Who solved the Bring quintic using the Rogers-Ramanujan continued fraction $R(q)$ and how to find all five roots?

Question

I. The octahedral group

Given the nome $q=e^{\pi i \tau}$, then the elliptic lambda function $\lambda(\tau)$ shown below with its Ramanujan–Selberg continued fraction,

\begin{align} \big(\lambda(\tau)\big)^{1/8} &= \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2}\\[8pt] &= \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}} \end{align}

was used by Hermite to solve the Bring quintic. This is a bit surprising since this continued fraction is more associated with the octahedral group (see Duke's "Continued Fractions and Modular Functions").

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II. The icosahedral group

Given the square of the nome, $q=e^{2\pi i \tau}$, then the Rogers–Ramanujan continued fraction,

$$R(q) = \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$$

can indeed be used to solve the Bring quintic. But I don't know who found the solution below, since I refined it from a single example anonymously added to Wikipedia in Oct 6, 2021.

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III. Solution

Let,

$$x^5-x+A=0$$

then a root is,

$$x = \frac{2-\big(1-R(q)\big)\big(1+R(q^2)\big)}{-5^{1/4}\sqrt{R(q)R(q^2)}\sqrt[4]{4\cot\big(4\arctan(\beta)\big)-3}}$$

where,

$$\beta = R(q)R^2(q^2)$$

$$q=e^{2\pi i\tau}$$

$$\tau = \frac{K'(k)}{K(k)}\sqrt{-\frac14}$$

$$k =\tan\left(\frac14\arcsin\Big(\frac{16}{25\sqrt5\, A^2}\Big)\right)$$

and $K(k)$ is the complete elliptic integral of the first kind. (The same $k$ appears in Hermite's method as in this post.)

Note: The function $\beta = R(q)R^2(q^2)$ is special enough that it was studied by Ramanujan and Shaun Cooper devoted an entire paper to it.

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IV. Example

Let,

$$x^5-x+1=0$$

then,

\begin{align} k &=\tfrac{-5^{5/4}+\sqrt{25\sqrt{5}+16}}{+5^{5/4}+\sqrt{25\sqrt{5}-16}}\approx 0.072696\\[6pt] \tau &=\,\frac{K'(k)}{K(k)}\sqrt{-\frac14}\ \,\approx\, 1.275286\sqrt{-1} \end{align}

Since $\tau$ can be derived from the quintic's single parameter $A$, a fast way to calculate $R(q)$ is to use its quadratic relationship to the Dedekind eta function $\eta(\tau)$,

$$\frac1{R(q)} - R(q) = \frac{\eta\big(\frac{\tau}5\big)}{\eta(5\tau)}+1$$

Then, using the formula, we find the quintic root $\color{blue}{x\approx-1.1673}$.

Note: The special case $A = 1+i = (-4)^{1/4}$ is tricky since the formula yields a complex number such that its absolute value $|a+bi|=\sqrt{a^2+b^2}$ is equal also to the absolute value of a complex quintic root.

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V. Monster

Recall that $\beta = R(q)R^2(q^2)$. Inspecting the original formula more closely, it turns out one expression is a square,

$$4\cot\big(4\arctan(\beta)\big)-3= \left(\frac{\eta(\tau)\,\eta(2\tau)}{\eta(5\tau)\,\eta(10\tau)}\right)^2$$

which, perhaps not surprisingly, is the McKay-Thompson series of class $10C$ for Monster, or A132041. Then using some equations from Cooper's paper, we can get rid of a $4$th root from the first formula below,

\begin{align} x &= \frac{2-\big(1-R(q)\big)\big(1+R(q^2)\big)}{-5^{1/4}\sqrt{R(q)R(q^2)}} \times \frac1{\sqrt[4]{4\cot\big(4\arctan(\beta)\big)-3}}\\[6pt] x &= \frac{2-\big(1-R(q)\big)\big(1+R(q^2)\big)}{-5^{1/4}\sqrt{R(q)R(q^2)}} \times \frac{\eta^2(10\tau)}{\eta^2(\tau)}\sqrt{\frac1{\beta}-\beta-4} \end{align}

Note: Can this be simplified further? And since $5^{1/4}$ has four complex solutions, then the correct one should be chosen, though one normally need only consider $\pm5^{1/4}$.

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VI. Questions

1. Who found this nice solution?
2. Without factoring, how do we find the other four roots of the quintic using $R(q)$? (Hermite's method gives all five roots by using five $\tau$.)

Answer 1

(Persistence pays off and updates below. The old answer has been moved to a more relevant MSE post about the Brioschi quintic, while this new answer goes straight to the Bring.)

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I. Answer to Question 1.

I found a different method (via the Brioschi) to solve the general quintic using $R(q)$ in this Sept 7, 2015 post. So the 2021 method (via the Bring) may be by:

1. Nikos Bagis from his Sept 30, 2015 paper. (Though I have read it and it is hard to find this there.)
2. Emil Jann Fiedler (aka Emil Jann Brahmeyer) who found in a Aug 12, 2022 post a different solution to the Bring quintic using $\vartheta_3(0,z)$. He is active in Wikipedia and both the 2021 and 2022 solutions involve a square root (and other similarities) which makes me wonder if he did both.
3. Or someone else.

Note: I've corresponded with Emil Jann. The method is his.

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II. Answer to Question 2.

The answer was literally at the tip of my nose the whole time. I forgot that $R(q)$ has the factor $q^{1/5}$ at the tip of the continued fraction, so we should affix the $5$th roots of unity. Let $\zeta = e^{2\pi i/5}$, then the Bring quintic's five roots $x_n$ for $n = 0,1,2,3,4$ are,

$$x_n = \frac{2-\big(1-\zeta^n R(q)\big)\big(1+\zeta^{2n} R(q^2)\big)}{\pm 5^{1/4}\sqrt{\zeta^{3n}R(q)R(q^2)}} \times M_{10}$$

where,

$$M_{10} = \frac1{\sqrt[4]{4\cot\big(4\arctan(\beta)\big)-3}} = \frac{\eta^2(10\tau)}{\eta^2(\tau)}\sqrt{\frac1{\beta}-\beta-4}$$

and,

$$\beta = \zeta^{n} R(q)\,\zeta^{4n} R^2(q^2) = R(q)R^2(q^2)$$

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III. Update: Recall that $\zeta = e^{2\pi i/5}$ and $q = e^{2\pi i \tau}.\,$ If we define the function,

$$U_n = \frac{2-\big(1-\zeta^n R(q)\big)\big(1+\zeta^{2n} R(q^2)\big)}{\pm\sqrt{\zeta^{3n}}}$$

then the method exploits a seemingly unknown (?) property of the Rogers-Ramanujan continued fraction $R(q)$ that, if the sign of $U_n$ is properly chosen, then,

$$U_0^m+U_1^m+U_2^m+U_3^m+U_4^m = 0, \quad \text{where}\; m = 1,2,3$$

a property also shared by the roots of the Bring quintic and which partly "explains" why the method works.

But we can define a slightly different function for the special case when $\tau = \sqrt{-v}\,$ (for $v$ a real number) to get rid of the sign ambiguity,

$$V_n = \frac{2-\big(1-\zeta^n R(q)\big)\big(1+\zeta^{2n} R(q^2)\big)}{\zeta^{4n}}$$

and,

$$V_0^m+V_1^m+V_2^m+V_3^m+V_4^m = 0, \quad \text{where}\; m = 1,2,3$$