**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. (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* $\eta(\tau)$, 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), 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.)

fourthone-parameter form the general quintic can be reduced to in this MO post. But it doesn't seem to have the correct discriminant $D_4$, nor is it solvable in radicals if its parameter is equated to some multiple of $j_4$. $\endgroup$