This solves [this post](https://math.stackexchange.com/questions/2090523/on-x3y3-z3-the-dixonian-elliptic-functions-and-the-borwein-cubic-theta-f) and is also related to this [MO post](https://mathoverflow.net/questions/259013/can-we-use-the-rogers-ramanujan-cfrac-to-parameterize-the-fermat-quintic-x5y) by involving $\tfrac12,\tfrac13,\tfrac14,\tfrac16$.



> 
> $p=2$
> 
> 
> 


The **singly**-periodic trigonometric functions and the **doubly**-periodic *Jacobi elliptic functions* parameterize the circle so, 


$$\sin^2 z+\cos^2 z=1\tag1$$
$$\mathrm{sn}^2(u)+\mathrm{cn}^2(u)=1\tag2$$


If we use these as arguments for hypergeometric ratios, then for $0<z<1$,


$$\frac{\_2F\_1\Big(\tfrac12,\tfrac12;\tfrac32; \cos^2 z \Big)}{ \_2F\_1\Big(\tfrac12,\tfrac12;\tfrac32; \sin^2 z \Big)}=\Big(\frac{\pi\_\color{blue}2}{2z}-1\Big) \frac{\sin z}{\cos z}\tag3$$


and [elliptic integral of the first kind](http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html) $u = F(\phi,k)$ for $\phi>0$ and *any* $k$,


$$\frac{\_2F\_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{cn}^2(u) \Big)}{ \_2F\_1\Big(\tfrac12,\tfrac12;\tfrac32; \mathrm{sn}^2(u) \Big)}=\Big(\frac{\pi\_\color{blue}2}{2\phi}-1\Big) \frac{\mathrm{sn}(u) }{\mathrm{cn}(u) }\tag4$$


where $\pi\_2 = B\big(\tfrac12,\tfrac12\big)=\pi$. It turns out this can be generalized.



> 
> $p=3$
> 
> 
> 


The **doubly**-periodic Dixonian elliptic functions $\mathrm{sm}(z), \mathrm{cm}(z)$ parameterize the *Fermat cubic* so


$$\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1\tag5$$


[In terms](https://math.stackexchange.com/questions/29935/f3-g3-1-for-two-meromorphic-functions) of the Weierstrass elliptic functions,


$$\mathrm{sm}(z)=\frac{-6\wp\left(z;0,\frac1{27}\right)}{3\wp^\prime\left(z;0,\frac1{27}\right)-1},\quad \mathrm{cm}(z)=\frac{3\wp^\prime\left(z;0,\frac1{27}\right)+1}{3\wp^\prime \left(z;0,\frac1{27}\right)-1}$$


After some experimentation with *Mathematica*, we find, for $0<z<1$, the analogous ratio,


$$\frac{\_2F\_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{cm}^3(z) \Big)}{\_2F\_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{sm}^3(z) \Big)}=\Big(\frac{\pi\_\color{blue}3}{3z}-1\Big) \frac{\mathrm{sm(z)}}{\mathrm{cm(z)}}\tag6$$


where $\pi\_3 = B\big(\tfrac13,\tfrac13\big)=\frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$.



> 
> **Question** for $p=4,6$
> 
> 
> 


What suitable functions $\alpha(z),\beta(z)$ with,
$$\alpha^k(z)+\beta^k(z)=1$$ 
will solve the hypergeometric ratios below in a similar manner?:


$$\frac{\_2F\_1\Big(\tfrac14,\tfrac34;\tfrac54;\beta^k(z)\Big)} {\_2F\_1\Big(\tfrac14,\tfrac34;\tfrac54;\alpha^k(z)\Big)}=\Big(\frac{\pi\_{\color{blue}4}}{4z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag7$$
***or***,


$$\frac{\_2F\_1\Big(\tfrac16,\tfrac56;\tfrac76;\beta^k(z)\Big)} {\_2F\_1\Big(\tfrac16,\tfrac56;\tfrac76;\alpha^k(z)\Big)}=\Big(\frac{\pi\_{\color{blue}6}}{6z}-1\Big) \frac{\alpha(z)}{\beta(z)}\tag8$$



