**I. Four quintics?**


The general quintic can be transformed in radicals to at least ***three*** one-parameter forms. For simplicity, assume this free parameter to be some generic "*alpha*". Hence,


$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$
$$x^5-5\alpha x -\alpha = 0\tag2$$
$$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$


which are the *Brioschi*, *Bring-Jerrard*, and *Bring-Euler* quintics, respectively. Naturally, these are $5T5$ with order $5!=120$. Their discriminants are,


\begin{align}
d\_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\
d\_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\
d\_3 &= 5^5\,(1-108\alpha)\,\alpha^2
\end{align}


We can do a minor transformation to get their respective variants,


$$y^3(y^2+5y+40) = j\_1\tag4$$
$$y(y-5)^4 = j\_2\tag5$$
$$y^3(y-5)^2 = j\_3\tag6$$


with discriminants,


\begin{align}
D\_1 &= 5^5\,(j\_1-1728)^2\,{j\_1}^2\\
D\_2 &= 5^5\,(j\_2-256)\,{j\_2}^3\\
D\_3 &= 5^5\,(j\_3-108)\,{j\_3}^3
\end{align}


However, it seems we are missing one quintic with discriminant $D\_4$,


$$D\_4 = 5^5(j\_4-64)^a\,{j\_4}^b\quad$$


which has level $p=6,7,8$ versions discussed [in this MO post](https://mathoverflow.net/q/446263/12905). (The octics in that post, after tedious manipulation, can be reduced to their deg-$7$ resolvents.)




---


**II. Eta quotients and the Monster**


Given [*Dedekind eta function*](https://en.wikipedia.org/wiki/Dedekind_eta_function#Eta_quotients) $\eta(\tau)$, define the four eta quotients which in fact are the first four *McKay-Thompson* series 1A, 2A, 3A, 4A of the Monster,


\begin{align}
\quad j\_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\
\quad j\_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\
\quad j\_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ 
\quad j\_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24} 
\end{align}


where $j\_1$ is just the *j-function*. Let $\tau = \sqrt{-d}$ or $\tau =\frac12+ \sqrt{-d}$ such that the $j\_i$ are **radicals**.




---


**III. Question 1**


Let $j\_i(\tau)$ be the radicals defined as above. Then is it true that for the quintics,


$$y^3(y^2+5y+40) = j\_1\tag4$$
$$y(y-5)^4 = j\_2\tag5$$
$$y^3(y-5)^2 = j\_3\tag6$$


the Galois group is **now** solvable, and the $y$ are solvable in radicals? For example, let,


$$j\_2\left(\tfrac{\sqrt{-232}}4\right)=396^4$$


which appears in Ramanujan's pi formula in the title. So, then a change of variable to $z$,


$$y(y-5)^4 = 396^4$$
$$z(z^4-5) = 396$$
$$z^5-5z-396 = 0$$


which is a solvable Bring-Jerrard quintic. (In fact, it factors).




---


**IV. Question 2**


However, there is still $j\_4$. To recall, for level $p=7$ ([in this MO post](https://mathoverflow.net/q/446263/12905)), the one-parameter formulas are complete for **all four**.


**Q:** So does this imply the general quintic can be reduced to a ***fourth*** one-parameter form (still unknown) and analogous to the three above?




---


**V. Sextic version?**


The "missing" quintic may have a sextic version (also with order $5! = 120$) and is given by,


$$j\_4 =\frac{(x + 1)^5 (x + 5)}x\tag7$$


with expected discriminant,


$$\text{Discrim}\_4 = 5^5\,(j\_4-64)^2\,{j\_4}^4$$


So what we're looking for might be its quintic subextension. But I am uncertain how to generate the *correct* quintic from this sextic. Ironically, the octics were easier. (*Note*: The correct quintic must have order $120$ for general $j\_4$ but be solvable when $j\_4 = j\_4(\tau)$ as defined in Section II.)



On Ramanujan's pi formula $\frac 1\pi=\sum_{k=0}^\infty\frac {(4k)!}{k!^4}\frac {Ak+B}{396^{4k}}$ and the solvable quintic $z^5-5z-396 = 0$?
(*This addresses Question 2*.)


**I.** In general, it asks if we can reduce the general quintic to a one-parameter form with a *specified* discriminant $D\_i$ that is different from the other well-known forms. These $D\_i$ involve the integers $\color{blue}{1728, 256, 108, 64}$, numbers which appear per level in *Ramanujan's theory of elliptic functions to alternative bases*, and his *pi formulas* with radii of convergence $\color{blue}{1/1728, 1/256, 1/108, 1/64}.$ ("*[Rational analogues of Ramanujan's series for 1/Pi](https://www.researchgate.net/publication/232016510_Rational_analogues_of_Ramanujan%27s_series_for_1I)*", Cooper and Chan, p. 23).


**II.** As a further constraint, we require its single parameter $j$, if equated to the appropriate eta quotient in Section II, must render the quintic **solvable** in radicals.


**III.** For the 4th form, it turns out one way is to start with the Brioschi quintic,


$$w^5 - 10 c w^3 + 45c^2 w - c^2 = 0\tag1$$


then using a rational Tschirnhausen transformation $x = P(w)/Q(w)$,


$$\left(w^2-3c\right)x - \left(\frac{j\,w}{j+512}-24c\right)=0\tag2$$


$$c= \frac{j^2}{(j-64)(j+512)^2}\tag3$$


Eliminating $w$ between $(1)$ and $(2)$ using resultants, then factoring, one gets the new form,


$$x^5 - 5j\, x^2 - 15j\, x - (8j + j^2) = 0\tag4$$


which, for general $j$, has order $5! = 120$. However, it has the desired property that if $j = j\_4(\tau)$, then it becomes solvable in radicals. The two quintics have the discriminants,


$$d\_1 = 5^5\,(1-\color{blue}{1728}c)^2\,c^8$$
$$D\_4 = 5^5\,(j−\color{blue}{64})^2(j-1)^2j^4$$


respectively. Note that $(3)$ is a just a cubic in $j$, so the general quintic has been transformed to the one-parameter form $(4)$ in radicals.


**IV.** As an afterword, just like the Bring quintic, the parameter of the Brioschi has infinitely many *rational* $c$ such that it is solvable in radicals, namely,


$$ c = \frac{n}{(n^2 - 10n + 5)^3 + \color{blue}{1728}n}$$


for rational $n$. And since $c$ in $(3)$ is a cubic in $j$, then the new quintic $(4)$ is solvable for infinitely many cubic radicals $j$, in addition to when $j = j\_4(\tau)$ are eta quotients.


