The space $M_8(SL_2(\mathbb Z))$ is one-dimensional. The two normalized weight-eight forms $E_8$ and $E_4^2$ therefore agree. Their expansions are
$$
E_8=1+480\sum_{n\geq1}\sigma_7(n)q^n
$$
and
$$
E_4^2
=1+480\sum_{n\geq1}\sigma_3(n)q^n
+57600\sum_{n\geq2}\left(\sum_{i=1}^{n-1}
\sigma_3(i)\sigma_3(n-i)\right)q^n.
$$
Equating the coefficient of $q^n$ and dividing by $480$ proves the <divisor-sum convolution identity of weights four and eight>
$$
\sigma_7(n)=\sigma_3(n)+120\sum_{i=1}^{n-1}
\sigma_3(i)\sigma_3(n-i).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 1 / d / Solution

The identity $E_8=E_4^2$ gives
$$
\sigma_7(n)=\sigma_3(n)+120\sum_{i=1}^{n-1}\sigma_3(i)\sigma_3(n-i).
$$


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 Incoming links: Divisor-sum convolution identity of weights four and eight