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With no details from Fiedler how the latter was derived, I've been trying to reverse-engineer it hoping it will lead to a clue for the last Platonic symmetry, namely tetrahedral symmetry, perhaps using Borwein cubic theta functions which obey $x^3+y^3=1$. (Update: The Huber quintic theta functions in this post which obey $x^5+y^5=1$ can solve the Bring as well.)

which is in principal quintic form and where,

where the last is the formula for the j-function in terms of $a=R(q)$. I thought resolving this would result in a horrible mess and it did end in a $120$-deg equation in $j.$ But factored, it was just a nice quadratic in j with a multiplicity of $60$ (the order of icosahedral symmetry),

with the eta quotient also a McKay-Thompson series, for class $10C$, or A132041.

The roots are,

with the first two by $\pm\sqrt{c^4-256}.$ The discriminant $D$ of the Bring $x^5-5x+c=0$ is,

$$D = 5^5(c^4-256)$$

so it is not surprising that expression appears.

where $(a,b)$ is related to Ramanujan's cubic continued fraction or the Borwein's cubic theta function to solve the Bring quintic? It would finally add the missing tetrahedral symmetry of the Platonic solids.

P.S. I've been perusing the following papers:

1. Solving the quintic by iteration (1989).
2. Icosahedral symmetry and the quintic equation (1992).
3. On Klein's icosahedral solution of the quintic (2014).

The first mentions a tetrahedral resolvent of the quintic, the second about "partitioning it into five objects of tetrahedral symmetry", and the third "inscibes a tetrahedron in an icosahedron". So it is possible. Now if we can only find an explicit function to do it.

With no details from Fiedler on the latter's derivation, I've been trying to reverse-engineer it hoping that, by using different functions such as Borwein cubic theta functions which obey $x^3+y^3=1$, it will lead to the last Platonic symmetry, the tetrahedral. (Update: The Huber quintic theta functions in this post which obey $x^5+y^5=1$ can solve the Bring as well.)

which is in principal quintic form and where,

where the last is the formula for the j-function in terms of $a=R(q)$. I thought resolving this would result in a horrible mess and a $120$-deg equation in $j$ did pop up. But factored, it was just a nice quadratic in j with a multiplicity of $60$ (the order of icosahedral symmetry),

with the eta quotient also a McKay-Thompson series, for class $10C$, or A132041.

The discriminant $D$ of $x^5-5x+c=0$ is $D = 5^5(c^4-256)$. The quartic roots are,

with the first two by $\sqrt{D/5^5}=\pm\sqrt{c^4-256}$ so it seems expected that $D$ appears.

where $(a,b)$ is related to Ramanujan's cubic continued fraction or the Borwein's cubic theta function to solve the Bring quintic? It would finally add the missing tetrahedral symmetry of the Platonic solids.

By two happy "coincidences", if $a=R(q)$ and $b=R(q^2)$, then $A = 0$ since it is the modular equation between them, while $B = \pmb\square$ is a square (as will be seen later). Since $A = 0$, the quintic will be in Bring form. Scale variables $x = B^{1/4}y$ to simplify further,

and we end up with a system of $3$ equations in $3$ unknowns $(a,b,j)$,

with the latter used in Ramanujan's famous $1/\pi$ formula. As was mentioned earlier, if $a=R(q)$ and $b=R(q^2)$, then $B$ in fact is a square,

with the eta quotient as the McKay-Thompson series of class $10C$, or A132041.

Substitute this into the relation between the two McKay-Thompson series $j_{1A}$ and $j_{2A}$, and it has three quartic factors in $m$. We assume $h = c^4$ and $m=k^2$ and it factors further into two quartics in $k$, the relevant one being,

$$4 (k^2 + 1)^2 + c^2 k (k^2 - 1) = 0$$

with the correct root being the semi-familiar,

$$k =\tan\left(\frac14\arcsin\Big(\frac{16}{c^2}\Big)\right)$$

$$\tau = \frac{K'(k)}{K(k)}\sqrt{-\frac14}$$

and $K(k)$ is the complete elliptic integral of the first kind. It is similar to Emil Jann Fiedler's solution but with the advantage we know how it was derived. (Plus an interesting detour through the Monster along the way.)

By two happy "coincidences", if $a=R(q)$ and $b=R(q^2)$, then $A = 0$ since it is the modular equation between them, while $B = \pmb\square$ is a square (to be seen later). Since $A = 0$, the quintic will be in Bring form. Scale variables $x = B^{1/4}y$ to simplify further,

and we get a system of $3$ equations in $3$ unknowns $(a,b,j)$,

with the latter used in Ramanujan's famous $1/\pi$ formula. And as was mentioned earlier, if $a=R(q)$ and $b=R(q^2)$, then $B$ in fact is a square,

with the eta quotient also a McKay-Thompson series, for class $10C$, or A132041.

