Ramanujan tau congruence modulo five
= Ramanujan tau congruence modulo five
{c}
{title2=$\tau(n)\equiv n\sigma_5(n)\pmod5$}
The <modular discriminant> satisfies $1728\Delta=E_4^3-E_6^2$, while the <Serre derivative> gives $2\Theta E_6=E_2E_6-E_4^2$. Their integral Fourier expansions have $E_4\equiv1$ and $E_2\equiv E_6\pmod5$, because $d^5\equiv d\pmod5$. Reducing the two identities gives $\Delta\equiv\Theta E_6\pmod5$. The coefficient of $q^n$ in $\Theta E_6$ is $-504n\sigma_5(n)\equiv n\sigma_5(n)$.