How to prove that $\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4)$? How can we prove that
$$\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4) \ ?$$
Here, $\eta(q)$ is the Dedekind Eta Function, which is defined by
$$\eta(q)=q^{1/24}\prod_{n=1}^\infty \left(1-q^{n} \right) $$
I looked up some identities of Dedekind Eta Function in the hope of simplifying the right hand expression but I failed. I would be grateful if any one could suggest a proof of this identity.
Source
I came across this identity in a paper. They refer to Entry $t_{8,12,48}$ and $t_{8,18,60a}$ of Dedekind eta function product identities. Unfortunately, I was unable to find these entries on that site.
 A: I will give another answer here. My idea is to use the dimension formula for a space of modular forms with given weight and level.
First, to find the weight and level of the left-hand side, I will use a result of Gordon, Hughes, and Newman for the modularity of some eta-quotients. I do not write down the statement here. You can find this result in Ken Ono's book 'The Web of Modularity' (cf. Theorem 1.64). According to this result, the eta-quotient in the left-hand side is a modular form of weight 5 for $\Gamma_{0}(8)$ with character $\chi$, where
$$\chi(n):=\begin{cases}
1 &\text{if}~n\equiv 1\mod{4},\\
-1 &\text{if}~n\equiv 3\mod{4}.
\end{cases}$$
Note that there is a simple formula for vanishing order of an eta-quotient at cusps (cf. Theorem 1.65 of 'The Web of Modularity'). Using this formula, we see that the left-hand side is a cusp form. We denote by $S_{5}(\Gamma_{0}(8),\chi)$ the space of cusp forms of weight 5 for $\Gamma_{0}(8)$ with character $\chi$ as usual.
Using the dimension formula given by Cohen and Oesterle, we obtain $\dim_{\mathbb{C}}S_{5}(\Gamma_{0}(8),\chi)=2$. From theorems 1.64 and 1.65 in Ken Ono's book, we have that $F_{1}:=\eta^{4}(q^{2})\eta^{2}(q^{4})\eta^{4}(q^{8})$ and $F_{2}:=\eta^{4}(q)\eta^{2}(q^{2})\eta^{4}(q^{4})$ are contained in $S_{5}(\Gamma_{0}(8),\chi)$. Note that $F_{1}$ and $F_{2}$ have Fourier expansions starting with $q^{2}$ and $q$, respectively. This follows easily from the infinite product expansion of $\eta$. Thus, $F_{1}$ and $F_{2}$ are linearly independent, hence forms a basis for $S_{5}(\Gamma_{0}(8),\chi)$. By comparing Fourier coefficients (we can compute a finite number of Fourier coefficients of an eta-quotients using Sage), we have
$$\frac{\eta^{14}(q^{4})}{\eta^{4}(q^{8})}=4F_{1}+F_{2}.$$
A: Dividing both sides of the proposed identity by a factor of $\eta(q^2)^{4}\eta(q^4)^{4}$ and rearranging we have the following expression
$$\eta(q)^{4}/\eta(q^2)^{2}=\eta(q^4)^{10}/\eta(q^2)^{4}\eta(q^8)^{4}-4\eta(q^8)^{4} \tag1$$
Now this is readily seen to be equivalent to the following theta function formula, namely
$$\theta_4(q)^{2}=\theta_3(q^2)^{2}-\theta_2(q^2)^{2} \tag2$$
Where each theta function ([1],[2] Chapter XXI,[6],[7]) corresponds to each successive eta function quotient in equation (1) ([3] Theorem 1.60, [4],[7]).
Equation (2) is actually equivalent to a specialization of an elegant theta function identity due to G. N. Watson ([5],[8]),
$$\theta_3(z,q)\theta_3(w,q)=\theta_3(z+w,q^2)\theta_3(z-w,q^2)+\theta_2(z+w,q^2)\theta_2(z-w,q^2)\tag3$$ where $z=w=\pi/2$, and $$\theta_3(\pi/2,q)=\theta_4(q),\theta_3(\pi,q^2)=\theta_3(q^2),\theta_2(\pi,q^2)=-\theta_2(q^2).$$
Q.E.D.
