Entry 200

There is a surprising relationship between solvable quintics and Pythagorean triples. Given

$$y^5+Ay^4+By^3+Cy^2+Dy+E=0$$

then there is an extremely broad solvable class with four free parameters $(A,B,C,D)$ and $E$ is only a quadratic with respect to the others. For simplicity, transform the quintic into depressed form,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0$$

If $c\neq0$ and the coefficients form the Pythagorean triple,

$$(c^3 - 25 c^4 - 16 c d^2 - c e + 10 c^2 e + d^2 - e^2)^2 + (2c^2 d - 2c f + 2d e)^2 = \\ (c^3 + 25 c^4 + 16 c d^2 - c e - 10 c^2 e + d^2 + e^2)^2$$ then the quintic is solvable in radicals. Expanding, the above becomes

$$(-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)\times c =W\times c = 0$$

where $W$ is simply the constant term of Watson's sextic resolvent equated to zero. (We just multiplied it by $c$ to get the Pythagorean form.)

Example. Let $(c,d,e)=(1,2,-3)$, then $W$ factors as $(f-28)(f+36)=0$. So another nice feature of these quintics is they come in pairs. Hence

$$x^5 + 10 x^3 + 20 x^2 - 15x +28 = 0\\ x^5 + 10 x^3 + 20 x^2 - 15x - 36 = 0$$

are irreducible but solvable in radicals. And using the formulas above, their $(c,d,e,f)$ yield the triple,

$$120^2+64^2 = 136^2$$

Entry 199

Combining the work of Felix Klein and Ramanujan, given the Ramanujan $G_m$ and $g_m$-functions. 

Conjecture: The following septics have a solvable Galois group

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= 4\left(\frac{4}{G_{m}^{16}}-G_{m}^{8}\right) \\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3x &= 4\left(\frac{4}{g_{m}^{16}}+g_{m}^{8}\right) \end{align}$$

This is a version of Entry 148. For example, $G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}$, $G_{13} = \left(\tfrac{3+\sqrt{13}}2\right)^{1/4}$, and $G_{37} = \left(6+\sqrt{37}\right)^{1/4}$  yields

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -2\left(-25+13\sqrt{5}\right)\\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -30\left(-31+9\sqrt{13}\right) \\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -60\left(-2837+468\sqrt{37}\right) \end{align}$$

while $g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}$ and  $g_{58} = \left(\tfrac{5+\sqrt{29}}2\right)^{1/2}$ yields

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -6\left(-65+27\sqrt5\right)\\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -30\left(-140989 + 26163\sqrt{29}\right) \end{align}$$

which are all solvable in radicals, and so on.

Entry 198

In Entry 190 and Entry 191, two relations between solvable quintics and $G_m$ and $g_m$ were proposed. Going higher,

Conjecture: The following sextics have a solvable Galois group

$$\begin{align}\frac{x^6+10x^3+5}x &= 4\left(\frac{4}{G_{m}^{16}}-G_{m}^{8}\right) \\ \frac{x^6+10x^3+5}x &= 4\left(\frac{4}{g_{m}^{16}}+g_{m}^{8}\right) \end{align}$$

For example, $G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}$ and $g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}$ yields

$$\begin{align}x^6+10x^3+5 &= -2\left(-25+13\sqrt5\right) x \\ x^6+10x^3+5 &= -6\left(-65+27\sqrt5\right) x \end{align}$$

which are solvable in radicals, and so on.

Entry 197

From Entry 195

$$2^{-1/4}G_{71} = x,\quad \text{where}\, x^7 - 2x^6 - x^5 + x^4 + x^3 + x^2 - x - 1=0$$

We will now give the radical solution to this septic using the well-known cubic $r^3+r^2-2r-1=0$ with roots $$r_1,\,r_2,\,r_3 = 2\cos\big(\tfrac{2\pi}7\big),\, 2\cos\big(\tfrac{4\pi}7\big),\, 2\cos\big(\tfrac{6\pi}7\big)$$ Define the function

$$\begin{align}y_n &= P(r_n) =\frac{29323 r^2 - 20538 r - 15494 + (1193 r^2 - 1048 r - 730)\sqrt{7\times71}}2\\ y_{n+3} &= Q(r_n) =\frac{29323 r^2 - 20538 r - 15494 - (1193 r^2 - 1048 r - 730)\sqrt{7\times71}}2\end{align}$$

For example, $y_1 = P(r_1) \approx 219.6454$. Then the real root of the septic is

$$\quad x = \frac{2 + y_1^{1/7}+y_2^{1/7}+y_3^{1/7}+y_4^{1/7}+y_5^{1/7}+y_6^{1/7}}7 = 2.1306068\dots$$

In fact, the $y_n$ are the roots of a sextic with rather large coefficients,

$$y^6+ay^5+by^4+cy^3+dy^2+ey-461^7 = 0$$

and where the constant term is a $7$th power. This sextic is special since it can factor either over the square root extension $\sqrt{7\times71}$ or the cubic extension $2\cos\big(\tfrac{2\pi}7\big)$.

