How are all Mersenne primes are of the form $27x^2+4y^2$ (except 3 and 7)? I noticed something interesting with Mersenne primes numbers: You can write it with the form $27x^2+4y^2$ except for 3 and 7 but it seems to work with all other Mersenne primes numbers and their associated perfect numbers.
Another something interesting: it doesn't works for composite Mersenne Numbers like 2047 or 8388607.
Is there a way to explain that? I am unsure of how to start and am hence looking for some tips on how to start.
 A: A Mersenne prime is one of the form $m = 2^p - 1$, looking mod $3$ we see that
$$m \equiv (-1)^p - 1 \equiv 1 \pmod 3$$
as soon as $p$ is odd. Thus all Mersenne primes (except 3) are congruent to 1 mod 3.
Now primes congruent to 1 mod 3 are precisely those of the form
$$x^2 + 3y^2.$$
But moreover cubic reciprocity gives that (https://www.math.uni-hamburg.de/home/charlton/teaching/primes_17/primes.pdf 9.3):
primes of the form
$$x^2 + 27y^2$$
are those which are 1 mod 3, and for which 2 is a cubic residue mod the prime.
So we have to show that $2 \equiv t^3 \pmod m$.
But we have
$$0 \equiv 2^p - 1 \pmod m$$
so
$$2^p \equiv 1 \pmod m$$
and $2$ has order coprime to $3$ as $p=3$ is excluded, hence is a third power (thanks to Erick Wong for this nicer way of saying this :)).
(Alternatively we can argue that:
$$2^{p+1} \equiv 2 \pmod m$$
if $p+ 1$ is divisible by 3 then we are done as 2 is the cube of $2^{(p+1)/3}$.
Otherwise $p \equiv 1 \pmod 3$ as the case of  $p = 3$ is the excluded case, in this case
$$(2^p)^2 \equiv 1^2 \equiv 1 \pmod m$$
so
$$ 2^{2p + 1} \equiv 2 \pmod m$$
and now $2p + 1$ is divisible by 3.)
So 2 is always a cubic residue modulo a Mersenne prime and $m$ is of the form
$$x^2 + 27y^2.$$
We just have to show in this expression that $x$ is even.
Lets look mod 4:
A Mersenne prime is always $-1 \pmod 4$ hence we have
$$x^2-y^2 \equiv -1 \pmod 4$$
The only squares mod 4 are $0$ and $1$ so we must have $x^2 \equiv 0 \pmod 4$ and $y^2 \equiv 1 \pmod 4$, hence $x$ is even and $y$ is odd.
So now we have that $$m = 4x^2 + 27y^2$$ for a different choice of $x$ (half of the original $x$).
