# Derivation of certain sums "the hard way"

It is a well-known fact, that one can derive some spectacular identities, e. g.

$\sum^{n-1}_{m=1}\sigma_3(m)\sigma_3(n-m)=\frac {\sigma_7(n)-\sigma_3(n)}{120}$
$\sum^{n-1}_{m=1}\sigma_3(m)\sigma_9(n-m)=\frac {\sigma_{13}(n)-11\sigma_9(n)+10\sigma_3(n)}{2640}$

just by equating several modular forms together. In the book The 1-2-3 of Modular Forms Don Zagier writes: "It is not easy to obtain any of these identities by direct number-theroretical reasoning (although in fact it can be done)"
Does anybody know how to derive these identities "the hard way" or at least point me to some discussion?

• I upvoted because I had this exact question in mind for a long time, but for some reason I never asked it on MO.
– user40023
Oct 8 '17 at 15:31
• The analytic meaning is $\mathbb{C}[E_4,E_6]$ contains all the modular forms. What would be the arithmetic meaning of those identities mixing convolution and Dirichlet convolution ? (the simplest example should be $1 \star \chi_4 =\frac{1}{4} \ 1_{\mathbb{Z}^2} \ast 1_{\mathbb{Z}^2}$ which tells about the factorization in $\mathbb{Z}[i]$) Oct 8 '17 at 19:18

Briefly, Ramanujan writes down the power series for $P$ (which is $E_2$), $Q$ (which is $E_4$) and $R$ (which is $E_6$), and shows that $$\Phi_{r,s}(x) = \sum_{n=1}^{\infty} x^n \Big(\sum_{ab=n} a^r b^s\Big)$$ can be expressed by polynomials in $Q$ and $R$. Then he multiplies $\Phi_{0,r}$ and $\Phi_{0,s}$ and compares the answer with $\Phi_{1,r+s}$ and $\Phi_{0,r+s+1}$.