TITLE: Prove $_4F_3(1/8,3/8,5/8,7/8;1/4,1/2,3/4;1/2)=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2}}}{2\,\sqrt2}$
QUESTION [14 upvotes]: How do I prove
$$_4F_3\left(\frac18,\frac38,\frac58,\frac78;\ \frac14,\frac12,\frac34;\ \frac12\right)\stackrel?=\frac{\sqrt{2-\sqrt2+\sqrt{2-\sqrt2\phantom{|}}}+\sqrt{2+\sqrt2+\sqrt{2+\sqrt2\phantom{|}}}}{2\,\sqrt2}$$

I encountered this conjecture when working with Fourier transforms of Kelvin functions.

REPLY [3 votes]: Some related values, based on Kirill's answer, given by radicals:
$$\begin{align}
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac59\right) &= \frac{1}{4}\sqrt{9+\sqrt{30}+2\sqrt{15+3\sqrt{30}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac34\right) &= \frac{1}{2}\sqrt{4+\sqrt{3}+\sqrt{6+4\sqrt{3}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,\frac89\right) &= \frac{1}{2}\sqrt{9+\sqrt{6}+\sqrt{12+6\sqrt{6}}}\\
{_4F_3}\left(\begin{array}c\tfrac18,\tfrac38,\tfrac58,\tfrac78\\\tfrac14,\tfrac12,\tfrac34\end{array}\middle|\,9\right) &= \frac{1}{4}\sqrt{\frac{1}{2}-i\sqrt{2}+\sqrt{-6-6i\sqrt{2}}}
\end{align}
$$
It seems that in general for any $z$ rational values the solution is in the form of $\sqrt{x}$ where $x$ is a root of an $8$th-degree polynomial with integer coefficients.