i am reading the book an introduction to non abelian class field theory by Seiken Saito and Toyokazu Hiramatsu in page 5 ,and i am stuck in the following part
let
$\eta(6z)\eta(18z)=\sum_{x,y \in \mathbb{Z}} (-1)^{n+m} q^{((6m+1)^2+3(6n+1)^2)/4)}$ Where $q=e^{2\pi iz}$
and let A(p) be the number of solution$(m,n)$ of $\begin{equation} (6m+1)^2+3(6n+1)^2=4p \quad(1) \end{equation} $
then
I)$A(p)=2$ and n+m is even if $p=x^2+27y^2$
II)$A(p)=1$ and n+m is odd if $p\neq x^2+27y^2$
Questions
i don't understand how they got the values of $A(p)$,and why they considered the binary quadratic form $x^2+27y^2 $?
my attempt from equation (1) by putting x=6m+1 and y=6n+1 the question is equivalent to
$x^2+3y^2=4p $
and by a familiar result from the theory of binary quadratic form ,the previous equation has a solution iff $z^2 \equiv -12 [4p]$ is solvable which is equivalent to $z^2 \equiv -3 mod p $ is solvable i,e $ \left( \frac{-3}{p} \right)=1$ it has solution when $p \equiv 1 [3]$ and we know that the number of equivalent class of binary quadratic forms of Discrimant $-12 $ is 1 since the class number of $\mathbb{Q}(\sqrt{-12})=\mathbb{Q}(\sqrt{-3}) $ is 1, so we can consider only the quadratic form $x^2+3y^2$ . but i don't know how to go further.
pages 80, 81 from Dickson (1929), Introduction to the Theory of Numbers