Consider one of the simplest non-abelian examples of modularity. Let $$\eta(6z)\eta(18z) = q\prod\_{n=1}^\infty (1 - q^{6n})(1 - q^{18n}) = q - q^7 - q^{13} -q^{19} + q^{25} + 2q^{31} - q^{37} + 2q^{43} - q^{61} - q^{67} - q^{73} - q^{79} + q^{91} - q^{97} - q^{103} \dots = \sum\_{n=1}^\infty a\_n q^n$$


Then $a\_p + 1 =$ the number of solutions of $\{x^3 - 2 = 0\}$ in $\mathbb{Z}/p\mathbb{Z}$, for prime $p > 3$.


1) What might be the best method to prove this? 


2) Is it possible to prove it without knowledge of modular forms? i.e. without using the dimension / basis of the relevant space of moduli forms of some weight and level and nebentypus. For example, can it be explained using Weyl–Kac character formula etc.?


3) Unfortunately [LMFDB](http://www.lmfdb.org) does not yet have a list of weight $1$ modular forms and their corresponding Artin representations. Where can we find more of such weight $1$ examples?



On $\eta(6z)\eta(18z)$ and the splitting / modularity of $x^3 - 2$
Here's an answer to 2. You can tell me if it's the "best method".


Euler's Pentagonal number theorem gives that
$$ \eta(24z) = \sum\_{n \in \mathbb{Z}} (-1)^{n} q^{(6n+1)^{2}}. $$
This yields the formula
$$ \eta(6z) \eta(18z) = \sum\_{m, n \in \mathbb{Z}} (-1)^{m+n} q^{\frac{(6n+1)^{2} + 3(6m+1)^{2}}{4}}. $$
If $m=x$ and $n = x+2y$, then $m+n$ is even and we get
$$ \sum\_{x, y \in \mathbb{Z}} q^{(6x+3y+1)^2 + 27y^{2}}. $$
On the other hand if $m=x$ and $n=x+2y+1$, then $m+n$ is odd and we get
$$ -\sum\_{x,y \in \mathbb{Z}} q^{4(3x+y+1)^{2} + 2(3x+y+1)(2y+1) + 7(2y+1)^{2}}.$$


The upshot (skipping a few small details) is that
$$ \eta(6z) \eta(18z) = \frac{1}{2} \left[ \sum\_{x,y \in \mathbb{Z}} q^{x^{2} + 27y^{2}} - q^{4x^{2} + 2xy + 7y^{2}} \right]. $$


Now, Gauss proved that if $p \equiv 1 \pmod{3}$ is prime, then there is some $z$ so that $z^{3} \equiv 2 \pmod{p}$ if and only if $p = x^{2} + 27y^{2}$ for some $x, y \in \mathbb{Z}$. It follows from this that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $2$ if $p \equiv 1 \pmod{3}$ and $2$ is a cube modulo $p$. 


If $p \equiv 1 \pmod{3}$, but $2$ is not a cube modulo $p$, then $p$ is represented by $4x^{2} + 2xy + 7y^{2}$ in two ways, and this means that the $p$th coefficient of $\eta(6z) \eta(18z)$ is $-1$.


If $p \not\equiv 1 \pmod{3}$, the $p$th coefficient of $\eta(6z) \eta(18z)$ is clearly $0$. 