Entry 196

In Entry 194

$$2^{-1/4}G_{47} = x,\quad \text{where}\; x^5 - x^3 - 2x^2 - 2x - 1 = 0$$ Ramanujan asked for the radical solution of this quintic. We give our version which partly uses the golden ratio $\phi$

$$\begin{align}y_1 &= \frac{1}{2\phi}\left(13+\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_2 &= \frac{1}{2\phi}\left(13-\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_3 &= \frac{\phi}{2}\left(13+\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right)\\ y_4 &= \frac{\phi}{2}\left(13-\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right) \end{align}$$

then the real root of quintic is

$$2^{-1/4}G_{47} = x = \frac{y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5}}{\sqrt5} = 1.73469134\dots$$

In fact, the $y_n$ are the four roots of the quartic

$$3125y^4 - 40625y^3 - 521250y^2 + 55250y - 1= 0$$

where the leading term is a $5$th power $3125 = 5^5$.

Entry 195

Continuing Entry 194 for $G_d$ with $d \equiv 7\,\text{mod}\,8$

IV. n = 7

There are only $d =71, 151, 223, 463, 487$ though, to prevent clutter, we include only the first two 

$$\begin{align}G_{71} &= 2^{1/4}x,\quad x^7 - 2x^6 - x^5 + x^4 + x^3 + x^2 - x - 1=0\\ G_{151} &= 2^{1/4}x,\quad x^7 - 3x^6 - x^5 - 3x^4 - x^2 - x - 1=0 \end{align}$$ It's remarkable how small the coefficients are. We will solve these in radicals in another entry.

V. n = 9

There are only $d =199, 367, 823, 1087, 1423$ though again the first two 

$$\begin{align}G_{199} &= 2^{1/4}x,\quad x^3 - (r^2 + 3r + 1)x^2 - x + r = 0,\quad\qquad r^3 + 4r^2 + r + 1 = 0\\ G_{367} &= 2^{1/4}x,\quad x^3 - (r^2 + 5r + 1)x^2 - (2r + 1)x + r =0,\quad r^3 + 7r^2 + 4r + 1 = 0 \end{align}$$

As nonics, these also have small coefficients and are easier solved in radicals as they can be factored over a cubic extension.

Entry 194

Previous entries discussed $G_d$ with $d\equiv 3\,\text{mod}\,8$. For $d\equiv 7\,\text{mod}\,8$ with odd class number $h(-d)=\color{blue}n$, then the formula is slightly different

$$G_d = 2^{1/4}x$$

where $x$ is a root of an algebraic equation of degree $\color{blue}n$ that is solvable in radicals and a unit constant term.

I. n = 1  $$G_7 = 2^{1/4}x,\quad x-1 = 0\quad$$ II. n = 3

$$\begin{align}G_{23} &= 2^{1/4}x,\quad x^3-x-1=0\\ G_{31} &= 2^{1/4}x,\quad x^3-x^2-1=0 \end{align}$$ III. n = 5

$$\begin{align}G_{47} &= 2^{1/4}x,\quad x^5 - x^3 - 2x^2 - 2x - 1 = 0\\ G_{79} &= 2^{1/4}x,\quad x^5 - 3x^4 + 2x^3 - x^2 + x -1=0\\G_{103} &= 2^{1/4}x,\quad x^5 - x^4 - 3x^3 - 3x^2 - 2x - 1 = 0\\ G_{127} &= 2^{1/4}x,\quad x^5 - 3x^4 - x^3 + 2x^2 + x - 1 = 0 \end{align}$$

$G_{23}$ and $G_{31}$ were known to Ramanujan and their $x$ involve the plastic ratio and supergolden ratio, respectively. In Bernt's "The Problems Submitted by Ramanujan to the JIMS" (p.17) Ramanujan asked about solving two quintics in radicals with $d = 47, 79$ so he knew $G_{47}$ and $G_{79}$ though I'm not sure for $G_{103}$ and $G_{127}$.  

