How to show an infinite number of algebraic numbers $\alpha$ and $\beta$ for $_2F_1\left(\frac13,\frac13;\frac56;-\alpha\right)=\beta\,$?

(Updated.) Courtesy of Nemo's answer, we finally find a simple closed-form solution to the equation, $$\,_2F_1\Big(\tfrac13,\tfrac13;\tfrac56;- \alpha\Big)=\beta\tag1 $$ in algebraic numbers $\alpha, \beta$ analogous to this post. Let $\lambda=\frac{\eta\big(\tfrac{\tau+1}{3}\big)}{\eta(\tau)}$, then, $$\begin{aligned} \alpha &=\alpha(\tau) =\frac{(u-1)^2}{4u} =\frac1{4\sqrt{27}}\big(\lambda^3-\sqrt{27}\,\lambda^{-3}\big)^2\\[2.5mm] \beta &= \beta(\tau) =\frac{1+N}{432^{1/4}}\,\color{blue}{\frac{\sqrt{-3}}{1+\tau}}\, \frac{u^{1/3}}{(2u^2-2)^{1/3}}\frac{_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{u^2}{u^2-1}\Big)}{\pi^{-1}\,K(k_3)}\end{aligned}$$ where, $$u=\frac{\lambda^6}{\sqrt{27}},\quad\tau= \tfrac{1+N\sqrt{-3}}2$$ The formulas for the equality $(1)$ holds for real $N>1,$ but $\alpha(\tau)$ and $\beta(\tau)$ are algebraic numbers for integer $N>1$. Example: $$\alpha\big(\tfrac{1+7\sqrt{-3}}2\big) = 27,\quad\quad\beta\big(\tfrac{1+7\sqrt{-3}}2\big) = \tfrac47$$ Also, note that, $$\frac{\,_2F_1\Big(\tfrac13,\tfrac23;1;\,1-\tfrac{u^2}{u^2-1}\Big)}{\,_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{u^2}{u^2-1}\Big)}=\color{blue}{\frac{\sqrt{-3}}{1+\tau}}$$


(Old answer) It seems,

$$_2F_1\left(\frac16,\frac13;\frac12;\,\gamma^2\right)=\delta\tag1$$ $$_2F_1\left(\frac13,\frac13;\frac56;-\alpha\right)=\beta\tag2$$ are complementary. Let $N$ be any positive integer.

I. If $\color{brown}{\tau =N\sqrt{-3}}\,$ and $\gamma$ is an appropriate root of,

$$\color{blue}{j(\tau) = \frac{432}{1+\gamma}\left(\frac{5+4\gamma}{1 - \gamma}\right)^3}$$ or alternatively, $$\frac{108}{1-\gamma^2}=\left(\frac{\eta^6\big(\tfrac{\tau}{3}\big)}{\eta^6(\tau)} +27\frac{\eta^6(\tau)}{\eta^6\big(\tfrac{\tau}{3}\big)} \right)^2$$ then $\gamma^2$ and $\delta$ of $(1)$ are algebraic numbers. Example, if $\tau =2\sqrt{-3}$, then $\gamma^2=\frac{25}{27}$ and $\delta = \frac34\sqrt{3}$.

II. If $\color{brown}{\tau =\frac{1+N\sqrt{-3}}2}\,$ and $\alpha$ is an appropriate root of,

$$\color{blue}{j(\tau) = \frac{432}{1+f}\left(\frac{5+4f}{1 - f}\right)^3},\quad \text{where}\quad f = \frac{2\alpha+1}{2\sqrt{\alpha(1+\alpha)}}$$ or alternatively, $$432\,\alpha(1+\alpha)=\left(\frac{\eta^6\big(\tfrac{\tau+1}{3}\big)}{\eta^6(\tau)} -27\frac{\eta^6(\tau)}{\eta^6\big(\tfrac{\tau+1}{3}\big)} \right)^2$$ then $\alpha$ and $\beta$ of $(2)$ are also algebraic numbers. Example, if $\tau =\frac{1+7\sqrt{-3}}2$ then $\alpha = 27$ and $\beta = \frac47$.

