How to prove $4\times{_2F_1}(-1/4,3/4;7/4;(2-\sqrt3)/4)-{_2F_1}(3/4,3/4;7/4;(2-\sqrt3)/4)\stackrel?=\frac{3\sqrt[4]{2+\sqrt3}}{\sqrt2}$

Question

I have the following conjecture, which is supported by numerical calculations up to at least $10^5$ decimal digits: $$4\times{_2F_1}\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)-{_2F_1}\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)\,\stackrel?=\,\frac{3\sqrt[4]{2+\sqrt{3}}}{\sqrt{2}},$$ where $_2F_1$ denotes the hypergeometric function.

Can you suggest any ideas how to prove it?

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The conjectural closed form was obtained using WolframAlpha query

ToRadicals[RootApproximant[2.94844576626425580599908814238570067699233]]

Answer 1

This is the identity 15.5.12 from DLMF, with $a=-1/4$, $b=3/4$, $c=7/4$ and the special form $$ F(b,a,a,x)=(1-x)^{-b}. $$ Is this how you got your identity?

Answer 2

Edited on request of moderator team

1. Gauss hypergeometric function satisfies a linear transformation formula $$ (c-a) _2F_1(a-1,b,c,z)+(2a-c-az+bz){}_2F_1(a,b,c,z)+a(z-1){}_2F_1(a+1,b,c,z)=0.$$ Setting $a=b$ makes $z$ disappear from the 2nd prefactor, and one obtains $$(c-a) _2F_1(a-1,a,c,z)+(2a-c){}_2F_1(a,a,c,z)+a(z-1){}_2F_1(a+1,a,c,z)=0.\tag{1}$$

2. Also, sometimes hypergeometric function simplifies to elementary functions. For example, $_2F_1(\alpha,\beta,\alpha,z)=(1-z)^{-\beta}$. Setting in this formula $\alpha=a+1$, $\beta=a$ and combining it with (1) with $c=a+1$, one obtains $$ _2F_1(a-1,a,a+1,z)+(a-1){}_2F_1(a,a,a+1,z)=a(1-z)^{1-a}.\tag{2}$$

3. Setting in (2) $\displaystyle a=\frac34$ and $\displaystyle z=\frac{2-\sqrt{3}}{4}$ gives the quoted relation.