References:
[1] Jacobian Theta Functions (http://dlmf.nist.gov/20.2#i)
[2] E. T. Whittaker & G. N. Watson A Course Of Modern Analysis (https://archive.org/details/courseofmodernan00whit)
[3] The Web of Modularity: Arithmetic of the Coefficients of Modular Forms and $q$-series, Ken Ono
http://www.ams.org/publications/authors/books/postpub/cbms-102
[4] ETA-QUOTIENTS AND THETA FUNCTIONS
ROBERT J. LEMKE OLIVER
http://math.tufts.edu/faculty/rlemkeoliver/papers/06-EtaTheta.pdf
[5] Watson's Identities
http://dlmf.nist.gov/20.7#v
[6] https://en.wikipedia.org/wiki/Jacobi_elliptic_functions
[7] http://mathworld.wolfram.com/JacobiEllipticFunctions.html
[8] §20.2(iii) Translation of the Argument by Half-Periods http://dlmf.nist.gov/20.2#iii
A: With the settings of the answer n.2 of this link (Direct proof of the level four eta identity) rearranging the equation
$$\frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4)$$
As
$$2\frac{\eta^{4}(q^{8})}{\eta^{4}(q^{4})}=\frac{1}{2}\frac{\eta^{8}(q^{4})}{\eta^{4}(q^{2})\eta^{4}(q^{8})}-\frac{1}{2}\frac{\eta^{4}(q)}{\eta^{2}(q^{2})\eta^{2}(q^{4})}$$
We have
$$\frac{1}{g_{64n}^{4}}=G_{16n}^{4}-\frac{g_{16n}^{6}}{\sqrt{2}G_{4n}^{4}}$$
This equation can be written as 
$$\frac{1}{g_{16n}^{4}}=G_{4n}^{4}-\frac{g_{4n}^{6}}{\sqrt{2}G_{n}^{4}}$$
Now we show how this equation is part of fundamental Ramanujan's equation (1).
We know from (1) of link above, that here we rewrite
$$4g_{4n}^8 G_{4n}^8\big(G_{4n}^8-g_{4n}^8\big)=1\tag1$$
That we can write as
$$\big(G_{4n}^4+g_{\frac{1}{4n}}^4\big) \big(G_{4n}^4-g_{\frac{1}{4n}}^4\big)= g_{4n}^8$$
Moreover
$$\big(G_{4n}^4+g_{\frac{1}{4n}}^4\big)=\frac{ g_{4n}^8}{\big(G_{4n}^4-g_{\frac{1}{4n}}^4\big)}= \frac{g_{16n}^{6}}{\sqrt{2}G_{4n}^{4}}=2g_{n}^2G_{n}^6$$
$$\big(G_{4n}^4-g_{\frac{1}{4n}}^4\big)=2g_{n}^6G_{n}^2$$
Multiplying the L.H.S. we have
$$\big(G_{4n}^8-g_{\frac{1}{4n}}^8\big)=g_{4n}^8$$
Q.D.E.
A: This Dedekind eta product identity
$$ \frac{\eta^{14}(q^4)}{\eta^{4}(q^8)}=4\eta^4(q^2)\eta^2(q^4)\eta^4(q^8)+\eta^4(q)\eta^2(q^2)\eta^4(q^4)$$
is equivalent to identity $\, t_{8,12,48} \,$  at
Dedekind eta product Identities. As mentioned there, it is also equivalent to, and given with proof, in
G. E. Andrews and B. C. Berndt, "Ramanujan's Lost Notebook", Part I, page 386, entry $17.3.8(c)$ as
$$ m(1-\alpha)^{1/4} + \beta^{1/2} = 1 $$
where $\, m=(\phi(q)/\phi(q^4))^2, \,$
$\, \alpha = \lambda(\tau), \,$
$\, \beta =  \lambda(4\tau), \,$
$\, \phi(q) \,$ is a Ramanujan theta function
and $\, \lambda(\tau) \,$ is the Modular lambda function.
There are a $13$ equivalent forms of the identity  $\, t_{8,12,48} \,$ listed
at the website mentioned earlier. The form
$\, \theta_4(q)^{2}=\theta_3(q^2)^{2}-\theta_2(q^2)^{2} \,$ mentioned in another answer is one of them. The one mentioned in the question comes from
Rogers, Wan, and Zucker, arXiv:1303.2259 page 4.
A: Usually, the Dedekind eta function is understood as a function of
$\tau\in\mathbb{H}$ (complex upper half plane) where
$q=\mathrm{e}^{2\pi\mathrm{i}\tau}$.
This way, the modular symmetries of $\eta$ can be expressed
easily and the branching issues with $q^{1/24}$ can be avoided.
In this answer, I will not make use of modular symmetries,
but perhaps some other answer will,
so let us stick to the convention of regarding $\eta$ as a function of $\tau$
and use another symbol when we regard it as a function of $q$.