Part 1, after some manipulation by the OP, can be derived a result in Zucker's and Joyce's, Special values of the hypergeometric series III. The fact that Part 2 shares tantalizing common forms suggest it may be amenable to a similar treatment.


The formula for $\beta={}_2F_1\Big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-\alpha\Big)$ in terms of elliptic integrals is $$\large\begin{align} &{}_2F_1\Big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};\tfrac{1}{2}+i\tfrac{(2+2p-p^2) (1-2 p -2p^2) (1+4p+p^2)}{6 \sqrt{3} ~p(p+2) (2 p+1)(1-p^2)}\Big)\\&=\tfrac{\sqrt[3]{p(2+p)(1-p^2)}}{K(k_3)3^{1/4} (2 p+1)^{1/6}}\Big(e^{-\frac{\pi i}{6}}K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)+\tfrac{e^{\frac{\pi i}{6}}}{\sqrt{3}}K\Big(\sqrt{1-\tfrac{p^3 (2+p)}{1+2 p}}\Big)\Big) \end{align}\tag1$$ valid for $0<p<1$ (the proof is given at the end of this post). It can be continued analytically in the vicinity of this range. One can easily see from this formula how $\alpha$ and $\beta$ are parametrized in terms of $p$. However more work needs to be done to show that they are both algebraic when suitable parameter $\tau$ takes values $\tau=\frac{1+n\sqrt{-3}}{2}$, $~n\in\mathbb{N}$. To do this one needs parametrization in terms of eta quotients. Such parametrization is given in chapter 33, of Ramanujan's Notebooks, Part V (referred in the following as V).

Define (according to Lemma 5.5 in V) $$p(v)=-2\,\frac{\eta\big(\tfrac{v}2\big)\,\eta^3\big(6v\big)}{\eta\big(2v\big)\,\eta^3\big(\tfrac{3v}2\big)}\tag2$$

with $\large v=\frac{\tau}{\tau+1}.$ Then $$ \alpha(\tau)=-\tfrac{1}{2}-i\tfrac{(2+2p-p^2) (1-2 p -2p^2) (1+4p+p^2)}{6 \sqrt{3} ~p(p+2) (2 p+1)(1-p^2)} $$ $$ \beta(\tau)=\tfrac{\sqrt[3]{p(2+p)(1-p^2)}}{K(k_3)3^{1/4} (2 p+1)^{1/6}}\Big(e^{-\frac{\pi i}{6}}K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)+\tfrac{e^{\frac{\pi i}{6}}}{\sqrt{3}}K\Big(\sqrt{1-\tfrac{p^3 (2+p)}{1+2 p}}\Big)\Big) $$

Example: If $\tau=\frac{1+7\sqrt{-3}}2$, then $v=\frac{7i}{26 \sqrt{3}}+\frac{25}{26}$, and $\alpha=27$, $~\beta=4/7$.

Proof of such a connection between $\tau$ and $v$ is equivalent to verification of an eta function identity, as shown in the sequel (a simpler example of analogous verification can be found in this answer). First, using the trivial identity $\eta \left(\frac{v+1}{2}\right)=\frac{\zeta_{48} \eta (v)^3}{\eta \left(\frac{v}{2}\right) \eta (2 v)}$ and modular relations for eta functions one obtains $$ p(v)=\frac{\eta^3 (4x) \eta^3 (6x) \eta^6 (x)}{\eta (12x) \eta^2 \left(3x\right) \eta^9 (2x)},\quad x=(\tau+1)/6 $$ $$ \lambda=\frac{\eta\big(\tfrac{\tau+1}{3}\big)}{\eta(\tau)}=\zeta_{24}\frac{\eta\big(\tfrac{\tau+1}{3}\big)}{\eta(\tau+1)}=\zeta_{24}\frac{\eta\big(2x\big)}{\eta(6x)},\quad x=(\tau+1)/6. $$ Then $$ \alpha(\tau)=-\tfrac{1}{2}-i\tfrac{(2+2p-p^2) (1-2 p -2p^2) (1+4p+p^2)}{6 \sqrt{3} ~p(p+2) (2 p+1)(1-p^2)}=\tfrac1{4\sqrt{27}}\big(\lambda^3-\sqrt{27}\,\lambda^{-3}\big)^2 $$ becomes an eta function identity, which can be verified algorithmically.

