Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 3 / c / Solution

From part (a), $\Theta F=E_{2}F$. Its product has integer coefficients and begins with $q$, so write $F={\sum }_{n\ge 1}\tau (n)q^n$, with $\tau (1)=1$. Comparing the coefficient of $q^n$ giveswhich rearranges toThis is a recurrence for the Ramanujan tau function; for example it gives $\tau (2)=-24$ and $\tau (3)=252$.

For the congruence, use the normalized Eisenstein expansionsThe modular form $E_4^3-E_6^2$ is a cusp form of weight twelve, and its leading coefficient is $720-(-1008)=1728$. Since $S_{12}$ is one-dimensional and $F$ starts with $q$,Also $D_6{E}_6$ is a weight-eight form, with constant coefficient $-1/2$. Since $M_8$ is one-dimensional and $E_4^2$ has constant coefficient one, $D_6{E}_6=-E_4^2/2$, orThese identities have rational coefficients whose only displayed denominators are prime to five, so coefficientwise reduction modulo five is legitimate. We have $E_4\equiv 1$. Furthermore $d^5\equiv d(\mathrm{mod}5)$ gives ${\sigma }_5(n)\equiv {\sigma }_1(n)$, while $-504\equiv -24\equiv 1$, so $E_6\equiv E_2$. ThereforeThe equality in the middle is in ${\mathbb{F}}_5$, where $-1/3=1/2$. Comparing coefficients in $\Theta E_6=-504{\sum }_{n\ge 1}n{\sigma }_5(n)q^n$ givesThis proves the Ramanujan tau congruence modulo five from the given small-dimensional modular-form spaces, without assuming that congruence as a theorem.