Entry 129

Given \(_2F_1(a,b;c;z)\) where \(a+b=c\). The previous four (Entries 125-128) discuss closed-forms and can be neatly summarized as type \(a+b=c=\color{blue}{\tfrac12}\)  $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\qquad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\qquad \tau_1=n\sqrt{-4}\\ &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\qquad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\qquad \tau_2=n\sqrt{-3}\\ &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\qquad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\qquad \tau_3=n\sqrt{-2}\\ &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad\; \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\qquad\tau_4=n\sqrt{-1}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. However, for the parameters \(c=\frac12, \frac23,\frac34. \frac56\), it seems only the first is easy. For type \(a+b=c=\color{blue}{\tfrac23}\), it's more challenging and will be discussed in the next entry.