On $x^k+y^k=1$ and the Dixonian elliptic functions
Define the generalized trigonometic functions (discussion of these functions is given in this [answer](https://mathoverflow.net/a/227030/82588))
$$
z=\int\_0^{\sin\_{pr}z}\frac{dt}{\sqrt[p]{1-t^r}},\qquad \cos\_{pr}z=\sqrt[r]{1-(\sin\_{pr}z)^r},\qquad \frac{1}{p}=1-\frac{1}{r}.
$$
It was shown by Edmunds,Gurka, and Lang (Properties of generalized trigonometric functions, J. Approx. Theory 164 (2012), no. 1, 47-56.) that when $r=4$
$$
\sin\_{\frac{4}{3},4}z=\sqrt{\frac{1-\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}{1+\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}}\tag{\*}
$$
Using formulas for the incomplete beta function <http://dlmf.nist.gov/8.17> one has
$$
B\Big(z;\tfrac14,\tfrac14\Big)=4\int\_0^{\sqrt[4]{z}}\frac{dt}{(1-t^4)^{3/4}}=4z^{1/4}{}\_2F\_1(1/4,3/4,5/4;z),
$$
$$
B\Big(x;\tfrac14,\tfrac14\Big)=\pi\_4-B\Big(1-x;\tfrac14,\tfrac14\Big).
$$
From these formulas we immediately see that
$$
4z=B\Big(\sin^4\_{\frac{4}{3}}z;\tfrac14,\tfrac14\Big)=4\sin\_{\frac{4}{3},4}z\cdot{}\_2F\_1\Big(1/4,3/4,5/4;\sin^4\_{\frac{4}{3},4}z\Big),
$$
$$
\pi\_4-4z=B\Big(\cos^4\_{\frac{4}{3},4}z;\tfrac14,\tfrac14\Big)=4\cos\_{\frac{4}{3},4}z\cdot{}\_2F\_1\Big(1/4,3/4,5/4;\cos^4\_{\frac{4}{3},4}z\Big).
$$
This gives the required parametrization
$$
\frac{\_2F\_1\Big(\tfrac14,\tfrac34;\tfrac54;\cos^4\_{\frac{4}{3},4}z\Big)} {\_2F\_1\Big(\tfrac14,\tfrac34;\tfrac54;\sin^4\_{\frac{4}{3},4}z\Big)}=\Big(\frac{\pi\_{\color{blue}4}}{4z}-1\Big) \frac{\sin\_{\frac{4}{3},4}z}{\cos\_{\frac{4}{3},4}z}\tag{7a},
$$
with $~\sin^4\_{\frac{4}{3},4}z+\cos^4\_{\frac{4}{3},4}z=1$ and (\*), or equivalently in terms of Weierstrass elliptic functions, 
$$\sin^2\_{\frac43,4}z=\frac{4\wp (z;1,0)}{4\wp^2(z;1,0)+1},\quad\cos^2\_{\frac43,4}z=\frac{4\wp^2 (z;1,0)-1}{4\wp^2(z;1,0)+1}\tag{i}$$


with (\*) implying,
$$
\sin\_{\frac{4}{3},4}z=\sqrt{\frac{4\wp (z;1,0)}{4\wp^2(z;1,0)+1}}=\sqrt{\frac{1-\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}{1+\text{cn}^4\left(z,\frac{1}{\sqrt{2}}\right)}}\tag{ii}
$$
with the $\mathrm{cn}(u)$ in *Mathematica* syntax as *JacobiCN[z,1/2]* since it uses the parameter $m=k^2$.




---


The case $r=6$ is considered similarly and the equation analogous to (\*) is given by
$$
\sin\_{\frac{6}{5},6}z=\left(\frac{1}{2}-\frac{1}{2}\sqrt{1-\left(1+\sqrt{3}\frac{1+\text{cn}\left(3^{1/4} 2^{4/3} z,k\right)}{1-\text{cn}\left(3^{1/4} 2^{4/3} z,k\right)}\right)^{-3}}\right)^{1/6},\tag{\*\*}
$$
where the modulus of elliptic function is $k=\sin\left(\frac{\pi }{12}\right)$. Also equivalent form of (\*\*) is $\Big(1-2\sin^6\_{\frac56,6}z\Big)^2=1-\frac{4}{\wp^3(z;0,16)}$. Then


$$
\frac{\_2F\_1\Big(\tfrac16,\tfrac56;\tfrac76;1-\sin^6\_{\frac{6}{5},6}z\Big)} {\_2F\_1\Big(\tfrac16,\tfrac56;\tfrac76;\sin^6\_{\frac{6}{5},6}z\Big)}=\Big(\frac{\pi\_{\color{blue}6}}{6z}-1\Big) \frac{\sin\_{\frac{6}{5},6}z}{\cos\_{\frac{6}{5},6}z}\tag{7b}.
$$


