Uniqueness up to sign of $L$ and $M$ in $p=\frac{1}{4}(L^2+27M^2)$. If $p\equiv 1\pmod{3}$, it's well know that $p$ can be expressed as
$$
p=\frac{1}{4}(L^2+27M^2).
$$
A handful of places, like wikipedia, say that in this representation $L$ and $M$ are unique up to sign, but none actually has a proof.
Is there a elementary proof that $L$ and $M$ are unique up to sign? Thanks.
 A: After your comment, the correct thing to say is that there is essentially one representation $$ p = x^2 + 3 y^2,  $$ where in this case the word essentially can be taken to mean up to $\pm$  sign.
Meanwhile, if you multiply the discriminant by 9, there is essentially one way to write 
$$  p = u^2 + 27 v^2,   $$
$$ p = 4 u^2 + 2 u v + 7 v^2 $$
 $$ p = 4 u^2 - 2 u v + 7 v^2. $$
It is the first of these if 2 is a cube $\pmod p,$ otherwise it is the second and third. We distinguish the $\pm$ cases in the second and third forms so as to get the correct group under Gauss composition. As always, "essentially" means "up to integral automorph." Note that
$$  4 (u^2 + 27 v^2) = (2u)^2 + 27 (2v)^2, $$
$$ 4( 4 u^2 + 2 u v + 7 v^2) = (4u+v)^2 + 27 v^2, $$
which leads to pretty much what you want.
This is treated in Ireland and Rosen, for one. An encyclopaedic treatment in David A. Cox, Primes of the form $x^2 + n y^2,$ in any way to describe the problem you might like. Still, I think your best bet is Hudson and Williams 1991 at MMMEEEEEEE
For, well, culture, see my question What numbers are integrally represented by $4 x^2 + 2 x y + 7 y^2 - z^3$ 
This all can be written in the language of imaginary quadratic fields. I've never cared enough to work that out.