Substitute this into the relation between the two McKay-Thompson series $j_{1A}$ and $j_{2A}$, and it has three quartic factors in $m$. We assume $h = c^4$ and $m=k^2$ and it factors further into two quartics in $k$, the relevant one being the near-palindromic,

$$k^4+\left(\tfrac{c}2\right)^2 k^3+2k^2-\left(\tfrac{c}2\right)^2 k+1 =0$$

The roots are,

$$k =\sqrt{\frac{\sqrt{2}\,c-\sqrt{c^2+\sqrt{c^4-256}}}{\sqrt{2}\,c+\sqrt{c^2+\sqrt{c^4-256}}}}$$

with the first two by $\pm\sqrt{c^4-256}.$ The discriminant $D$ of the Bring $x^5-5x+c=0$ is,

$$D = 5^5(c^4-256)$$

so it is not surprising that expression appears.

$$\tau = \frac{K'(k)}{K(k)}\sqrt{-\frac14}=\frac{_2F_1\big(\tfrac12,\tfrac12,1,\,1-k^2\big)}{_2F_1\big(\tfrac12,\tfrac12,1,k^2\big)}\sqrt{-\frac14}$$

and $K(k)$ is the complete elliptic integral of the first kind for real $c$. (For complex $c$, one may have to use the other roots of the quartic.) It is similar to Emil Jann Fiedler's solution but with the advantage we know how it was derived. (Plus an interesting detour through the Monster along the way.)

With no details from Fiedler how the latter was derived, I've been trying to reverse-engineer it hoping it will lead to a clue for the last Platonic symmetry, namely tetrahedral symmetry (perhaps by using the Borwein cubic theta functions).

\begin{align} A &= a^2 - b + a^3 b^2 + a b^3\\[4pt] B &= a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4 = \color{red}\square\\[4pt] C &= 1 + a^5 + 10 a^3 b + 10 a b^2 - 10 a^4 b^3 + 10 a^2 b^4 - b^5 + a^5 b^5 \end{align}

By two happy "coincidences", if $a=R(q)$ and $b=R(q^2)$, then $A = 0$ since it is the modular equation between them, while $B = \color{red}\square$ is a square (as will be seen later). Since $A = 0$, the quintic will be in Bring form. Scale variables $x = B^{1/4}y$ to simplify further,

where the last is the formula for the j-function in terms of $a=R(q)$. I thought resolving this would result in a horrible mess and it did end in a $120$-deg equation in j. But factored, it was just a nice quadratic in j with a multiplicity of $60$ (the order of icosahedral symmetry),

with the latter used in Ramanujan's famous $1/\pi$ formula. Furthermore, as was mentioned earlier, if $a=R(q)$ and $b=R(q^2)$, then $B$ in fact is a square,

and $K(k)$ is the complete elliptic integral of the first kind. It is similar to Emil Jann Fiedler's solution with the advantage we know how it was derived. (Plus an interesting detour through the Monster along the way.)

P.S. I've been reading the following papers.

With no details from Fiedler how the latter was derived, I've been trying to reverse-engineer it hoping it will lead to a clue for the last Platonic symmetry, namely tetrahedral symmetry, perhaps using Borwein cubic theta functions which obey $x^3+y^3=1$. (Update: The Huber quintic theta functions in this post which obey $x^5+y^5=1$ can solve the Bring as well.)

\begin{align} A &= a^2 - b + a^3 b^2 + a b^3\\[4pt] B &= a + a^4 b - 3 a^2 b^2 - b^3 - a^3 b^4\\[4pt] C &= 1 + a^5 + 10 a^3 b + 10 a b^2 - 10 a^4 b^3 + 10 a^2 b^4 - b^5 + a^5 b^5 \end{align}

By two happy "coincidences", if $a=R(q)$ and $b=R(q^2)$, then $A = 0$ since it is the modular equation between them, while $B = \pmb\square$ is a square (as will be seen later). Since $A = 0$, the quintic will be in Bring form. Scale variables $x = B^{1/4}y$ to simplify further,

where the last is the formula for the j-function in terms of $a=R(q)$. I thought resolving this would result in a horrible mess and it did end in a $120$-deg equation in $j.$ But factored, it was just a nice quadratic in j with a multiplicity of $60$ (the order of icosahedral symmetry),

with the latter used in Ramanujan's famous $1/\pi$ formula. As was mentioned earlier, if $a=R(q)$ and $b=R(q^2)$, then $B$ in fact is a square,

and $K(k)$ is the complete elliptic integral of the first kind. It is similar to Emil Jann Fiedler's solution but with the advantage we know how it was derived. (Plus an interesting detour through the Monster along the way.)

P.S. I've been perusing the following papers: