# Finding a sextic analogue to the solvable octic $\frac{(x + 1)^6(x^2 + x + 7)}x = -k^3$ where $e^{(\pi/3)\sqrt{d}}\approx k^3+41.999999999999\dots$

I. Degree 8

Assume the $$j_i$$ to be free parameters. The following octics in $$x$$ belong to $$8T43,$$ have group $$\text{PGL}(2,7)$$, and order $$2\times168 = 336.$$

\begin{align} {j_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\ {j_2}^2 &=\frac{j_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\ {j_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\ {j_4}^2 &=\frac{7 j_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x \end{align}

Expanded out, some of the coefficients are quadratic in $$j_i$$. They also have nice discriminants $$D_i$$. (Note: $$D_2$$ has another square factor I've suppressed for brevity. I have a feeling the second octic may have a simpler version.)

\begin{align} D_1 &= -7^7(j_1-1728)^4\,{j_1}^4\\ D_2 &= -7^7(j_2-256)^4\,{j_2}^6 \\ D_3 &= -7^7(j_3-108)^3\,{j_3}^5\\ D_4 &= -7^7(j_4-64)^4\,{j_4}^{12} \end{align}

For general $$j_i$$, these octics are not solvable in radicals. However, if we substitute the following integers,

\begin{align} j_1 &= -640320^3\quad \\ j_2 &= -63^2,\, 396^4\quad \\ j_3 &= -300^3\quad \\ j_4 &= -2^9,\, 2^{12}\quad \end{align}

then they become solvable. The first two $$j_i$$ are famous being in the Chudnovsky and Ramanujan pi formulas, and the last two can be used in Ramanujan-Sato pi formulas as well.

II. Eta quotients

There are infinitely many such radical $$j_i$$ given by,

\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 $$\eta(\tau)$$ is the Dedekind eta function and $$j_1$$ is just the j-function. (For examples of $$\tau$$, click on the link.)

Thus, the octics in $$x$$ are also formulas for the $$j_i$$ and where $$x$$ involve eta quotients (though I didn't include them to prevent clutter).

III. Degree 6

So far, I've only found two with group $$\text{PGL}(2,5)$$ hence $$6T14$$ and order $$2\times60 = 120$$. (Edit: For instant comparison, we add the other two $$6T14$$ from Joachim König's answer, in red.)

\begin{align} j_1 &=\frac{(x^2 + 10x + 5)^3}x\\ \color{red}{j_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\ \color{red}{{j_3}^2} &= \frac{j_3\,(2 y^6 + 29y^5 + 85y^4 + 50y^3) + 5^4 y^6}{(2y - 1)} \\ j_4 &=\frac{(x + 1)^5 (x + 5)}x \end{align}

with discriminants,

\begin{align} D_1 &= 5^5\,(j_1-1728)^2\,{j_1}^4\\ D_2 &= 5^5\,(j_2-256)^3\,{j_2}^3 \\ D_3 &= 5^5(j_3-108)^3\,{j_3}^{11}\\ D_4 &= 5^5\,(j_4-64)^2\,{j_4}^4 \end{align}

It is preferred the constant factor is only $$5^5$$. Note: $$D_3$$ actually has the extra square factor $$(32 j_3+84375)^2$$, but I truncated it for aesthetics.

IV. Question:

So are there analogous sextic $$6T14$$ formulas in $$x$$ for $$j_2, j_3$$, where the $$j_i$$ are linear or quadratic, and discriminants similar to the others? (Note: Has been answered in the affirmative.)

• For another example, $j_3 = -(2\sqrt3\,r)^6$ such that, $$e^{\pi\sqrt{929/3}} \approx (2\sqrt3\,r)^6 +41.99999999999999999999923\dots$$ and $r$ is a root of the simple cubic $r^3 - 2898r^2 + 2346r - 583 = 0.$ Commented May 6, 2023 at 11:11
• 1) There is a $d$ in the title that is never defined or referred to in the body of the question. 2) $j_1$ and $j_3$ are not polynomials, so it's not clear how they are octics. 3) The formula for $j_2^2$ expresses it in terms of $j_2$; similarly for $j_4^2$; what is going on there? 4) "for some integers $a,b,c,d$" but there is no other mention of those integers. This whole post is very confusing. Commented May 7, 2023 at 0:44
• It may help to add a more combinatorial viewpoint: these four all define three-point ramified covers (i. e., essentially Belyi maps) of the line parametrized by $j_i$ with inertia groups of orders $(2,3,p)$, $(2,4,p)$, $(2,6,p)$ and $(2,p,p)$ in the four respective cases. That makes them essentially unique (in the respective groups $PGL_2(p)$), and changing from $p=7$ to $p=5$ gives the "analoga". Not all of them are genus zero in the action on $p+1$ points though, some are genus $1$ (as easily read off from the cycle types of the permutations involved), and then one needs degree 2 in $j_i$. Commented May 7, 2023 at 8:57
• @JoachimKönig Thanks for the novel analysis. Do you suppose u can find an alternative octic formula for $j_2$? I find it overly complex compared to the others (and its discriminant has an extra polynomial factor I suppressed). I still have to re-edit the post anyway. Commented May 7, 2023 at 9:09
• @GerryMyerson I've edited the post to address the points you mentioned. And I've included an addendum to explain the role of $d$, among other things. Commented May 7, 2023 at 15:23

Didn't think about eta quotients, but if the suggested analogies of ramification types are anything to go by, then $$j_2 = \frac{(x+1)^4(x^2+6x+25)}{x}$$ (which matches the prescribed discriminant exactly), and $$j_3^2 = \frac{2(x+1)^3(x^3+32x^2+231x+400) j_3 + 5^4(x+1)^6}{64x}$$ (or possibly some reparametrization of it). The last equation has additional divisors a $$2$$-power as well as a square of a linear factor (however not corresponding to a branch point of the function field extension); on the other hand, the quintic subextension of the Galois closure could be parameterized by $$j_3 =y^3(y-5)^2$$, which does give exactly matching discriminant. Not sure if there's any direct reason why reparametrization should allow for extra factors to vanish, just like for general number fields it's not always possible to find a polynomial whose discriminant matches the field discriminant.

• I just noticed there seems to be something wrong with your formula for $j_3$. Magma says the order is $36$ (not $120$). And it factors as two cubics, $$(j_3 x - 16 - 48 x - 48 x^2 - 16 x^3) (j_3 + 4 + 12 x + 12 x^2 + 4 x^3) = 0$$ So it is a not sextic with group $\text{PGL}(2,5)$, unlike the other three. Commented May 9, 2023 at 15:59
• I'm sorry, it seems I picked the wrong solution somewhere in a Groebner basis and then didn't check again. Fixed now with slight edits. Commented May 9, 2023 at 16:44
• Thanks for the correction. Magma says your new formula for $j_3$ has group $\text{PGL}(2,5)$ for generic $j_3$, but is solvable for $j_3 = j_3(\tau) = -300^3$ and others. And it has discriminant, $$\text{Discrim} = 2^{30}\,5^5(j_3 - 108)^3{j_3}^{11}(32j_3 - 27\times 5^5)^2$$ It has an added factor, but at least now we know why $5^5$ didn't appear with the old formula. Commented May 9, 2023 at 17:00

(Note: This addendum to address some comments has been moved here to prevent clutter in the main post.)

These $$j_i = j_i(\tau)$$ have an interesting property best known for $$j_1(\tau)$$. Let $$|z|$$ be the absolute value of $$z$$, then,

$$e^{(\pi/1)\sqrt{d}} \approx \left|j_1\left(\tfrac{1+\sqrt{-d}}2\right)\right|+744$$ $$e^{(\pi/2)\sqrt{d}} \approx \left|j_2\left(\tfrac{2+\sqrt{-d}}4\right)\right|+104$$ $$e^{(\pi/3)\sqrt{d}} \approx \left|j_3\left(\tfrac{3+\sqrt{-d}}6\right)\right|+42$$ $$e^{(\pi/4)\sqrt{d}} \approx \left|j_4\left(\tfrac{4+\sqrt{-d}}8\right)\right|+24$$

as well as,

$$e^{(\pi/1)\sqrt{d}} \approx j_1\left(\tfrac{\sqrt{-d}}2\right)-744$$ $$e^{(\pi/2)\sqrt{d}} \approx j_2\left(\tfrac{\sqrt{-d}}4\right)-104$$ $$e^{(\pi/3)\sqrt{d}} \approx j_3\left(\tfrac{\sqrt{-d}}6\right)-42$$ $$e^{(\pi/4)\sqrt{d}} \approx j_4\left(\tfrac{\sqrt{-d}}8\right)-24$$

When $$d$$ are positive integers, then $$j_i$$ are radicals. And the larger the $$d$$, the better the approximation.

For example, the largest $$d$$ with class number $$h(-d) = 1$$ is $$d=163$$. However, for class number $$h(-d) = 2$$, its largest $$d=2m$$ is $$d = 232 =4\times58,$$ while its largest $$d=3m$$ is $$d = 267 =3\times89.$$ So,

$$j_2\left(\tfrac{\sqrt{-232}}4\right)=396^4$$ $$j_3\left(\tfrac{3+\sqrt{-267}}6\right)=-300^3$$

and,

$$e^{(\pi/2)\sqrt{232}} = e^{\pi\sqrt{58}} \approx 396^4-104$$ $$e^{(\pi/3)\sqrt{267}} = e^{\pi\sqrt{89/3}} \approx 300^3+42$$

Furthermore, and this is the main goal of the post,

$$\frac{(x+1)^4(x^2+6x+25)}{x} = 396^4$$ $$\frac{(x + 1)^6(x^2 + x + 7)}x = -300^3$$

is a solvable sextic and octic, respectively, as well as for infinitely many other radical $$j_2, j_3$$.