Following Michael Somos's example at the product identity website
linked in the question, I will write $h(q)$ for $\eta(\tau)$.
Dividing by $h^4(q^4)\,h^4(q^2)$ and isolating the rightmost summand,
your identity in question becomes
$$\frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)} -
\frac{4\,h^4(q^8)}{h^2(q^4)} = \frac{h^4(q)}{h^2(q^2)}
\tag{*}$$
Apart from having some power of $q$ as overall multiplier,
the factors of eta products (and quotients) can be written as
$1+a_n$ with $a_n=\operatorname{O}(q^{kn})$ for some $k>0$
with $k$ and the implied $\operatorname{O}$ constant not depending on $n$,
therefore $\sum_{n=1}^\infty |a_n|$ converges absolutely for $|q|<1$,
and we can reorder the factors accordingly. Using rules like
$$\begin{aligned}
 \prod_{n=1}^\infty (1-x^{2n}w)(1-x^{2n-1}w)
 &= \prod_{n=1}^\infty (1-x^n w)
\\ \prod_{n=1}^\infty (1+x^n)(1-x^{2n-1}) &= 1
\end{aligned}$$
for $x=q^k$ and applying Jacobi's
triple product identity,
we find
$$\begin{aligned}
 \frac{h^{10}(q^4)}{h^4(q^2)\,h^4(q^8)}
 &= \prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m-2})^4
 = \left(\sum_{n\in\mathbb{Z}} q^{2n^2}\right)^2
 = \vartheta_3^2(q^2)
\\ \frac{4\,h^4(q^8)}{h^2(q^4)}
 &= q\prod_{m=1}^\infty (1-q^{4m})^2\,(1+q^{4m})^2\,(1+q^{4m-4})^2
 = q\left(\sum_{n\in\mathbb{Z}} q^{2n(n+1)}\right)^2
 = \vartheta_2^2(q^2)
\\ \frac{h^4(q)}{h^2(q^2)}
 &= \prod_{m=1}^\infty (1-q^{2m})^2\,(1-q^{2m-1})^4
 = \left(\sum_{n\in\mathbb{Z}} (-q)^{n^2}\right)^2
 = \vartheta_4^2(q)
\end{aligned}$$
where $\vartheta_2, \vartheta_3, \vartheta_4$ are Jacobi thetanull functions.
Therefore (*) is equivalent to the identity
$$\vartheta_3^2(q^2) - \vartheta_2^2(q^2) = \vartheta_4^2(q) \tag{**}$$
This is one of six formulae that relate theta nullvalues with those
corresponding to halved or doubled period ratios.
It can be shown by reordering the nested sum for $\vartheta_4^2(q)$
in a chessboard-like manner:
$$\begin{aligned}
 \vartheta_4^2(q)
 &= \sum_{u,v\in\mathbb{Z}}
 (-q)^{u^2+v^2}
 = \sum_{\substack{r,s\in\mathbb{Z}\\r+s\text{ even}}} q^{r^2+s^2}
 - \sum_{\substack{r,t\in\mathbb{Z}\\r+t\text{ odd}}} q^{r^2+t^2}
\\ &\stackrel{\substack{r = k - m\\s = k + m\\t = k + m + 1}}{=}
 \sum_{k,m\in\mathbb{Z}} q^{2(k^2+m^2)}
 - \sum_{k,m\in\mathbb{Z}} q^{\left((2k+1)^2+(2m+1)^2\right)/2}
 = \vartheta_3^2(q^2) - \vartheta_2^2(q^2)
\end{aligned}$$
You may be interested in writing down the other period-ratio doubling
and halving formulae and transforming them to eta product identities
that involve power products of $h(q), h(q^2), h(q^4), h(q^8)$ only.
A: If $q\in(0,1)$ and $k\in(0,1)$ is the elliptic modulus corresponding to nome $q$ and $K=K(k) $ is complete elliptic integral of first kind then we have the relation $$\eta(q) =2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{1}$$ Using Landen transformation we can replace $q$ by $q^2$ and in this process $k$ gets replaced by $(1-k')/(1+k')$ and $K$ is replaced $K(1+k')/2$. In this manner we get the following set of identities
\begin{align} \eta(q^2)&=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{2}\\
\eta(q^4)&=2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{3}\\
\eta(q^8)&=2^{-13/12}\sqrt{\frac{2K}{\pi}}\frac{k^{2/3}k'^{1/24}}{(1+k')^{1/4}}\tag{4}
\end{align}
The problem is now reduced to algebraic manipulation and can be done without much effort. 