It is known that if $z_1,z_2\in\mathfrak{H}$belong to an imaginary quadratic field then $\eta(z_1)/\eta(z_2)$ is algebraic. Since $\tau$ and also $v$ belong to $\mathbb{Q}[\sqrt{-3}]$ one obtains that $p(v)$ is algebraic. This proves that if $\tau=\frac{1+\sqrt{-3}}2$,$~n\in\mathbb{N}$ then $\alpha$ is algebraic. To show that $\beta$ is algebraic one needs to consider only the ratios $$ \frac{K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)}{K(k_3)},~~\frac{K\Big(\sqrt{1-\tfrac{p^3 (2+p)}{1+2 p}}\Big)}{K(k_3)}\tag3 $$ but since the elliptic integrals $K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)$, $K\Big(\sqrt{1-\tfrac{p^3 (2+p)}{1+2 p}}\Big)$ have complementary moduli algebraicity of one of the ratios would automatically imply algebraicity of the other ratio. More specifically there is the formula (provided by OP) $$ \frac{_2F_1\Big(\tfrac{1}{2},\tfrac{1}{2};1;1-\tfrac{p^3 (2+p)}{1+2 p}\Big)}{_2F_1\Big(\tfrac{1}{2},\tfrac{1}{2};1;\tfrac{p^3 (2+p)}{1+2 p}\Big)}=3(1-v)\sqrt{-1}=\frac{3\sqrt{-1}}{1+\tau}.\tag{4} $$

To prove it note that (5.1-5.14 in V) $$ p(v)+2=2\frac{\eta^2 (3 v) \eta \left({v}/{2}\right) \eta^3 (2 v)}{\eta^2 (v) \eta^3 \left({3 v}/{2}\right) \eta (6 v)} $$ $$ 2p(v)+1=\frac{\eta^2 (3 v) \eta^4 \left({v}/{2}\right) }{\eta^2 (v) \eta^4 \left({3 v}/{2}\right) }. $$ These formulas give $$ \frac{p^3 (2+p)}{1+2 p}=-\frac{16 \eta (6 v)^8}{\eta \left(\frac{3 v}{2}\right)^8}. $$ Further simplification by means of $\eta \left(\frac{v+1}{2}\right)=\frac{\zeta_{48} \eta (v)^3}{\eta \left(\frac{v}{2}\right) \eta (2 v)}$ and modular relation for eta function allows one to write $$ \frac{p^3 (2+p)}{1+2 p}=\left(\frac{\eta ((\tau+1)/6)^2 \eta (2(\tau+1)/3)}{\eta ((\tau+1)/3)^3}\right)^8.\tag{5} $$ From the theory of Jacobi elliptic functions it is known that $$ \omega=i\frac{K'}{K},~K=K(k),~K'=K(k'),~k'=\sqrt{1-k^2},~k'=\large \Big(\tfrac{\eta (2 \omega) \eta^2 (\omega/2)}{\eta^3 (\omega)}\Big)^4. $$ Comparing with (5) one gets $$ \frac{\tau+1}{3}=i\frac{K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)}{K\Big(\sqrt{1-\tfrac{p^3 (2+p)}{1+2 p}}\Big)} $$ equivalent to (4).

