I was unable to solve some questions asked in my mid term of number theory exam and so I am asking it here.
If $\tau \in H$ and $x= e^{2\pi i \tau}$, prove that $[{504 \sum_{n=0}^{\infty}\sigma_{5}(n) x^n}]^2 = [j(\tau) -{12}^3 ] \sum_{n=1}^{\infty} \tau(n) x^n$.
Here $j(\tau)={12}^3 J(\tau)$ . I used $j(\tau) = 1/x + 744 + \sum_{n=1}^{\infty} c(n) e^{2\pi in \tau}$ but it doesn't have $\sigma$ term in it . The problem I am facing is that I don't think that we have been thaught a result that has both $\sigma$ and $J(\tau)$ in it.
Please help me in proving this question.
Ty
This is simply a reformulation of the identity $$j=1728 \frac{E_4^3}{E_4^3-E_6^2},$$ where $$E_{k}(\tau)=1-\frac{2k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)\,q^n$$ and $\Delta(\tau)=E_4(\tau)^3-E_6(\tau)^2=\sum_{n=1}^\infty \tau(n)q^n$.