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The product defining the modular discriminant converges locally uniformly for $|q|<1$ and has integral Fourier coefficients. Its logarithmic derivative isTherefore $D\Delta =E_2\Delta$. Comparing the coefficient of $q^n$ gives $n\tau (n)=\tau (n)-24{\sum }_{j=1}^{n-1}{\sigma }_1(j)\tau (n-j)$, orThis also fixes the initial value $\tau (1)=1$.

For the Ramanujan tau congruence modulo five, first justify the link between this product and $E_4,E_6$. Question 1 constructs ${\Delta }_0=(E_4^3-E_6^2)/1728\in S_{12}$ with first coefficient one. Its Serre derivative is in $S_{14}$, which vanishes by the valence formula for the modular group: a nonzero such cusp form would contribute at least $1+1/2>14/12$, since it vanishes at infinity and at $i$. Hence $D{\Delta }_0=E_2{\Delta }_0$. The product and ${\Delta }_0$ have the same differential equation and leading coefficient. Equivalently their quotient has derivative zero near $q=0$ and limit one; the identity theorem givesAlso $D_6{E}_6\in M_8=\mathbb{C}{E}_4^2$. Its constant coefficient is $-1/2$, so the Serre derivative gives $2D{E}_6=E_2{E}_6-E_4^2$.

Work now in the formal power series ring ${\mathbb{F}}_5[[q]]$. The integral expansions give $E_4\equiv 1$ and $E_2\equiv E_6(\mathrm{mod}5)$, because $d^5\equiv d(\mathrm{mod}5)$, $-24\equiv -504\equiv 1$, and $240\equiv 0$. The two identities reduce toThus $3\Delta \equiv -2D{E}_6\equiv 3D{E}_6$, and division by the nonzero element $3$ gives $\Delta \equiv D{E}_6$. Since the coefficient of $q^n$ in $D{E}_6$ is $-504n{\sigma }_5(n)$, we conclude

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.