Equivalency between a "mixed modular equation" of Gauss and a later theorem of Ramanujan. In p. 476 of volume 3 of Gauss's collected works, appear several interesting identities on Jacobi theta functions which were used by the czech mathematician Karel Petr in an 1904 article "Bemerkung zur einer Gausschen Formel über die Thetafunktionen"  to derive relations giving the number of representations of a number $N$ by three quaternary quadratic forms: $x^2 + y^2 + 9z^2 + 9u^2$, $x^2+y^2+z^2+9u^2$, $x^2+9y^2+9z^2+9u^2$. The identities by Gauss are:
$$(\frac{3P^2-P^0\cdot P^0}{2})^2= p^4-4(\frac{pqr}{2})^{\frac{4}{3}}$$
$$(\frac{3Q^2-Q^0\cdot Q^0}{2})^2=q^4+4(\frac {pqr}{2})^{\frac{4}{3}}$$
where the relevant quantities are defined to be:
$$P(x^3,1)=P , P(x,1)=p, P^0=P(x^{\frac{1}{3}},1)$$
$$Q(x^3,1)=Q , Q(x,1)=q, Q^0 = Q(x^{\frac{1}{3}},1)$$
$$R(x^3,1)=R , R(x,1)= r$$
and the three functions $P(x,y),Q(x,y),R(x,y)$ are equivalent to Jacobi's theta functions $\vartheta_3,\vartheta_4,\vartheta_2$ and defined to be:
$$P(x,y)=1+x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})+x^9(y^3+\frac{1}{y^3})+...$$
$$Q(x,y)= 1-x(y+\frac{1}{y})+x^4(y^2+\frac{1}{y^2})-x^9(y^3+\frac{1}{y^3})+...$$
$$R(x,y)=x^{\frac{1}{4}}(y^{\frac{1}{2}}+y^{-\frac{1}{2}})+x^{\frac{9}{4}}(y^{\frac{3}{2}}+y^{-\frac{3}{2}})+x^{\frac{25}{4}}(y^{\frac{5}{2}}+y^{-\frac{5}{2}})+...$$
To see the equivalency between Gauss's notation and Jacobi's theta functions, look at the post Interpretation of a certain general theorem used by Gauss in his work on theta functions.. User Paramanand Singh helped me understand the meaning of Gauss's identities and remarked that they are essentialy a "mixed modular equation" which connects theta functions of $x^{\frac{1}{3}},x,x^3$ (or, equivalently, $\tau,\tau^3,\tau^9$), and said that Ramanujan also gave this modular equation.
According to Paramanand Singh's comments, the following identity (from p. 142 of the book "pi and the AGM") of Ramanujan is equivalent to Gauss's:
$$\frac{\theta_3(q)}{\theta_3(q^9)} - 1 = (\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1)^{\frac{1}{3}}$$
However, he didn't provide proof of equivalency. Therefore, it's not certain this is the desired identity of Ramanujan, and it's not even certain that Ramanujan stated an equivalent identity at all.
Therefore, my questions are:

*

*Can anyone familiar with Ramanujan's results on mixed modular equations say which of Ramanujan's theorems is equivalent to Gauss's identities? and where can i find the relevant fragment of Ramanujan's writings on the internet?

*Can someone also give a proof of equivalency between Ramanujan's theorem and Gauss's identities? this doesn't have to be a proof of correctness of Gauss's identities (which can be quite complicated), just a proof of equivalency to Ramanujan's theorem.

