TITLE: Some convincing reasoning to show that to prove that Ramanujan tau function is multiplicative is very difficult
QUESTION [13 upvotes]: I am curious to know (some words or reasoning  about how to justify) why to prove that the so-called Ramanujan $\tau$ is a multiplicative function is (was) very difficult.
In this Wikipedia is showed the conjecture due to Ramanujan, and I see that the function $\tau(n)$ is defined from the first formula of the Wikipedia's article. Then I suspect that it was the difficulty to prove that such arithmetic function is multiplicative. But I would like to know a reasoning, if it is possible, to know why is very difficult to prove that this function is multiplicative (since the definition of this arithmetic function ins't explicit in term of prime powers).

Question. How do you convince me that from the/a definition of the $\tau(n)$ function is very difficult to show that is multiplicative? Many thanks. 

Feel free to refer the literature if you need to answer invoking the literature (I can to search and read those literature that you need to refer).

REPLY [5 votes]: Based on the answer from Markus Scheuer I present the proof for multiplicative property of Ramanujan's Tau function. This is not supposed to be a direct answer to the question here, but it lets the readers judge for themselves the level of difficulty of the multiplicative property of Ramanujan's Tau function.

Let $z\in\mathbb{C}$ be such that imaginary part of $z$ is positive and then the variable $q=e^{2\pi i z} $ satisfies $|q|<1$ and thus we can define a function $f$ by $$f(z) =q\prod_{n=1}^{\infty}(1-q^n)^{24}=\sum_{n=1}^{\infty}\tau(n)q^n\tag{1}$$ The function $f$ is connected to the famous Dedekind Eta function by the relation $f(z) =\eta^{24}(z)$ and consequently satisfies the functional equations $$f(z+1)=f(z),f(-1/z)=z^{12}f(z)\tag{2}$$ Since the transformations $z\to z+1$ and $z\to - 1/z$ generate the full modular group it follows that $$f\left(\frac{az+b} {cz+d} \right) =(cz+d) ^{12}f(z)\tag{3}$$ where $a, b, c, d$ are integers with $ad-bc=1$.
The key is to study another related function $F$ given by $$F(z) =p^{12}f(pz)+\sum_{k=0}^{p-1}f\left(\frac{z+k}{p}\right)\tag{4}$$ where $p$ is a prime number and show that $F$ satisfies exactly the same functional equations given in $(2)$ as satisfied by $f$. The relation $F(z+1)=F(z)$ is easy to verify and we establish that $F(-1/z)=z^{12}F(z)$. Note that expression for $F$ consists of $(p+1)$ terms and replacing $z$ by $-1/z$ just multiplies the first two terms with $z^{12}$. It is bit tricky to show that the same happens for remaining $(p-1)$ terms corresponding to $k=1,2,\dots,p-1$.
For each term in the  sum $$\sum_{k=1}^{p-1}f\left(\frac{kz-1}{pz}\right)$$ we can find integers $l, m$ such that $kl+mp=-1$ where $l$ runs through $1,2,\dots,p-1$ modulo $p$ as $k$ runs through $1,2,\dots,p-1$ (this is possible because $k$ is coprime to $p$, see Bezout's identity). Let $z'=(z+l) /p$ and $a=k, b=m, c=p, d=-l$ so that $ad-bc=1$ and $$\frac{az'+b} {cz'+d} =\frac{az'+b} {z} =\frac {az+al+bp} {pz} =\frac{kz-1}{pz}$$ and hence from $(3)$ we get $$f\left(\frac{kz-1}{pz}\right)=f\left (\frac{az'+b} {cz'+d} \right) =(cz'+d) ^{12}f(z')=z^{12}f\left(\frac{z+l}{p}\right)$$ and thus we have proved that $$\sum_{k=1}^{p-1}f\left(\frac{(-1/z)+k}{p}\right)=\sum_{k=1}^{p-1}f\left(\frac{kz-1}{pz}\right)=z^{12}\sum_{l=1}^{p-1}f\left(\frac{z+l}{p}\right)$$ and therefore $F$ satisfies the functional equations as described in $(2)$.
It therefore follows that the function $g$ given by $g(z) =F(z) /f(z) $ satisfies $$g(z+1)=g(z),g(-1/z)=g(z)\tag{5}$$ Next we have
\begin{align}
F(z) &=p^{12}\sum_{n=1}^{\infty} \tau(n) q^{pn} +\sum_{k=0}^{p-1}\sum_{n=1}^{\infty}\tau(n)q^{n/p}e^{2\pi ikn/p} \notag\\
&=p^{12}\sum_{n=1}^{\infty}\tau(n)q^{pn}+\sum_{n=1}^{\infty}\tau(n)q^{n/p}\sum_{k=0}^{p-1}e^{2\pi ikn/p} \notag\\
&=p^{12}\sum_{n=1}^{\infty}\tau(n)q^{pn}+p\sum_{n=1}^{\infty} \tau(pn) q^n\tag{6}
\end{align}
and thus $$g(z) =p\tau(p) +c_1q+c_2q^2+\cdots$$ so that $g$ is a modular form of weight $0$ and hence a constant and obviously equal to $p\tau(p) $. We have thus $$F(z) =p\tau(p) f(z) $$ which gives us the identity $$\tau(p) \sum_{n=1}^{\infty}\tau(n)q^n=p^{11}\sum_{n=1}^{\infty}\tau(n)q^{pn}+\sum_{n=1}^{\infty}\tau(pn)q^n\tag{7}$$ The multiplicative property of Ramanujan's tau function is equivalent to the following $$\tau(p^kn) =\tau(p^k) \tau(n) \tag{8}$$ where $p$ is a prime which does not divide $n$. In what follows we show how to get $(8)$ from $(7)$.
The strategy is to apply induction on $k$. For $k=0$ the result is obvious and for $k=1$ the result follows by equating coefficients of $q^n$ in $(7)$. If we equate the coefficients of $q^{p^{k-1}n} $ on both sides of $(7)$ we get $$\tau(p) \tau(p^{k-1}n) =p^{11}\tau(p^{k-2}n)+\tau(p^{k}n)\tag{9}$$ Putting $n=1$ we get $$\tau(p) \tau(p^{k-1})=p^{11}\tau(p^{k-2})+\tau(p^{k})\tag{10}$$ and if we set $$a_k=\tau(p^kn)-\tau(p^k) \tau(n) $$ then the equations $(9), (10)$ lead us to the recursion $$a_{k}=\tau(p)a_{k-1}- p^{11}a_{k-2}$$ Since $a_0=a_1=0$ it now follows by induction that $a_{k} =0$ for all $k$ and the proof is complete.