Ramanujan's tau function identity While studying Ramanujan's tau function, I observed that the function satisfies a beautiful identity that I had not seen previously in the literature.
Let $\tau(n)$ be Ramanujan's tau function, such that
$$q\prod_{n=1}^\infty (1-q^n)^{24} = \sum_{n=1}^\infty \tau(n)\,q^n,$$
then 
$$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}}$$ 
I believe it is well known due to the popularity of the function. Can anyone prove the identity?
The identity implies the congruence property $$\tau(2)\equiv0\pmod{8}$$ 
Thus proving the identity is akin to proving the congruence property.
 A: The identity follows immediately from the fact that 1728 can be expressed in terms of ramanujan tau function in two different ways,namely
$$2\tau(2){(\tau(3)-2^{11})}-\tau(8)=12^{3}$$
And 
$$-\dfrac{\tau(2){(\tau(4)+2^{11})}}{8}=12^{3}$$
After equating the two identities and using the multiplicative property of the function,
$\tau(mn)=\tau(m)\tau(n)$ for $\gcd(m,n)=1$,
Then we formally obtain  the identity after some simple algebraic manipulation.
A: This answer is meant to demonstrate some ways of calculating $\tau(n)$,
ranging from elegant via straight to weird.
Elegant:
Ramanujan's $\tau(n)$ is a multiplicative arithmetic function;
so $\tau(mn) = \tau(m)\,\tau(n)$ whenever $\gcd(m,n)=1$.
However, its multiplicativity properties are also known for non-coprime $m,n$:
$$\begin{align}
\tau(m)\,\tau(n) &= \sum_{d\mid\gcd(m,n)}
d^{11}\tau\!\left(\frac{mn}{d^2}\right) \\
\tau(mn) &= \sum_{d\mid\gcd(m,n)}
\mu(d)\,d^{11}\tau\!\left(\frac{m}{d}\right)\tau\!\left(\frac{n}{d}\right)
\end{align}$$
where $\mu$ is the Möbius function.
Since $\tau$ is not the zero function, its multiplicativity implies $\tau(1)=1$,
which I will implicitly use from now on.
From the special multiplicativity properties above we get
$$\begin{align}
\tau(4) &= \tau(2\cdot 2) = \tau(2)^2 - 2^{11} \\
\tau(6) &= \tau(2)\,\tau(3) \\
\tau(8) &= \tau(2\cdot 4) = \tau(2)\,\tau(4) - 2^{11}\tau(2)
= \tau(2)^3 - 2^{12}\tau(2)
\end{align}$$
Plugging in and assuming $\tau(2)\neq0$, which will be justified later,
we find that your claim is equivalent to
$$7\,\tau(2)^2 = 16\,\tau(3) \tag{*}$$
Straight:
So let us work out $\tau(2),\tau(3)$ now. Denoting
$$\begin{align}
\operatorname{D}(q) &= q\prod_{n=1}^\infty(1-q^n)^{24} =
\sum_{n=1}^\infty\tau(n)\,q^n && \text{(modular discriminant)} \\
\operatorname{P}(q) &= 1 - 24\sum_{n=1}^\infty \sigma(n)\,q^n &&
\text{(quasi-modular Eisensein series)} \\
\text{where}\quad \sigma(n) &= \sum_{d\mid n}d && \text{(divisor sum)}
\end{align}$$
we have
$$q\frac{\mathrm{d}\operatorname{D}(q)}{\mathrm{d}q} =
\operatorname{P}(q) \operatorname{D}(q)$$
which after expansion into $q$-series and comparing coefficients
yields the recurrence relation
$$(n-1)\,\tau(n) = -24\sum_{k=1}^{n-1}\sigma(k)\,\tau(n-k)$$
In particular,
$$\begin{align}
\tau(2) &= -24 \\
\tau(3) &= -12\left(3 + \tau(2)\right) = 2^2 3^2 7 = 252
\end{align}$$
and that fulfills $(*)$.
Weird:
In fact, $(*)$ is a direct consequence of another general recurrence which
I had worked out years ago just for fun, without finding any use for it.
Namely, $\operatorname{P}(q)$ fulfills a differential equation of order 3
and Chazy type III:
$$2\operatorname{P}'''\!{} - 2\operatorname{P}\operatorname{P}''\!{}
+ 3\operatorname{P}'^2 = 0 \quad\text{where}\quad
(\,)' = q\frac{\mathrm{d}(\,)}{\mathrm{d}q}$$
Plugging in $\operatorname{P} = \frac{\operatorname{D}'}{\operatorname{D}}$
gives a fourth-order differential equation for $\operatorname{D}$:
$$2\operatorname{D}^3\operatorname{D}''''\!{}
- 10\operatorname{D}^2\operatorname{D}'\operatorname{D}'''\!{}
- 3\operatorname{D}^2\operatorname{D}''^2\!{}
+ 24\operatorname{D}\operatorname{D}'^2\operatorname{D}''\!{}
- 13\operatorname{D}'^4 = 0$$
Plugging in the $q$-series for $\operatorname{D}$ then results in
a nonlinear recurrence relation for $\tau(n)$ that does not use
other arithmetic functions:
$$\begin{align}
2(n-2)(n-1)^3\tau(n) &=
\sum_{\substack{a,b,c,d=1,\ldots,n-1\\a+b+c+d=n+3}}
f(a,b,c,d)\,\tau(a)\,\tau(b)\,\tau(c)\,\tau(d) \\
\text{with}\quad
f(a,b,c,d) &= -2 a^4 + 10 a^3 b + 3 a^2 b^2 - 24 a^2 b c + 13 a b c d
\end{align}$$
Surely you want to quit, but we are now ready to apply this for $n=3$.
Then $(a,b,c,d)$ runs through the six distinct permutations of $(1,1,2,2)$,
and we get
$$16\,\tau(3) = (15+34+16-14-80+36)\,\tau(2)^2 = 7\,\tau(2)^2$$
which confirms $(*)$ without even needing an explicit value for $\tau(2)$.
As mentioned before, I do not believe such relations to be of great interest;
but if you are in a playful mood, working out such stuff might be fun.
By the way, one can work out an order-$3$ differential equation for
$\operatorname{D}$ (which unfortunately is even more complex) and
try to work out a recurrence relation for $\tau(n)$ from that.
Interested?
A: Looking at the link,
$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}}
$
is
$-24
=8\frac{84480-2\ (-6048)}{-1472-15\cdot 2048}
=-24
$
so it checks.
How did you find it?
