I have a question related to modular forms (in particular, cusp forms) that I have managed to get stuck on. Here is the exercise.
Let $\Delta (z)=q\prod_{n=1}^{\infty}(1-q^n)^{24} := \sum_{n=1}^{\infty}\tau(n)q^n\in S_{12}(SL_2(\mathbb{Z}))$ be the cusp form of weight 12. Show that $\tau(n)$ is odd if and only if $n=(2m+1)^2$ for some m.
Hint: Use Jacobi's formula: $$\prod_{n=1}^\infty (1-q^n)^3=\sum_{k=0}^\infty(-1)^k(2k+1)q^{\frac{k(k+1)}{2}}$$
This isn't meant to be a case of "please do my homework for me" - I don't just want somebody to post the actual solution, I would just like some ideas about how I should proceed or some things that I should go and read up on to better understand the problem.
My first issue is what is $q$? Or is this a typo and should this be $z$ (Edit: After some more reading around, I am lead to believe that $q=e^{2\pi iz}$.)
I have started to use Jacobi's formula, and obtained
$$\Delta(z) = q\left(\prod_{n=1}^{\infty}(1-q^n)^3\right)^8=q\left(\sum_{k=0}^{\infty}(-1)^k(2k+1)q^{\frac{k(k+1)}{2}}\right)^8=\left(\sum_{k=0}^{\infty}(-1)^k(2k+1)q^{\frac{k(k+1)}{2}-\frac{1}{8}}\right)^8=\left(\sum_{k=0}^{\infty}(-1)^k(2k+1)q^{\frac{4k(k+1)-1}{8}}\right)^8.$$
Im unsure whether I have gone down the correct route here, as I have now hit a roadblock. I am trying to obtain a formula for $\tau(n)$ so I can try to start proving the statement. Maybe there's another way to go about this, but I am not seeing it - I am fairly new to modular forms and my complex analysis is rather rusty. I have an itching feeling that I should be using the definition of a cusp form to help solve this...?
Thank you in advance.
Andy.
Hint: modulo $2$, we have $x^8-y^8 = (x-y)^8$. So
$$\Delta(q) = q \left(\prod_{n=1}^\infty (1-q^n)^3\right)^8 = q \prod_{n=1}^\infty (1-q^{8n})^3 \pmod 2$$
Then use Jacobi's formula with $q^8$ in place of $q$.