 A: The Broweins' pi and the AGM identity
$$ \frac{\theta_3(q)}{\theta_3(q^9)} - 1 =
 \left(\frac{\theta_3^4(q^3)}{\theta_3^4(q^9)}-1\right)^{\frac{1}{3}} \tag{1} $$
appears in a slightly modified form in
Bruce Berndt, Ramanujan's Notebooks, Part III, page 218, equation (24.29)
$$  \frac{\phi(q^\frac13)}{\phi(q^3)} =
 1 + \left(\frac{\phi^4(q)}{\phi^4(q^3)} - 1\right)^\frac13. \tag{2}$$
The Gauss identity (with a typo fixed)
$$ \left(\frac{3P^2-P^0\cdot P^0}{2}\right)^2= p^4-4\left(\frac{p\,q\,r}2\right)^{\frac{4}{3}} \tag{3} $$
after translating the theta functions into the
Ramanujan notation
$$ q = x^{\frac13}, \quad \phi(q) := f(q,q), \quad f(-q) := f(-q,-q^2) \tag{4} $$
where the general Ramanujan theta function is
$$ f(a,b) := 1 +(a+b) +(a^2+b^2)(ab) +(a^3+b^3)(ab)^3 +\dots \tag{5} $$
and using the identity
$$ \frac{p\,q\,r}2 = \left(q^\frac14 f(-q^6)\right)^3 \tag{6} $$
becomes
$$ \left(\frac{3\phi(q^9)^2 - \phi(q)^2}2\right)^2 = \phi(q^3)^4
 - 4\,q\,f(-q^6)^4. \tag{7} $$
However, the identity in equation $(2)$ is very different than
this equation $(7)$. In the notation of my eta product identity
database, equation $(2)$ is $\texttt{q36_44_420}$ while equation
$(7)$ is not in the database but would be $\texttt{x36_52_520}$.
However, it is possible that several Ramanujan results combined
would imply the Gauss identity.
A: The identity of Gauss has a typo and the correct form is $$\left(\frac{3P^2-P^0P^0}{2}\right)^2=p^4-4\left(\frac{pqr}{2}\right)^{4/3}\tag{1}$$ As in Somos's answer this translates to $$(3\phi^2(q^9)-\phi^2(q))^2=4\phi^4(q^3)-16qf^{4}(-q^6)\tag{2}$$ where $$\phi(q) =\vartheta_3(q)=\sum_{n\in\mathbb {Z}} q^{n^2},f(-q)=\prod_{n=1}^{\infty} (1-q^n)\tag{3}$$ We will use Ramanujan's notation and the proof borrows heavily from work of Dr. Bruce C. Berndt and his collaborators.
Given the nome $q$, the corresponding elliptic modulus $k$ and elliptic integral $K$ (or Ramanujan's $z$) are given in terms of theta functions $$\vartheta_2(q)=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2},k=\frac{\vartheta_2^2(q)} {\vartheta_3^2(q)}, z=\frac{2K}{\pi}=\phi^2(q)\tag{4}$$ Let the parameter $k^2$ for nomes $q, q^3,q^9$ be denoted by $\alpha, \beta, \gamma $ respectively (so that for example $\beta=\vartheta_2^4(q^3)/\vartheta_3^4(q^3)$). Further let the parameter $z$ corresponding to these nomes be denoted by $z_1,z_3,z_9$ (so that $z_3=\phi^2(q^3)$).
The modular equations of degree 3 allow us to express $\alpha, \beta, \gamma $ in terms of multiplier $m$. Let us define $$m=\frac{z_1}{z_3},m'=\frac{z_3}{z_9}\tag{5}$$ Berndt gives us $$\alpha = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}, \beta =\frac{(m - 1)^{3}(3 + m)}{16m}\tag{6}$$ and $$1-\alpha=\frac{(3 - m)^{3}(m + 1)}{16m^{3}}, 1-\beta =\frac{(m + 1)^{3}(3 - m)}{16m}\tag{7}$$ If we replace $m$ by $m'$ in above equations then $\alpha$ changes to $\beta$ and $\beta$ changes to $\gamma$. It should also be clear that if $0<q<1$ then we have $1<m,m'<3$.
Next Berndt uses another set of parameters $$u=\frac{m-1}{2}+\left(\frac{m-1}{2}\right)^2, v=\frac{m'-1}{2}+\left(\frac{m'-1}{2}\right)^2\tag{8}$$ (Berndt actually uses intermediate parameters $p, p'$ with $m=1+2p,m'=1+2p'$ and $q=p+p^2, q'=p'+p'^2$ instead of $u, v $, but this will conflict with the usage of $q$ for nome and hence we avoid it).
We have the obvious relations $$m=\sqrt {1+4u},m'=\sqrt{1+4v}\tag{9}$$ and the more difficult relations $$\alpha(1-\alpha)=u\left(\frac {2-u}{1+4u}\right)^3,\beta(1-\beta)=u^3\left(\frac{2-u}{1+4u}\right)\tag{10}$$ and $$\beta(1-\beta)=v\left(\frac{2-v}{1+4v}\right)^3,\gamma(1-\gamma)=v^3\left(\frac{2-v}{1+4v}\right)\tag{11}$$ Using the two expressions for $\beta(1-\beta)$ one gets a polynomial equation between $u, v$ of degree $4$ which Berndt successfully (!) solves using another parameter $t$ given by $v=2t^3$.
The algebra involved is tedious and I reproduce the necessary results directly from Berndt's Ramanujan's Notebooks Vol 3, page 354: $$m=\frac{(1+2t)^2}{\sqrt{1+8t^3}},m'=\sqrt{1+8t^3}\tag{12}$$ and $$\beta(1-\beta)=16t^3\left(\frac{1-t^3}{1+8t^3}\right)^3\tag{13}$$ The relation $(2)$ can be transcribed easily as $$\left(\frac{3}{m'}-m\right)^2=4-2^{8/3}(\beta(1-\beta))^{1/3}\tag{14}$$ which is easily verified using $(12),(13)$.

