I am trying to prove the the well-known result:
$a_{p}=0$ if $\left( \frac{p}{23} \right)=-1$
$a_{p}=2 $ if $\left( \frac{p}{23} \right)=1$ and $p$ is represented by $x^2+xy+6y^2$
$a_{p}=-1 $ if $\left( \frac{p}{23} \right)=1$ and $p$ is represented by $ 2x^2+xy+3y^2 $
where $a_{p}$ is the coefficient of $q^{p}$ in
$\eta(\tau)\eta(23\tau)=q\prod_{k=1}^{\infty}(1-q^k)(1-q^{23k}) $
I know that $N_{p}(f)=a_{p}+1$ where $N_{p}$ is the number of roots of the polynomial $f(x)=x^3-x-1 \in \mathbb{F}_p[x]$, but I am trying to compute $ a_{p}$ from the $q$-expansion.
Here is my attempt.
Using the Euler pentagonal number theorem, we have
$$\eta(\tau)\eta(23\tau)=\prod_{k=1}^{\infty}(1-q^k)(1-q^{23k})= $$ $$\sum_{x,y \in \mathbb{Z}} (-1)^{n+m} q^{(3n^2+n+69m^2+23m+2)/2} = $$ $$ \sum_{x,y \in \mathbb{Z}} (-1)^{n+m} q^{((6m+1)^2+23(6n+1)^2)/24}. $$ It remains to determine when p is represented by the quadratic form $Q(x,y)=x^2+23y^2$. We know that p must verify $ \left( \frac{p}{23}\right)=1$, but I could not go any further.