<s>[INST] Closed form of hypergeometric $\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)$ How can we prove $$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\frac{\sqrt{k(z)}}{2}\right) K\left(\frac{1}{2}-\frac{1}{2 \sqrt{k(z)}}\right), |z|\le1$$
Where $k(z)=-2 \sqrt{z^2-z}-2 z+1$ (thanks to @maxim for the correct form)?

Update: A proof is given by verifying that both LHS and RHS of the following are solutions of certain order $4$ ODE, satisfying same initial conditions (Mathematica command DifferentialRootReduce should do it well). The rest are trivial by analytic continuation. Thus
$${_4 F_3} {\left(  \frac 1 4, \frac 1 4, \frac 3 4, \frac 3 4;  \frac 1 2, 1, 1;  -\left( \frac w 2 - \frac 1 {2 w} \right)^{\! 2} \right)} = \frac 4 {\pi^2}  K {\left( \frac 1 2 - \frac w 2 \right)}  K {\left( \frac 1 2 - \frac 1 {2 w} \right)}$$
 [/INST] This is not an answer.
For small values of $z$, we have
$$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=1+\frac{9 z}{128}+\frac{3675 z^2}{131072}+\frac{266805
   z^3}{16777216}+O\left(z^4\right)$$ while, with $ k(z)=-2 \sqrt{z^2-z}-2 z+1$,$$\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\frac{k(z)}{2}\right) K\left(\frac{1}{2}-\frac{1}{2k(z)}\right)=1+\frac{9 z}{32}+\frac{1371 z^2}{8192}+\frac{31605
   z^3}{262144}+O\left(z^{4}\right)$$
To be different, they are !
Edit
After @Maxim comment
$$w_\pm = \sqrt {1 - 2 z \pm 2 \sqrt {z^2 - z}}\implies -\left( \frac {w_\pm} 2 - \frac 1 {2 w_\pm} \right)^2=z$$
$$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=\frac 4 {\pi^2}  K {\left( \frac 1 2 - \frac {w_\pm} 2 \right)}  K {\left( \frac 1 2 - \frac 1 {2 w_\pm} \right)}$$ seems to hold for positive and negative values of $z$.
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