Given the Dedekind eta function $\eta(\tau)$, define,


$$\alpha(\tau) =\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}$$


$$\beta(\tau) =\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\quad\;$$


$$\quad\gamma(\tau) =\frac{\alpha(\tau)}{\beta(\tau)} =\frac{\sqrt2\,\eta(4\tau)}{\eta(\tau)}$$


Note that $\alpha^8(\tau)=\lambda(2\tau)$ with *[modular lambda function](https://en.wikipedia.org/wiki/Modular_lambda_function)* $\lambda(\tau)$. We have the well-known,


$$\alpha^8(\tau)+\beta^8(\tau) = 1\tag1$$


as well as the nice,


$$\frac{1}{\beta^2(\tau)}-\beta^2(\tau) =\left(2^{1/4}\,\gamma(2\tau)\right)^4\tag2$$


$$\frac{1}{\alpha^2(2\tau)}-\alpha^2(2\tau) =\left(\frac{2^{1/4}}{\gamma(\tau)}\right)^4\tag3$$


and,


$$\frac{1}{\alpha^2(2\tau)}+\alpha^2(2\tau) =\left(\frac{2^{1/4}}{\alpha(\tau)}\right)^4\tag4$$


$$\frac{1}{\beta^2(\tau)}+\beta^2(\tau) =\left(\frac{2^{1/4}}{\beta(2\tau)}\right)^4\tag5$$


As eta functions (not as quotients $\alpha$ and $\beta$), these 5 are in [Somos' database](http://eta.math.georgetown.edu/) (as *t4\_24\_48, t8\_12\_24, t8\_12\_48, t8\_18\_60a, t8\_18\_60b*). After some algebraic manipulation, I realized these can be expressed in more aesthetic forms. (Of course, it is easy to transform these to the form $x^2+y^2=1$.) 




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**Questions:**


1. For the next step, can we express
$$\frac{1}{\beta^4(\tau)}+\beta^4(\tau) =t\_1^2$$
$$\frac{1}{\alpha^4(2\tau)}+\alpha^4(2\tau) =t\_2^2$$
where the $t\_i$ are eta quotients?
2. Are there are other eta quotient parameterizations **strictly of form**,
$$\left(m\_1\prod\eta(a\_1\tau)^{c\_1}\,\eta(a\_2\tau)^{c\_2}\dots\right)^2 + \left(m\_2\prod\eta(b\_1\tau)^{d\_1}\,\eta(b\_2\tau)^{d\_2}\dots\right)^2 = 1$$
where $a\_i, b\_i, c\_i, d\_i$ are **integers** (the exponents $c\_i,d\_i$ may be negative) and $m\_i$ are algebraic similar to the above 5?



Are these 5 the only eta quotients that parameterize $x^2+y^2 = 1$?
For your question $2$ all that is needed is a quick search of 'eta07.gp' for three terms identities with even rank and with all even exponents. I turned
up $t\_{12,12,40}$ and the other $5$ members of the $6$ cluster. They are: $$t\_{12,12,40},\; t\_{12,12,56},\; t\_{12,24,72},\; t\_{12,24,90},\; t\_{12,24,120},\; t\_{12,24,126}.$$ For the first, the identity is equivalent to
$$ \left(\sqrt{\frac14}\,\frac{u\_1^3\,u\_{12}}{u\_3\,u\_4^3}\right)^2 + \left(\sqrt{\frac34}\,\frac{u\_2^4\,u\_6^2}{u\_1\,u\_3\,u\_4^4}\right)^2 = 1 $$
where $u\_k = \eta(k\tau)$, and similarly for the other $5$ identities. Probably other similar clusters can be found.


Note the two transforms. First, given an eta identity, replacing $q$ with $-q$ gives an eta identity. Second, the same with replacing $q^d$ with $q^{N/d}$ where $N$ is the level. By iterating these two transforms we get an eta identity cluster. There can be up to $12$ identities in a cluster.


