(Courtesy of a comment by Nemo who suggested Huber's paper.)

Part I. $x^5+y^5 = 1$

In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ then Huber's four functions in simplified form are,

$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ is the Rogers-Ramanujan cfrac, $$\frac{a(\tau)}{b(\tau)} = q^{1/5}\frac{H(q)}{G(q)}=R(q)\tag3$$ and $R(q)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(q^5)}{\phi+R(q^5)}\tag4$$

Part II. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. Given, $$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then we have the simple,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag3$$

where,

 $$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ which can be derived by combining $(1),(2)$ and knowing $(3),(4)$.

(Courtesy of a comment by Nemo who suggested Huber's paper.)

Part I. $x^5+y^5 = 1$

In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ and the Rogers-Ramanujan continued fraction, $$R(z)=q^{1/5}\frac{H(q)}{G(q)}=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then Huber's four functions in simplified form are,

$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ as well as $c,d$ are Rogers-Ramanujan cfracs, $$\frac{a(\tau)}{b(\tau)} = R(\tau)\tag3$$ $$\frac{c(\tau)}{d(\tau)} = \phi\, R\big(\tfrac{-1}{5\tau}\big)\tag4$$ so $R(\tau)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(5\tau)}{\phi+R(5\tau)}\tag5$$

Part II. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. This is given by,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag6$$

where,  $$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ and which can be derived by combining $(1),(2)$.

(This is a partial answer. Courtesy of a comment by Nemo.)

Part I. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. Given, $$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then we have the simple,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag1$$

where,

$$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ and golden ratio $\phi$. This was derived by yours truly using Tim Huber's "A Theory of Theta Functions to the Quintic base".

Part II. $x^5+y^5 = 1$

Huber defines four functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ then Huber's four functions in simplified form are,  $$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ which obeys, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag2$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag3$$ It then follows that the ratio of $a,b$ is the Rogers-Ramanujan cfrac, $$\frac{a(\tau)}{b(\tau)} = R(q)$$ and $R(q)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(q^5)}{\phi+R(q^5)}$$ Thus, combining $(2),(3)$ will yield $(1)$.

(Courtesy of a comment by Nemo who suggested Huber's paper.)

Part I. $x^5+y^5 = 1$

In "A Theory of Theta Functions to the Quintic base", Tim Huber defines four theta functions which can be ultimately expressed in terms of the Rogers-Ramanujan identities. Define $q=e^{2\pi i z}$ and, $$P(z):=q^{11/60}H(q)=q^{11/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}$$ $$Q(z):=q^{-1/60}G(q)=q^{-1/60}\prod_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}$$ then Huber's four functions in simplified form are, 

$$\begin{aligned} a(\tau) &=\eta^{2/5}(\tau)\;P(\tau)\\[2mm] b(\tau) &=\eta^{2/5}(\tau)\;Q(\tau)\\[2mm] c(\tau) &= 5^{1/4}\phi^{1/2}\,\eta^{2/5}(5\tau)\;P\big(\tfrac{-1}{5\tau}\big)\\[2mm] d(\tau) &= \frac{5^{1/4}}{\phi^{1/2}}\,\eta^{2/5}(5\tau)\;Q\big(\tfrac{-1}{5\tau}\big)\end{aligned}$$ with golden ratio $\phi$. These obey, $$\Big(\frac{a\,\phi}{b}\Big)^5+\Big(\frac{c}{b}\Big)^5 = 1\tag1$$ $$-\Big(\frac{a}{b\,\phi}\Big)^5+\Big(\frac{d}{b}\Big)^5 = 1\tag2$$ It then follows that the ratio of $a,b$ is the Rogers-Ramanujan cfrac, $$\frac{a(\tau)}{b(\tau)} = q^{1/5}\frac{H(q)}{G(q)}=R(q)\tag3$$ and $R(q)$ in a way can parameterize the Fermat quintic. Some manipulation will also show that, $$\frac{c(\tau)}{d(\tau)} = \phi\,\frac{1-\phi\, R(q^5)}{\phi+R(q^5)}\tag4$$

Part II. $x^5+y^5+z^5 = 1$

It turns out that just like, $$x^3+y^3=1$$ can be solved by the cubic continued fraction $C(q)$ and $C(q^3)$, its quintic analogue $$x^5+y^5+z^5 = 1$$ has a beautiful solution using the Rogers-Ramanujan cfrac $R(q)$ and $R(q^5)$. Given, $$R(q)=\cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

then we have the simple,

$$\alpha^5\phi^5+\beta^5\phi^5+\alpha^5\beta^5 = 1\tag3$$

where,

$$\alpha = R(q),\quad\beta = \frac{1-\phi R(q^5)}{\phi+R(q^5)}$$ which can be derived by combining $(1),(2)$ and knowing $(3),(4)$.