theorem of Fermat every prime $p=1 \bmod 3$ can be written as $p=c^2+3 \cdot d^2$

then $$(1+6 \cdot p)=a^2+3 \cdot b^2$$

if $b=0 \bmod 3$ for a properties of the cubic residues $\left( \frac{2}{(1+6 \cdot p)} \right)_3=1$

then $$2^p-1=0\mod(1+6 \cdot p)$$

Example

$p=37$ , $(1+6 \cdot p)=223=14^2+3\cdot 3^2$

$(2^{37}-1)=0 \bmod 223$

Question

Is this result always true?

Answer 1

In $\Bbb{Z}[\zeta_3]=\Bbb{Z}[x]/(x^2+x+1)$, if $q \equiv 1\bmod 6$ is a prime number then $\Bbb{F}_q[x]/(x^2+x+1)$ isn't an integral domain thus $(q)$ isn't a prime ideal, and since $\Bbb{Z}[\zeta_3]$ is a PID it means $(q)= (a+b\sqrt{-3})(a-b\sqrt{-3})$ and $\Bbb{Z}[\zeta_3]/(a+b\sqrt{-3}) \cong \Bbb{F}_q$.

Iff $2$ is a cube and a square in $\Bbb{Z}[\zeta_3]/(a+b\sqrt{-3})$ then it is a $6$-th power in $\Bbb{F}_q$ so that $2^{(q-1)/6}\equiv 1\bmod q $ and $q\ | \ 2^{(q-1)/6}-1$.

Quadratic reciprocity says $2$ is a square when $q\equiv \pm 1\bmod 8$,

Cubic reciprocity says $2$ is a cube when $3|b$.