In a previous post, we mentioned $4+2=6$ methods to solve the Bring quintic. The first four uses the same quartic to find the elliptic modulus $k$ and the last two uses the same octic. For balance, we wondered if there were two missing methods that also use an octic. It turns out there are, and this is the first one. (Hermite's solution will be included for comparison.)
I. The $j$-function
Hermite's approach employs the modular lambda function $\lambda(\tau)$ which appears in a j-function formula,
$$j(\tau) = 256\frac{(x^2-x+1)^3}{(x^2-x)^2}$$
The function $\lambda(\tau)$ is a highly symmetric function with remarkable properties, one of which is all the 6 roots of the formula are $8$th powers $x=y^8$,
$$y_n =\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)},\; \frac{\eta^2\big(\tfrac{\tau}2\big)\,\eta(2\tau)}{\eta^3(\tau)},\; \frac{(-1)^8\sqrt2\,\eta(2\tau)}{\;\eta\big(\tau/2\big)},\; \frac1{y_1},\;\,\frac1{y_2},\;\,\frac1{y_3}$$
and they obey,
$$y_1^8\,+\,y_2^8=1\\ \frac1{y_1^8}+\frac1{y_3^8}=1\;\\ \frac1{y_2^8}+y_3^8=1$$
To solve the Bring quintic, Hermite used $y_1$ and $y_2$ and a quartic for the elliptic modulus $k$. We'll use $y_3$ and an octic.
II. Modular equations
Define,
\begin{align} a(\tau) &=\frac{\sqrt2\,\eta\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}{\eta^3(\tau)}\\[6pt] b(\tau) &=\frac{\sqrt2\,\eta(2\tau)}{\;\eta(\tau/2)}\\[6pt] c(\tau) &=\frac{\eta^2\big(\tfrac{\tau}2\big)\,\eta(2\tau)}{\eta^3(\tau)} \end{align}
such that,
$$a^8+c^8 = 1\\ -b^8+\frac1{c^8} = 1\quad$$
The modular relation between $a(\tau)$ and $a(5\tau)$ is,
$$\frac{u^3}{v^3}-\frac{v^3}{u^3}+4\left(\frac1{u^2v^2}-u^2v^2\right)+5\left(\frac{u}{v}-\frac{v}{u}\right)=0\tag1$$
while that between $b(\tau)$ and $b(5\tau)$ differs only by sign,
$$\frac{u^3}{v^3}+\frac{v^3}{u^3}-4\left(\frac1{u^2v^2}+u^2v^2\right)-5\left(\frac{u}{v}+\frac{v}{u}\right)=0\tag2$$
For the 1st, if $v = a(\tau)$, then the six roots $u$ are,
$$-u = -a(5\tau),\;a\big(\tfrac{\tau}{5}\big),\;a\big(\tfrac{\tau+m}{5}\big),\;a\big(\tfrac{\tau+2m}{5}\big),\;a\big(\tfrac{\tau+3m}{5}\big),\;a\big(\tfrac{\tau+4m}{5}\big)$$
For the 2nd, if $v = b(\tau)$, then,
$$\; u = \;b(5\tau),\,\;b\big(\tfrac{\tau}{5}\big),\,\;b\big(\tfrac{\tau+m}{5}\big),\,\;b\big(\tfrac{\tau+2m}{5}\big),\,\;b\big(\tfrac{\tau+3m}{5}\big),\,\;b\big(\tfrac{\tau+4m}{5}\big)$$
both for $\color{blue}{m=16}$. Define the functions,
$$A(\tau) = \left(\frac{-a(5\tau)-a(\tau/5)}{2\sqrt5\, a(\tau)\sqrt{1-a^8(\tau)}}\right) \left(a\big(\tfrac{\tau+m}{5}\big)-a\big(\tfrac{\tau+4m}{5}\big)\right) \left(a\big(\tfrac{\tau+2m}{5}\big)-a\big(\tfrac{\tau+3m}{5}\big)\right)$$
$$B(\tau) = \left(\frac{\color{red}+b(5\tau)-b(\tau/5)}{2\sqrt5\, b(\tau)\sqrt{1\color{red}+b^8(\tau)}}\right) \left(b\big(\tfrac{\tau+m}{5}\big)-b\big(\tfrac{\tau+4m}{5}\big)\right) \left(b\big(\tfrac{\tau+2m}{5}\big)-b\big(\tfrac{\tau+3m}{5}\big)\right)$$
Thus, all we need is to find $\tau$.
III. Bring quintic
The complete elliptic integral of the first kind $K(k)$ and,
$$\tau =\frac{K'(k)}{K(k)}\sqrt{-1} \,=\, \frac{_2F_1\big(\tfrac12,\tfrac12,1,\,1-k^2\big)}{_2F_1\big(\tfrac12,\tfrac12,1,k^2\big)}\sqrt{-1}$$
yield solutions to,
$$x^5-5x-c = 0\\ y^5+5y+c = 0$$
A. For the 1st, then $x_n = A(\tau+16n)$ where $k$ is any appropriate root of the quartic,
$$4k^4 + c^2k^3 + 8k^2 - c^2k + 4=0$$
B. For the 2nd, then $y_n = B(\tau+16n)$ where $k$ is any appropriate root of the octic,
$$256k^8 - 512k^6 + (384 + c^4)k^4 - (128 + c^4)k^2 + 16=0$$
for $n = 0,1,2,3,4.$
V. Question
This is the third time the same octic has appeared (and there is actually a fourth one). But why would one method yield a quartic and a second yield an octic when the procedures are so analogous?
