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Let \begin{align*} \Delta(q)=q\prod_{n=1}^{\infty}(1-q^n)^{24}, \end{align*} where $|q|<1$, then we can write it as $$\Delta(q)=\sum_{n=1}^{\infty}\tau(n)q^n.$$ Then Ramanujan proves that for any prime number $p$, we have \begin{align*} \tau(p)\equiv 1+p^{11}. \end{align*}


I've found a proof using modular forms. More generally, we have $$\tau(n)\equiv \sigma_{11} \mod 691,$$ where $\sigma_{k}(n)=\sum_{d\mid n}d^k.$ In the proof, he uses Eisenstein series $E_k$ and proves $E_{12}-E_6^2=\frac{c}{691}\Delta.$ This proof is great, but I have the following questions:

  • Is there other proof of this congruent identity?
  • What the advantages of modular forms?

I want explain my question further. At first glance, I think it is a question about $q-$series. And we may try a algebraic or combinational proof. Thus I seek for a such proof. Besides, what's the advantages of modular forms in dealing with congruent problem?

I am looking forward an answer, so any help will be appreciated!

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  • $\begingroup$ I was wondering if you could tell me where you found the proof using Modular Forms? Was it Haruzo Hida's notes? $\endgroup$ Commented May 17, 2018 at 18:11
  • $\begingroup$ It is in Kazuya Kato's book "Number Theory: Iwasawa theory and modular forms". $\endgroup$ Commented May 25, 2018 at 14:18

1 Answer 1

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Ramanujan used his functions $P, Q, R$ defined by \begin{align} P(q) &= 1-24\sum_{i=1}^{\infty}\frac{iq^{i}}{1-q^{i}}\tag{1}\\ Q(q)&=1+240\sum_{i=1}^{\infty} \frac{i^{3}q^{i}}{1-q^{i}}\tag{2}\\ R(q)&=1-504\sum_{i=1}^{\infty}\frac{i^{5}q^{i}}{1-q^{i}}\tag{3} \end{align} to prove the desired congruence. Using some trigonometry and algebraic manipulation he established the fundamental differential equations satisfied by $P, Q, R$: \begin{align} q\frac{dP} {dq} &=\frac{P^{2}-Q}{12}\tag{4}\\ q\frac{dQ}{dq}&=\frac{PQ-R} {3}\tag{5}\\ q\frac{dR}{dq}&=\frac{PR-Q^{2}}{2}\tag{6} \end{align} Using these differential equations it is easy to prove that $$q\frac{d} {dq} (Q^{3}-R^{2})=P(Q^{3}-R^{2})$$ or $$q\frac{d} {dq} \log(Q^{3}-R^{2})=P\tag{7}$$ Also from the definitions of $\Delta$ (see question) and $P$ it is clear that $$P=q\frac{d} {dq} \log\Delta(q) \tag{8}$$ and from equations $(7),(8)$ it follows that $Q^{3}-R^{2}$ is a constant multiple of $\Delta $. And comparing coefficients we see that $$Q^{3}-R^{2}=1728\Delta\tag{9}$$ Using his algebraic manipulation techniques Ramanujan further proved that $$691+65520\sum_{i=1}^{\infty}\frac{i^{11}q^{i}}{1-q^{i}}=441Q^{3}+250R^{2}$$ which can be rewritten using equation $(9)$ as $$691+65520\sum_{i=1}^{\infty}\sigma_{11}(i)q^{i}=441\cdot 1728\Delta + 691R^{2}\tag{10}$$ This equation alongwith $(3)$ can be used to prove the congruence $$\tau(i) \equiv \sigma_{11}(i)\pmod{691}\tag{11}$$ with a little amount of algebraic manipulation.


The above is more of a sketch of the proof Ramanujan offered for the congruence $(11)$ and the details can be found in my blog posts here and here. The advantage of Ramanujan's approach is that it is far simpler compared to the difficult and deep theory of modular forms and with patience it can be understood by anybody with a basic knowledge of algebra and calculus.

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