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(Note: This is the case $a=\frac14$ of ${_2F_1\left(a ,a ;a +\tfrac12;-u\right)}=2^{a}\frac{\Gamma\big(a+\tfrac12\big)}{\sqrt\pi\,\Gamma(a)}\int_0^\infty\frac{dx}{(1+2u+\cosh x)^a}.\,$ There is also $a=\frac13$ and $a=\frac16$.)

In a post, Reshetnikov considered some integrals and their equivalent forms, $$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{3}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}3\big) = \frac{2}{3^{3/4}}\tag1$$ $$\frac{1}{2\sqrt2\,K(k_1)}\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+\color{blue}{80}x^4}}=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-\color{blue}{80}\big) = \frac35\tag2$$

Just like for the case $a=\tfrac13$, we postulate these are just the first of an infinite family of algebraic numbers $\alpha$ and $\beta$ such that,
$$_2F_1\left(\frac14,\frac14;\frac34;-\alpha\right)=\beta\tag3$$

Conjecture: Let $\tau = \frac{1+p\sqrt{-1}}{2}$. To find $\alpha$, define the eta quotient $\displaystyle\lambda=\frac{\sqrt2\,\eta(2\tau)}{\zeta_{48}\,\eta(\tau)}$ with $\zeta_{48} = e^{2\pi i/48}$. Then, like the case $a=\tfrac13$, $\alpha$ also is a quadratic of similar form, $$16\cdot64\,\alpha(1+\alpha)=\left( \lambda^{12} -64\, \lambda^{-12} \right)^2$$ with analogous root, $$\alpha= \frac1{4\sqrt{64}}\big(\lambda^6-\sqrt{64}\,\lambda^{-6}\big)^2\tag4$$ And if $p=4k\pm1$ is a prime, then $\alpha$ and $\beta^4$ of $(3)$ are algebraic numbers of degree $k$.

The table below summarizes results and show some trends,

$$\begin{array}{|c|c|c|c|c|} \hline p&\tau&\alpha(\tau)&\beta(\tau)&\text{Deg}\\ \hline 3&\frac{1+3\sqrt{-1}}2&4\cdot1^4-1=3& \large\frac2{3^{3/4}} &1\\ 5&\frac{1+5\sqrt{-1}}2&3^4-1=80& \large\frac35 &1\\ 7&\frac{1+7\sqrt{-1}}2&4(2+\sqrt7)^4-1& \large\frac4{7^{7/8}} \frac1{U_7^{1/4}} &2\\ 11&\frac{1+11\sqrt{-1}}2&4u^4 - 1 &\large \frac6{11^{11/12}}\Big(\frac{11^{2/3}-(14-6\sqrt3)^{1/3}-(14+6\sqrt3)^{1/3}}{3}\Big) &3 \\ 13&\frac{1+13\sqrt{-1}}2& v^4-1 &\large \frac7{13^{2/3}}\Big(\frac{13^{-1/3}+(-2+6\sqrt3)^{1/3}-(2+6\sqrt3)^{1/3}}{3}\Big) &3 \\ 17&\frac{1+17\sqrt{-1}}2& x_1^4 - 1 & \large \frac9{17}x_2 &4 \\ 19&\frac{1+19\sqrt{-1}}2& 4y_1^4 - 1 & \large \frac{10}{19}y_2^{1/4} &5 \\ \hline \end{array}$$ and fundamental unit $U_7=8+3\sqrt7$, with $u,v$ as the real root of the cubics, $$u^3 - 23u^2 + 15u - 9=0\\v^3 - 67v^2 - 159v - 99=0$$ and $x_i$ and $y_i$ as roots of quartics and quintics, and so on. $\text{Deg}$ is degree of $\alpha(\tau)$ and $\beta^4(\tau)$.

Questions:

  1. How do we prove the conjecture? (And the analogous one in the other post?)
  2. Is there a third family of this group with infinitely many algebraic numbers $\alpha_3$ and $\beta_3$? (See case $a=\frac16$.)

