Prove $_2F_1\!\left(\frac76,\frac12;\,\frac13;\,-\phi^2\right)=0$

Please help me to prove the identity $$_2F_1\!\left(\frac76,\frac12;\,\frac13;\,-\phi^2\right)=0,$$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio.

Solution 1:

As the hypergeometric function $$_2F_1\left(\frac16,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)=\frac{\sqrt{1-r^2}}{2r+1}$$ is algebraic, the hypergeometric function with first parameter shifted by $1$ is algebraic as well. To compute its explicit expression, it suffices to note that $${}_2F_1\left(a+1,b;c;z\right)=\frac{z^{1-a}}{a}\frac{d}{dz}\Bigl(z^a{}_2F_1\left(a,b;c;z\right)\Bigr).$$ One then finds \begin{align}_2F_1\left(\frac76,\frac12;\frac13;-\frac{r(r+2)^3}{(r+1)(1-r)^3}\right)&=\frac{\sqrt{1-r^2}}{\left(2r+1\right)^4}\left(5r^3-6r+1\right)=\\ &=-\frac{\left(\sqrt{5}\,r-\phi^{-2}\right)\left(\sqrt{5}\,r+\phi^{2}\right)\left(1+r\right)^{\frac12}\left(1-r\right)^{\frac32}}{\left(2r+1\right)^4}, \end{align} where the vicinity of $r=0$ corresponds to principal branches of the square roots on the right.

Now the result is obtained by setting $r=5^{-\frac12}\phi^{-2}$ in the last formula.