I am sure that Gauss would have provided a simpler proof using his relations between theta functions.
A: I just want to describe the arithmetic information yielded by Gauss's identities. As in Somos's and Paramanand Singh's answers, Gauss's identity translates into the equation:
$$(3\phi^2(q^9)-\phi^2(q))^2 = 4\phi^4(q^3)-16qf^4(-q^6)$$
After opening the parentheses, this equation becomes:
$$9\phi^4(q^9)- 6\phi^2(q^9)\phi^2(q) +\phi^4(q) = 4\phi^4(q^3)-16qf^4(-q^6)\implies 6\phi^2(q^9)\phi^2(q) = 9\phi^4(q^9)+\phi^4(q)-4\phi^4(q^3)+16qf^4(-q^6)$$
Now, the coefficient of the $N$ power in the series expansion of the left side of the second equation, $6\phi^2(q^9)\phi^2(q)$, is none other than $6$ times the number of representations of $N$ by the quaternary quadratic form $x^2+y^2+9z^2+9u^2$. Denote this number as $\varphi(N)$. In addition the coefficient of the $N$ power in the series expansion of the term $\phi^4(q)$ is none other than the usual sum of four squares function, $r_4(N)$ (number of representations of $N$ by $x^2+y^2+z^2+u^2$), so let us denote this as $\chi(N)$.
For $N$ not divisible by $3$, the term $9\phi^4(q^9)-4\phi^4(q^3)$ contribute nothing to $\varphi(N)$ (these theta powers generate only powers of $q$ with exponents divisible by $3$). In addition, the coefficient of the $N$ power in the series expansion of the term $16qf^4(-q^6)$ can be interpreted as $16$ times the number of partitions of $\frac{N-1}{6}$ with at most $4$ repetitions of the same summand, with each partition counted as $(-1)$ or $(+1)$ if the partition is into an odd or even number of parts, respectively. Denote this as $16\gamma(\frac{N-1}{6})$.
Than we get:
$$\varphi(N)=\frac{1}{6}(\chi(N)+16\gamma(\frac{N-1}{6})), N\ne 0 \pmod 3$$
In p.300 of volume 2 of L. Dickson "history of the theory of numbers" you can find Karel Petr's exhaustive result for all the cases of $N$, including a different form for the case $N\ne 0 \pmod 3$ (which I was not able to prove from my interpretation of $qf^4(-q^6)$). Unfortunately, I was not able to find Karel Petr's original article about his interpretation of Gauss's equation, nor I am able to understand what led Gauss to this particular equation.
