I. Reduced quintics
The general quintic can be reduced to the one-parameter forms,
$$x^5+5x+\alpha=0\\[5pt] x^5+5\alpha x^2-\alpha=0$$
for some generic alpha. The first is the Bring form and there are at least 10 methods to solve it. The second is the Euler form and seems to have been slightly ignored. Its discriminant is $D=5^5\alpha^4(1-108\alpha^2)$.
The integer $\color{blue}{108=4\times27}$ plays a significant role in Ramanujan's theory of elliptic functions to the alternative base of signature 3 and that is what we will use to solve the Euler quintic.
II. Eta quotients
Let $\eta(\tau)$ be the Dedekind eta function. Surprisingly, the general quintic can be solved by simple eta quotients:
- $a(\tau)=2^{-1/4}\dfrac{\eta(\tau)}{\eta(2\tau)}\,$ via Ramanujan's $g_n$-function here
- $b(\tau)=3^{-1/4}\dfrac{\eta(\tau)}{\eta(3\tau)}\,$ in this post
- $c(\tau)=4^{-1/4}\dfrac{\eta(\tau)}{\eta(4\tau)}\,$ via a solution to $x^8+y^8=1$ here
- $d(\tau)=7^{-1/4}\dfrac{\eta(\tau)}{\eta(7\tau)}\,$ (still unknown)
III. Modular equations
The modular equation between $f(\tau)$ and $f(5\tau)$ for these functions are beautifully similar,
\begin{align} &\frac{u^3}{v^3}-\frac{v^3}{u^3}-2\left(\frac1{u^2v^2}+u^2v^2\right)=0\\[5pt] &\frac{u^3}{v^3}-\frac{v^3}{u^3}-3\left(\frac1{u^2v^2}+u^2v^2\right)-5=0\\[5pt] &\frac{u^3}{v^3}+\frac{v^3}{u^3}-4\left(\frac1{u^2v^2}+u^2v^2\right)-5\left(\frac{u}{v}+\frac{v}{u}\right)=0\\[5pt] &\frac{u^3}{v^3}-\frac{v^3}{u^3}-7\left(\frac1{u^2v^2}+u^2v^2\right)-5\left(\frac{u^2}{v^2}+\frac{v^2}{u^2}\right)+5=0 \end{align}
where respectively,
$$v=a(\tau),\quad u=a(5\tau)\quad\\ v=b(\tau),\quad u=b(5\tau)\quad\\ v=c(\tau),\quad u=c(5\tau)\quad\\ v=d(\tau),\quad u=d(5\tau)\quad$$
IV. Definitions
Recall that $b(\tau)=3^{-1/4}\dfrac{\eta(\tau)}{\eta(3\tau)}$. Using the six roots $u$ of its modular equation, define,
$$X(\tau)\,\small{=\left(\frac{-b(5\tau)-b(\tau/5)}{\sqrt5\, b^3(\tau)}\right) \left(b\Big(\frac{\tau+6}5\Big)+ b\Big(\frac{\tau+24}5\Big)\right) \left(b\Big(\frac{\tau+12}5\Big)+ b\Big(\frac{\tau+18}5\Big)\right)}$$
Then the roots $x_n$ of the Euler quintic form,
$$x^5+5\alpha x^2-\alpha = 0$$
are given by $n=(0,1,2,3,4)$,
$$x_n=\pm\sqrt{\frac1{X(\tau+n)}}$$
$$\tau = \frac{_2F_1\big(\tfrac13,\tfrac23,1,\,1-\beta)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\sqrt{-\frac13}$$
$$\beta = \frac{1-\sqrt{1-108\alpha^2}}2$$
Like the other methods, $x_n$ involves a square root so one has to be careful with signs.
V. Question
In Noriko Yui and Imin Chen's paper, “Singular Values of Thompson Series”, there is a table with eta quotients $\dfrac{\eta(\tau)}{\eta(m\tau)}$ for $m=(2,3,4,5,7,9,13,25),$ integers $m$ such that $\dfrac{24}{m-1}$ is also an integer. After much effort, there are quintic formulas for all but two of these.
Q: Can there be a quintic formula using $m=7$ or $m=13$?
