In this MO post, it was discussed how the Bring quintic $x^5-x+A=0$ can be solved in two ways by $q$-continued fractions, namely the Rogers-Ramanujan $R(q)$ and the Ramanujan-Selberg $S(q)$. It seems there is a third way.
I. Jacobi theta functions
For brevity of notation, let Jacobi theta functions $\vartheta_n(0,q) = \vartheta_n(q)$. These functions obey,
$$\vartheta_2(q)^4+\vartheta_4(q)^4 = \vartheta_3(q)^4$$
and have beautiful $q$-continued fractions,
$$\vartheta_2(q) = \cfrac{2q^{1/4}\quad}{1 - \cfrac{q^2}{1 + q^2 - \cfrac{q^4}{1 + q^4 - \cfrac{q^6}{1 + q^6 -\ddots}}}}\qquad$$
$$\vartheta_3(q) = -1+\cfrac{2\qquad}{1 - \cfrac{q}{1 + q - \cfrac{q^3}{1 + q^3 - \cfrac{q^5}{1 + q^5 -\ddots}}}}$$
$$\vartheta_4(q) = 1 - \cfrac{2q\qquad}{1 + \cfrac{q^3}{1 - q^3 + \cfrac{q^5}{1 - q^5 + \cfrac{q^7}{1 - q^7 +\ddots}}}}$$
It's long known since Hermite that ratios of theta functions like $\frac{\vartheta_2(q)}{\vartheta_3(q)}$ can solve the quintic. But the method below uses only $\vartheta_3(q)$. Is it "special", or can all three be used singly?
II. Solution to Bring quintic
Note: This solution was rescued from a deleted MSE answer by Lion Emil Jann Fiedler (in Aug 2022) which was deleted by the community. I've refined it so it yields all five roots, simplified its denominator, and found that an expression inside the radical is in the OEIS.
Express the Bring quintic in the form,
$$x^5+5x-4c=0$$
Define the following,
\begin{align} k &= \frac{c+\sqrt{1+\sqrt{1+c^4}}}{\sqrt{2+2c^2+2\sqrt{1+c^4}}}\\[5pt] \tau &= \frac{K'(k)}{K(k)}\sqrt{-1}\\[5pt] q &= e^{\pi i\tau}\\[5pt] \zeta &= e^{2\pi i/5} \end{align}
with complete elliptic integral of the first kind $K(k)$ and nome $q$ (the traditional argument of the Jacobi theta functions). Using only $\vartheta_3(z)$ and the Dedekind eta function $\eta(\tau)$, then all five roots $x_n$ of the quintic are,
$$x_n=\pm\frac{\vartheta_3^2(\zeta^n q^{1/5})-5\vartheta_3^2(q^5)}{8\eta^3(\tau)}\sqrt{-4\vartheta_3^2(q)+4\vartheta_3^2(q^5)+\Big(\vartheta_3(\zeta^n q^{1/5})-\vartheta_3(q^5)\Big)^2\,}$$
for $n=0,1,2,3,4$ (with the $\pm$ sign chosen appropriately).
Note 1: The original denominator of Fiedler had the factor $\vartheta_2(q)\,\vartheta_3(q)\,\vartheta_3(q) = 2\eta^3(\tau)$. It is simpler to use the RHS.
Note 2: For $z = e^{2\pi i\tau}$, one expression inside the radical has an alternative form,
$$-\vartheta_3^2(z)+\vartheta_3^2(z^5) = \frac{-4\eta^2(2\tau)\,\eta(5\tau)\,\eta(20\tau)}{\eta(\tau)\,\eta(4\tau)}$$
which is A053694 (the number of self-conjugate $5$-core partitions of $n$).
III. Example
Let $x^5+5x-4=0,\,$ so $c=1$. Then,
\begin{align} k &= \frac{1+\sqrt{1+\sqrt2}}{\sqrt{4+2\sqrt2}}\\[5pt] \tau &= \frac{K'(k)}{K(k)}\sqrt{-1} \approx 0.536767\sqrt{-1}\\[5pt] q &= e^{\pi i\tau} \approx 0.1852028 \end{align}
Substituting these into the modified formula, we get all five roots $x_n$ with $x_0 \approx 0.751926$.
IV. Question
Q: The formula uses only $\vartheta_3(q)$. Are there corresponding formulas for $\vartheta_2(q)$ and $\vartheta_4(q)$?
