The post has been divided into sections to show some patterns.

I. Parameter $s=\frac12$

Given the nome $q = e^{\pi i\tau}$ and the Jacobi theta functions $\vartheta_n(0,q)$. Define the modular lambda function,

$$\alpha = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$

Then we propose for appropriate $\tau$ the ratios below are algebraic numbers,

\begin{align} \left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=\alpha\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1 \end{align}

Note that adding the first two implies the third. Hence,

$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$

which is known to be true. As eta quotients in the same order of $\vartheta_n(0,q)$ above,

$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4 \overset{\color{red}?}= \left(\frac{\eta^4\big(\tfrac{\tau}2\big)+4\eta^4(2\tau)}{2\eta^3(\tau)}\right)^4$$

where the sum has a nice alternative form to be consistent with the cubic version below.

II. Parameter $s=\frac13$

Given the square of the nome, so $q = e^{2\pi i\tau}$ and the Borwein cubic theta functions $a(q),b(q),c(q)$. Define,

$$\beta = \left(\frac{3}{\left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3=\left(\frac{c(q)}{a(q)}\right)^3$$

Then we propose,

\begin{align} \left(\frac{c(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=\beta\\ \left(\frac{b(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1-\beta\\ \left(\frac{a(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1 \end{align}

Adding the first two implies the third,

$$\big(c(q)\big)^3+\big(b(q)\big)^3=\big(a(q)\big)^3$$

which is also known to be true. As eta quotients,

$$\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3+\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$

III. Parameter $s=\frac14$

Given the square of the nome $q = e^{2\pi i\tau}$. Define, $$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$

Then,

\begin{align} \left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1 \end{align}

The first two implies the third,

$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$

where $C(q), B(q), A(q)$ are defined by the eta quotients,

$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$

Note that,

$$\frac{24}{A(q)-1}= \frac1{q} -1 - 3q + 6q^2 + q^3 - 20q^4 + 24q^5 + 38q^6 - 132q^7 + \dots$$

which seems to be A335227.

IV. Parameter $s=\frac16$

(Unfinished.)

V. Context

These observations arose from evaluations of the complete elliptic integral of the first kind, $K(k)$. For ex., given the tribonacci constant $T$, the real root of $T^3-T^2-T-1=0$, then,

$$K(k_{11}) = \frac{\pi\,(2T)^{2/3}}{2\;}\times \frac{\Gamma\big(\tfrac1{11}\big) \Gamma\big(\tfrac3{11}\big) \Gamma\big(\tfrac4{11}\big) \Gamma\big(\tfrac5{11}\big) \Gamma\big(\tfrac9{11}\big)}{11^{1/4}(2\pi)^3}$$

Manipulating the $s=\frac12$ relations above, we can have a much shorter version,

$$K(k_{11}) = \frac{\pi\,(2T)^{4/3}}{2\;}\times\eta^2\Big(\tfrac{1+\sqrt{-11}}2\Big)\quad\quad$$

Equating the two formulas, this also gives the explicit evaluation of $\eta(\tau)$.

VI. Question

Q: Are the proposed relations $M\overset{\color{red}?}=N$ with the red question marks in fact true?