(This extends this post.) Given fixed rationals $a,b,c,$ the problem of determining, $$_2F_1\big(a,b;c;u) =v $$ such that both $u,v$ are algebraic numbers may be solved by appealing to modular functions/forms like the j-function $j(\tau)$ and Dedekind eta function $\eta(\tau)$.
In the tables below, I derived the formulas empirically using data from Zucker and Joyce in "Special values of the hypergeometric series II, III". However, I'm missing five examples and their formulas.
I. Type $a+b=c=\color{blue}{\tfrac12}$.
$$\begin{aligned} &\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\quad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\quad \tau_1=N\sqrt{-4}\\[2mm] &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\quad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\quad \tau_2=N\sqrt{-3}\\[2mm] &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\quad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\quad \tau_3=N\sqrt{-2}\\[2mm] &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\quad\tau_4=N\sqrt{-1}\end{aligned}$$
though the argument $z_4$ of the fourth can be found more simply as $z_4 = (1-2\alpha_4)^2 = \frac{j(\tau)-1728}{j(\tau)}$.
Examples: $$_2F_1\left(\tfrac14,\tfrac14;\tfrac12;\,\tfrac{(3-6\sqrt2)^2}{(1+\sqrt2)^4}\right)=\tfrac{3}{4\sqrt2}(1+\sqrt2)$$ $$_2F_1\left(\tfrac16,\tfrac13;\tfrac12;\,\tfrac{25}{27}\right)=\tfrac{3\sqrt3}{4}$$ $$_2F_1\left(\tfrac18,\tfrac38;\tfrac12;\tfrac{2400}{2401}\right)=\tfrac{2\sqrt7}{3}$$ $$_2F_1\left(\tfrac1{12},\tfrac5{12};\tfrac12;\tfrac{1323}{1331}\right)=\tfrac{3\,\sqrt[4]{11}}{4}$$ using $\tau_1=2\sqrt{-4},\;\tau_2=2\sqrt{-3},\;\tau_3=3\sqrt{-2},\;\tau_4=2\sqrt{-1}$, respectively. And so on for any integer $N>1$.
II. Type $a+b=c=\color{blue}{\tfrac23}$.
$$\begin{aligned} &\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;(1-2\beta_1)^2\big),\quad \color{red}{\beta_1 =\,?} \\[2mm] &\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;(1-2\beta_2)^2\big),\quad \frac{1}{\beta_2}-1=\sqrt{\frac{-1}{1728}\big(2k+\sqrt{4k^2-1728^2}\big)}\\[2mm] &\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;(1-2\beta_3)^2\big),\quad \color{red}{\beta_3 =\,?} \\[2mm] &\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;(1-2\beta_4)^2\big),\quad \frac{1}{\beta_4}-1=\frac{-1}{1728}\big(2k+\sqrt{4k^2-1728^2}\big)\end{aligned}$$
and where $k=j(\tau)-864$.
Examples: Use $\tau = \frac{1+N\sqrt{-3}}2$, like $\tau = \frac{1+3\sqrt{-3}}2$,
$$_2F_1\big(\tfrac16,\tfrac12;\tfrac23;\tfrac{125}{128}\big) =\tfrac43\,2^{1/6}$$ $$_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;\tfrac{64000}{64009}\big) =\tfrac23\,253^{1/6}$$
III. Type $a+b=c=\color{blue}{\tfrac34}$.
$$\begin{aligned} &\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\sqrt{\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}\Big)^{24}}\\[2mm] &\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\[2mm] &\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1=\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}\Big)^{24}\\[2mm] &\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?} \end{aligned}$$ where $\zeta_{48} = e^{2\pi i/48}$.
Examples: Use $\tau = \frac{1+N\sqrt{-1}}2$, like $\tau = \frac{1+5\sqrt{-1}}2$,
$$_2F_1\big(\tfrac14,\tfrac12;\tfrac34;\tfrac{80}{81}\big)=\tfrac95$$ $$_2F_1\big(\tfrac18,\tfrac58;\tfrac34;\tfrac{25920}{25921}\big)=\tfrac35\,161^{1/4}$$
IV. Type $a+b=c=\color{blue}{\tfrac56}$.
$$\begin{aligned} &\,_2F_1\big(\tfrac12,\tfrac13;\tfrac56;(1-2\delta_1)^2\big),\quad\;\delta_1 = \delta_2\\[2mm] &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\sqrt{\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}}\\[2mm] &\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\[2mm] &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1=\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12} \end{aligned}$$ Note: Of course, $_2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big)\,$ so the first form is superfluous.
Examples: Use $\tau = \frac{1+N\sqrt{-3}}2$, like $\tau = \frac{1+5\sqrt{-3}}2$,
$$\begin{aligned} &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;\tfrac45\big)=\;\tfrac35\sqrt5 \\[2mm] &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{80}{81}\big)=\tfrac35\,(9\sqrt5)^{1/3}\end{aligned}$$ with the last one discussed in this post.
Q: How do we find the five missing formulas (?) for the variables in red?
