Ramanujan tau congruence modulo five ($\tau (n)\equiv n{\sigma }_5(n)(\mathrm{mod}5)$)

The modular discriminant satisfies $1728\Delta =E_4^3-E_6^2$, while the Serre derivative gives $2\Theta E_6=E_2{E}_6-E_4^2$. Their integral Fourier expansions have $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$. Reducing the two identities gives $\Delta \equiv \Theta E_6(\mathrm{mod}5)$. The coefficient of $q^n$ in $\Theta E_6$ is $-504n{\sigma }_5(n)\equiv n{\sigma }_5(n)$.