Entry 193

Continuing from Entry 192, some more $G_d$,

$$G_{59} = 2^{-1/4}x,\quad x^3 - 2r x^2 +2 (r^2 - r)x - 2 = 0,\quad r^3-2r^2-1=0\quad$$

and the shared pairs

$$\begin{align}G_{19} &= 2^{-1/4}x,\quad x^3 - 2r x^2 + (r^2 - 5r-2)x - 2 = 0,\quad r=0 \\ G_{379} &= 2^{-1/4}x,\quad x^3 - 2r x^2 + (r^2 - 5r-2)x - 2 = 0,\quad r^3-6r^2-5r-2=0\end{align}$$

and

$$\begin{align}G_{107} &= 2^{-1/4}x,\quad x^3 - 2r x^2 + (r^2 -r-2)x - 2 = 0,\quad r^3-r-4=0\\ G_{139} &= 2^{-1/4}x,\quad x^3 - 2r x^2 + (r^2 - r-2)x - 2 = 0,\quad r^3-r^2-2r-4=0\end{align}$$

The first one has additional context since the real root of $r^3-2r^2-1=0$ is the supersilver ratio, a cubic analogue of the silver ratio $1+\sqrt2$.

Entry 192

From Entry 191,

$$G_{11} = 2^{-1/4}x, \quad x^3 - 2x^2 + 2 x - 2=0\qquad$$

so $x$ is the real root of a cubic $x^3-2ax^2+2bx-2=0$ where $a=b=1$. The discriminant $d=11$ has class number $n=h(-d)=1$. For $G_d$, one can observe that if $d\equiv 3\,\text{mod}\,8$, then $(a,b)$ are algebraic numbers at most of degree $n$. Thus if $n=3$, then $(a,b)$ are roots of cubics. 

There are 16 fundamental discriminants $d$ with class number $n=3$ and the largest is $d=907$. The smallest two $d = 23,31$ have different form $d\equiv 7\,\text{mod}\,8$, and will be discussed in another entry while the rest are $d\equiv 3\,\text{mod}\,8$,

$$\begin{align}G_{11} &= 2^{-1/4}x, \quad x^3 - 2rx^2 + 2 x - 2 = 0,\quad r = 1\\ G_{331} &= 2^{-1/4}x, \quad x^3 - 2rx^2 + 2 x - 2 = 0,\quad r^3-7r^2+9r-4=0\end{align}$$

and

$$\begin{align}G_{43} &= 2^{-1/4}x, \quad x^3 - 2x^2 + 2r x - 2 = 0,\quad r = 0\\ G_{83} &= 2^{-1/4}x, \quad x^3 - 2x^2 + 2r x - 2 = 0,\quad r^3-r^2-3r+4=0\end{align}$$

and

$$\begin{align}G_{67} &= 2^{-1/4}x, \quad x^3 - 2r x^2 - 2r x - 2 = 0,\quad r = 1\\ G_{211} &= 2^{-1/4}x, \quad x^3 - 2rx^2 - 2r x - 2 = 0,\quad r^3-3r^2+r-2=0\\ G_{283} &= 2^{-1/4}x, \quad x^3 - 2rx^2 - 2r x - 2 = 0,\quad r^3-4r^2-1=0\end{align}$$

and

$$\begin{align}\quad G_{163} &= 2^{-1/4}x, \quad x^3 - 2r x^2 +4 x - 2 = 0,\quad r = 3\\ G_{907} &= 2^{-1/4}x, \quad x^3 - 2rx^2 +4 x - 2 = 0,\quad r^3-29r^2+85r-66=0\end{align}$$

P.S. These are the simplest cubic "templates" and it seems interesting the largest discriminants for class numbers $1$ and $3$ have a $G_d$ that share the same template. For class number $5$, they get more complicated though. 

Entry 191

We propose another nice relation between quintics and $G_m$ and $g_m$.

