Define $\color{blue}{q=e^{2\pi i \tau}}$ and Dedekind eta function $\eta(\tau)$. Note: I found these relations empirically, but their consistent forms suggest they can be rigorously proven.

I.$p=2$

We have,

$$\left(\frac{8}{\alpha^8+8}\right)^2+\left(\frac{\beta^8}{\beta^8+32}\right)^2=1\tag1$$ $$\alpha=\frac{\eta(\tau/2)}{\eta(2\tau)},\quad\beta=\frac{\eta(\tau)}{\eta(4\tau)}$$ If $\tau=\sqrt{-n}$, then, $$\frac{_2F_1\Big(\tfrac14,\tfrac34;1;\,1-\big(\tfrac{8}{\alpha^2+8}\big)^2\Big)}{_2F_1\Big(\tfrac14,\tfrac34;1;\,\big(\tfrac{8}{\alpha^8+8}\big)^2\Big)}=\color{red}{\sqrt{2n}}$$

II.$p=3$

We have,

$$\left(\frac{3}{\gamma^3+3}\right)^3+\left(\frac{\delta^3}{\delta^3+9}\right)^3=1\tag2$$ $$\gamma=\frac{\eta(\tau/3)}{\eta(3\tau)},\quad\delta=\frac{\eta(\tau)}{\eta(9\tau)}$$ If $\tau=\sqrt{-n}$, then, $$\frac{_2F_1\Big(\tfrac13,\tfrac23;1;\,1-\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\,\big(\tfrac{3}{\gamma^3+3}\big)^3\Big)}=\color{red}{\sqrt{3n}}$$

Note: The identity,
$$\left(\frac{x^3y + y}{x y^3 + x}\right)^3+\left(\frac{x^3 - y^3}{x y^3 + x}\right)^3=1,\quad \text{if}\; x^3+y^3=1$$
should guarantee *infinitely* many eta parametrizations to $(2)$. Note also that,
$$\gamma^3+3=4C^2(q)+\frac1{C(q)}\\
\delta^3+3=4C^2(q^3)+\frac1{C(q^3)}$$
with ** cubic continued fraction**,
$$C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}=\cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}$$

$\color{blue}{Update:}$
I just realized that the cubic *eta* identity $(2)$ is equivalent to the Borweins' cubic *theta* identity,
$$\frac{c^3(q)}{a^3(q)}+\frac{b^3(q)}{a^3(q)} = 1$$
where,
$$a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$
$$b(q) = \tfrac{1}{2}\big(3a(q^3)-a(q)\big)$$
$$c(q) = \tfrac{1}{2}\big(a(q^{1/3})-a(q)\big)$$
from Ramanujan's Notebooks Vol. V, p.93.

III.$p=4,8$

$$\left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4+\left(\frac{\vartheta_4(0,q)}{\vartheta_3(0,q)}\right)^4 = 1\tag{3a}$$ with Jacobi theta function $\vartheta_n(0,q)$. If $q=e^{2\pi i \tau}$, then equivalently,

$$\left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8+\left(\frac{\eta^2(\tau)\,\eta(4\tau)}{\eta^3(2\tau)}\right)^8 = 1\tag{3b}$$

hence the addends of $(3a)$ and $(3b)$ are equal. Expressed as, $$U^8(\tau)+V^8(\tau) =1$$

If $\tau=\sqrt{-n}$, then,

$$\frac{_2F_1\big(\tfrac12,\tfrac12;1;\,1-U^8(\tau)\big)}{_2F_1\big(\tfrac12,\tfrac12;1;\,U^8(\tau)\big)}=\color{red}{\sqrt{4n}}$$

This has a beautiful ** octic continued fraction** studied by Ramanujan,
$$U(\tau)=\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}= \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$
and can solve the general quintic.

IV. Question

**Q:** Are there analogous eta quotients to parameterize the Fermat quintic $x^5+y^5=1$? And can we also use the ** Rogers-Ramanujan cfrac** to do so?

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