$\color{green}{Update}:$

It seems there is a third family. The obvious choices are $a=\frac12$ and $a=\frac16$. For the latter, and using a better search method, I found,

$$_2F_1\Big(\frac{1}{6},\frac{1}{6};\frac{2}{3};-\frac{125}3\Big) = \frac{2}{3^{5/6}}$$

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  • $\begingroup$ Whew. One killer question. Just to clarify though, you don't desire proofs of the first two conjectures (in this post at least), nor do you require a proof of a third member in this family of simplified Hypergeometrics? From what I get, you just want another conjectured form (if one exists). $\endgroup$ – Brevan Ellefsen Dec 23 '16 at 16:58
  • $\begingroup$ @BrevanEllefsen: After using a better search, I found a third family, so I asked it as a separate question. $\endgroup$ – Tito Piezas III Dec 26 '16 at 13:50
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Essentially, the answer is given in Zucker and Joyce, "Special values of the hypergeometric functions II". Starting fromenter image description here they getenter image description here and using $K(1/\sqrt{2})=\frac{1}{4\sqrt{\pi}}\Gamma^2(1/4)$ rewrite it in alternative formenter image description here Then they apply the quadratic transformation with $a=1/4$enter image description here and obtain $$ \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\left(\frac{1-2 k \sqrt{1-k^2}}{2 k\sqrt{1-k^2}+1}\right)^2\right)=\sqrt{k\sqrt{1-k^2}+\frac{1}{2}}\frac{K\left(k\right)+K\left(k'\right)}{2 K\left(\frac{1}{2}\right)}. $$ Now applying Pfaff's transformation to the lhs converts it into $$ \, _2F_1\left(\frac{1}{4},\frac{1}{4};\frac{3}{4};-\tfrac{1}{8} \left(\sqrt[4]{k^3/k'}-\sqrt[4]{k'^3/k}\right)^4\right)=\frac{K\left(k\right)}{K\left(1/\sqrt{2}\right)}\frac{\sqrt[4]{kk'}}{2^{3/4}}(1+K(k')/K(k)), $$ which gives the parametrization $$ \alpha=\tfrac{1}{8} \left(\sqrt[4]{k^3/k'}-\sqrt[4]{k'^3/k}\right)^4,~\beta=\tfrac{K\left(k\right) }{K(1/\sqrt{2})}\frac{(1-iv) \sqrt[4]{kk'}}{2^{3/4}},~q=e^{\pi iv},~k=\frac{\vartheta _2(q){}^2}{\vartheta _3(q){}^2}.\tag{*} $$ It is known that when $v/i=n\in \mathbb{N}$ the ratio $K(k_{n^2})K(1/\sqrt{2})$ is algebraic. This settles the question of algebraicity of both $\alpha$ and $\beta$ when $v=in$,$~n\in\mathbb{N}$. From this one might guess that $\tau$ is related to $v$ in (*) via $$ \tau=\frac{1+v}{2} $$ and indeed numerical calculation indicates that $$ \alpha =\large \tfrac18\left(\sqrt[4]{k^3/k'}-\sqrt[4]{k'^3/k}\right)^4=\tfrac1{4\sqrt{64}}\left(\Big(\tfrac{\sqrt{2} \eta (2 \tau)}{\zeta_{48} \eta (\tau)}\Big)^6-\sqrt{64}\Big(\tfrac{\sqrt{2} \eta (2 \tau)}{\zeta_{48} \eta (\tau)}\Big)^{-6}\right)^2,\tag{**} $$ with $k' = \sqrt{1-k^2},\; k = \large \Big(\frac{\sqrt2\, \eta(v/2)\, \eta^2(2v)}{\eta^3(v)}\Big)^4,\; k'=\large \Big(\frac{\eta (2 v) \eta^2 (v/2)}{\eta^3 (v)}\Big)^4,\; \tau=\large \frac{1+v}2$.

Using the trivial identity $\eta \left(\frac{v+1}{2}\right)=\frac{\zeta_{48} \eta (v)^3}{\eta \left(\frac{v}{2}\right) \eta (2 v)}$ (**) can be written in equivalent form (just simple rearrangements and cancellations of terms) $$ 16 \eta^8 (v/2) \eta^{16} (2v)+\eta^{16} (v/2) \eta^8 (2v)=\eta^{24} (v). $$ This is well known identity due to Jacobi.

Similarly the case $a=\frac16$ is considered in "Special values of the hypergeometric functions III".

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  • $\begingroup$ I guess the crucial point is to rigorous show the equality $(**)$ as the LHS contains eta quotients of form $\eta(v)$ while the RHS has $\eta(\tfrac{1+v}2)$. It would be good to know a similar one for $a=\tfrac13$ as such equalities were not discussed in the Zucker and Joyce paper. $\endgroup$ – Tito Piezas III Dec 29 '16 at 11:38
  • $\begingroup$ Would you know the corresponding identity for $a=\frac13$? It would be nice to group them together. $\endgroup$ – Tito Piezas III Dec 29 '16 at 13:28
  • $\begingroup$ I don't know, it seemed to me that Zucker and Joyce had prooved it, but it turns out they didn't. $\endgroup$ – Nemo Dec 29 '16 at 14:21

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