TITLE: Evaluation of Rogers-Ramanujan continued fraction $R(e^{-2\pi/5})$
QUESTION [4 upvotes]: Let $A = \{(\sqrt{5} + 1)/2\}^{5}$ and let $\alpha,\beta$ be positive reals such that $\alpha\beta = \pi^{2}/5$. Then it is known that $$\left\{A + R^{5}(e^{-2\alpha})\right\}\left\{A + R^{5}(e^{-2\beta})\right\} = 5\sqrt{5}A$$ where $$R(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^{2}}{1 + \cfrac{q^{3}}{1 + \cdots}}}}$$ is the Rogers-Ramanujan continued fraction. A proof of the above result is available on my blog http://paramanands.blogspot.com/2013/09/values-of-rogers-ramanujan-continued-fraction-part-2.html
Using the above I am trying to evaluate $R(e^{-2\pi/5})$ using the already known value of $R(e^{-2\pi})$ given by $$R(e^{-2\pi}) = \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}$$
My calculation yields the value $$R(e^{-2\pi/5}) = \frac{\sqrt{5} + 1}{2}\left(3\sqrt{85 - 38\sqrt{5}} - \frac{17\sqrt{5} - 37}{2}\right)^{1/5}$$ and I have checked using calculator that both the values of $R(e^{-2\pi})$ and $R(e^{-2\pi/5})$ shown above satisfy the $\alpha,\beta$ relation mentioned above.
My problem is that the value of $R(e^{-2\pi/5})$ does not look like a unit. On the other hand the value $R(e^{-2\pi})$ is a unit and there is a general theorem that all values of $R(q)$ are units for $q = e^{-\pi\sqrt{n}}$ where $n$ is positive rational. Probably I am making some calculation mistakes or may be I am not able to figure out how the expression for $R(e^{-2\pi/5})$ is a unit. Unfortunately I have not been able to find any online reference to check this value of $R(e^{-2\pi/5})$ and I am stuck with these crazy radical manipulations. Please help me out.

REPLY [3 votes]: It turns out that if I move the factor $(\sqrt{5} + 1)/2$ inside the 5th root things look much better and then $$R(e^{-2\pi/5}) = \left(3\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{9 + \sqrt{5}}{2}\right)^{1/5}$$
From this form it is easy to see the unitness. Clearly we have $$\left(3\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{9 + \sqrt{5}}{2}\right)\left(3\sqrt{\frac{5 + \sqrt{5}}{2}} + \frac{9 + \sqrt{5}}{2}\right) = 1$$
It was silly of me not to have thought of merging the factor $(\sqrt{5} + 1)/2$ inside the 5th root.
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