It is seen that $\frac{\tau+1}{3}$ is obtained from $\tau_0=n\sqrt{-3}$ by a shift $\tau_0\to\tau_0+1$, duplication, then another analogous shift and triplication. This means that the ratios (3) are algebraic.


Proof of equation (1). First step is to rewrite $\beta$ in terms of hypergeometric functions with the third parameter $1/2$ and $3/2$ according to eq. 2.11(3) from Erdelyi et. al. vol I

$${}_2F_1\Big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};\tfrac12+\tfrac{z}{2}\Big)=\tfrac{\sqrt{\pi } ~\Gamma \left(\frac{5}{6}\right)}{\Gamma \left(\frac{2}{3}\right)^2}{}_2F_1\Big(\tfrac{1}{6},\tfrac{1}{6};\tfrac{1}{2};z^2\Big)-z\tfrac{2\sqrt{\pi } ~\Gamma \left(\frac{5}{6}\right)}{\Gamma \left(\frac{1}{6}\right)^2}{}_2F_1\Big(\tfrac{2}{3},\tfrac{2}{3};\tfrac{3}{2};z^2\Big).\tag7$$

Then the first hypergeometric on the lhs is converted to ${}_2F_1\Big(\tfrac{1}{6},\tfrac{1}{3};\tfrac{1}{2};\frac{z^2}{z^2-1}\Big)$ via Pfaff's transformation for which Zucker and Joyce in their third paper show thatenter image description here whereenter image description here The second hypergeometric function on the lhs of (7) is converted to ${}_2F_1\Big(\tfrac{2}{3},\tfrac{5}{6};\tfrac{3}{2};\frac{z^2}{z^2-1}\Big)$ by Pfaff's transformation, and subsequently to a sum of hypergeometric functions with the third parameter equal to $1$ with the help of eq. 2.11(9) from Erdelyi et. al. vol I $$ \sqrt{\tfrac{z^2}{z^2-1}}{}_2F_1\Big(\tfrac{2}{3},\tfrac{5}{6};\tfrac{3}{2};\tfrac{z^2}{z^2-1}\Big)={\tfrac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{1}{3}\right)}{4z \sqrt{\pi }\Gamma \left(\tfrac{5}{6}\right)} \left(\, _2F_1\left(\tfrac{1}{3},\tfrac{2}{3};1;\tfrac{1}{2}-\tfrac{1}{2}\sqrt{\tfrac{z^2}{z^2-1}}\right)-\, _2F_1\left(\tfrac{1}{3},\tfrac{2}{3};1;\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{z^2}{z^2-1}}\right)\right)}. $$ Equation 5.17 in V enables one to write${}_2F_1\left(\tfrac{1}{3},\tfrac{2}{3};1;...\right)$ in terms of elliptic integrals. The required parametrization is $$ z=-i\frac{((2-p) p+2) (1-2 p (p+1)) (p (p+4)+1)}{3 \sqrt{3} p \left(2 p^4+5 p^3-5 p-2\right)}. $$ Combining all these formulas one eventually arrives at $$ \large\begin{align} \tfrac{2K(k_3)3^{1/4} (2 p+1)^{1/6}}{\pi \sqrt[3]{p(2+p)(1-p^2)}}{}_2F_1\Big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};\tfrac{1}{2}+i\tfrac{(2+2p-p^2) (1-2 p -2p^2) (1+4p+p^2)}{6 \sqrt{3} ~p(p+2) (2 p+1)(1-p^2)}\Big)\\=e^{-\frac{\pi i}{6}}{}_2F_1\Big(\tfrac{1}{2},\tfrac{1}{2};1;\tfrac{p^3 (2+p)}{1+2 p}\Big)+\tfrac{e^{\frac{\pi i}{6}}}{\sqrt{3}}{}_2F_1\Big(\tfrac{1}{2},\tfrac{1}{2};1;1-\tfrac{p^3 (2+p)}{1+2 p}\Big) \end{align}$$