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.

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    $\begingroup$ I really hope someone is able to provide nice proofs for the various hypergeometric identities you've been conjuring. It'd be a shame if all we can do is just shrug at them. $\endgroup$ Jul 30, 2014 at 0:54

1 Answer 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.

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    $\begingroup$ You should really do a MSE blog post discussing the monodromy/geometric approach to hypergeometrics. I've seen the methods quoted before, but they remain rather esoteric. $\endgroup$ Jul 30, 2014 at 12:50
  • $\begingroup$ @Semiclassical Maybe some day. It is very straightforward to check whether for given parameter values the hypergeometric function is algebraic - I tried to outline the main ideas in the post I am referring to. However (if it is not), specific hypergeometric values are a more subtle and random science. $\endgroup$ Jul 30, 2014 at 17:03
  • $\begingroup$ Thanks! It looks like there should be $a+1$ rather than $a$ in the left hand side of the second formula. $\endgroup$ Jul 30, 2014 at 20:37
  • $\begingroup$ @VladimirReshetnikov oops, thanks! corrected. $\endgroup$ Jul 30, 2014 at 20:52

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