Conjecture $_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]4}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}$

Question

Using a numerical search on my computer I discovered the following inequality: $$\left|\,{_2F_1}\left(\frac14,\frac34;\,\frac23;\,\frac13\right)-\rho\,\right|<10^{-20000},\tag1$$ where $\rho$ is the positive root of the polynomial equation $$12\,\rho^8-12\,\rho^4-8\,\rho^2-1=0,\tag2$$ that can be expressed in radicals: $$\rho=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]{4\vphantom{\large1}}}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}.\tag3$$ Based on this inequality I conjecture that the actual difference is the exact zero, i.e. $$\color{#808080}{_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\rho}.\tag4$$ I looked up in DLMF and MathWorld, but did not find a known special value with exactly these parameters. It also appears that CAS like Maple or Mathematica do not know this identity.

---

Could you please suggest any ideas how to prove the conjecture $(4)$?

---

Update: I can propose even more general conjecture: $$\color{#808080}{27\,(x-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^8+18\,(x-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^2=1}$$

Answer 1

Let's start with this Pfaff transformation for $\,a=\frac 14,b=\frac34,c=\frac23$ : $$\tag{1}_2F_1\left(a,b\;;c\;;z\right)=(1-z)^{-a}\;_2F_1\left(a,c-b\;;c\;;\frac z{z-1}\right)$$

The 'Darboux evaluation' $(42)$ of Vidunas' "Transformations of algebraic Gauss hypergeometric functions" is : $$\tag{2}_2F_1\left(\frac14,-\frac 1{12};\,\frac23;\,\frac {x(x+4)^3}{4(2x-1)^3}\right)=(1-2x)^{-1/4}$$

Solving $\,\displaystyle\frac {x(x+4)^3}{4(2x-1)^3}=\frac z{z-1}\,$ gives : $$\tag{3}z=\frac{x(x+4)^3}{(x^2-10x-2)^2} $$ that we will use as : $$\tag{4}z-1=\frac {4(2x-1)^3}{(x^2-10x-2)^2}$$

while $(1)$ and $(2)$ return : $$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[(z-1)(2x-1)\right]^{-1/4}$$

so that :

$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[\frac{4\,(2x-1)^4}{(x^2-10x-2)^2}\right]^{-1/4}$$ and (up to a minus sign) : $$\tag{5}_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)^2=-\frac{x^2-10x-2}{2\,(2x-1)^2}$$ and indeed the substitution of $(z-1)$ and $_2F_1()^2$ with $(4)$ and $(5)$ in your formula gives : $$27\,(z-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^8+18\,(z-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^2=1$$

Many other formulae of this kind may be deduced using Vidunas' paper.

Answer 2

The January issue of the Notices has an article by Frits Beukers. There is a criterion to decide whether certain hypergeometric functions are algebraic. Checking these numbers in Theorem 2 of that article, we find that $$ {}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right) $$ is an algebraic function of $z$.$^*$ He also mentions H. A. Schwarz's list of 1873; I did not check, but this is probably in it.

$^*$ Of course Raymond found (much more than) that in his solution.