Conjecture: The following quintics have a solvable Galois group

$$\begin{align}x^3(x^2+5x+40) &= 4^3\left(\frac{4}{G_{m}^{16}}-G_{m}^{8}\right)^3\\ x^3(x^2+5x+40) &= 4^3\left(\frac{4}{g_{m}^{16}}+g_{m}^{8}\right)^3\end{align}$$

For example, $G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}$ and $g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}$ yields

$$\begin{align}x^3(x^2+5x+40) &= -2^3\left(-25+13\sqrt5\right)^3\\ x^3(x^2+5x+40) &= -6^3\left(-65+27\sqrt5\right)^3\end{align}$$

which indeed are solvable in radicals, and so on. 

Ramanujan tabulated a lot of explicit values for $G_m$ and $g_m$, with odd and even $m$, respectively. Most odd $m$ were $m \equiv 1\,\text{mod}\,4$ like $m=5,13,37$ which have class number $2$. But for $m \equiv 3\,\text{mod}\,4$ like $m=11,19,43,67,163$ which have class number $1$, it seems he found

$$G_m = 2^{-1/4}x_m$$

where $x_m$ is the real root of the following cubics in Entry 160

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$ and in Part V of Ramanujan's Notebooks. The last leads to Ramanujan's constant,

$$\quad e^{\pi\sqrt{163}} \approx x_{163}^{24}-24 \approx 640320^3-743.99999999999925\dots$$

but I'm unsure if he missed the equality

$$4^3\left(\frac{4}{G_{163}^{16}}-G_{163}^{8}\right)^3 = -640320^3$$

Entry 190

From Entry 184, the quintic $y(y-5)^4 =  j_2(\tau)$ has a solvable Galois group. For example

$$\begin{align}y(y-5)^4 &= -(4\sqrt2)^4\\ y(y-5)^4 &=  -(12\sqrt2)^4\\ y(y-5)^4 &= 12^4\\ y(y-5)^4 &= 396^4\qquad\end{align}$$

But these can also be expressed as

$$\begin{align}x^4(x+5) &= (4\sqrt2)i\\ x^4(x+5) &=  (12\sqrt2)i \\ x^4(x-5) &= 12\\ x^4(x-5) &= 396\qquad\end{align}$$

In general, since there is a relationship between $j_2(\alpha)$ and $j_4(\beta)$ where the latter is expressible by Ramanujan's $G_m$-function, then,

Conjecture: The special Bring-Jerrard quintics below are solvable

$$\begin{align}x^5+5x &= \sqrt{2^3\left(\frac1{G_m^{12}}-G_m^{12}\right)}\\ x^5-5x &= \sqrt{2^3\left(\frac1{g_m^{12}}+g_m^{12}\right)}\end{align}$$

For example, $G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}\,$ and $g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}$ yields

$$\begin{align}x^5+5x &= (4\sqrt2)i\\ x^5-5x &= 12\qquad\end{align}$$

respectively, where the latter is quite well-known as an example of a solvable quintic with small integer coefficients. And so on for all Ramanujan $G$ and $g$-functions, with explicit examples in the previous entries.

Entry 189

Ramanujan's $g_m$-function has a consistent form for fundamental discriminants $d=4m$ with class number $2$ and $4$, where $m=2p$ for primes $p\equiv 3\,\text{mod}\,4$. For example, for $p=3,11$ and $p=7,23,71$ then

$$\begin{align}g_{6}&= \sqrt[6]{1+\sqrt2}\\ g_{22}&= \sqrt{1+\sqrt2}\\ g_{14}&= \sqrt{\frac{-1+\sqrt2}4}+  \sqrt{\frac{4+(-1+\sqrt2)}4}\\ g_{46}&= \sqrt{\frac{1+\sqrt2}4}+  \sqrt{\frac{4+(1+\sqrt2)}4}\\ g_{142}&= \sqrt{\frac{(1+\sqrt2)^3}4}+  \sqrt{\frac{4+(1+\sqrt2)^3}4}\end{align}$$

It seems $\sqrt{p}$ is not needed but only $\sqrt{2}$, or the fundamental unit $U_2 = 1+\sqrt{2}$ also known as the silver ratio. For the modular lambda function $\lambda(\sqrt{-m})$ for these same $m$, see also Entry 134 where

$$\frac1{\lambda(\sqrt{-14})} = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 ,\quad n=2(1+\sqrt2)$$

Entry 188

Ramanujan g-function $g_m$ and $\tau = \sqrt{-m}$ is,

$$2^{1/4}g_m = \frac{\eta\big(\frac12\tau\big)}{\eta(\tau)}$$ Ramanujan tabulated $G_m$ and $g_m$ for odd and even $m$, respectively. For the latter and $m\geq4$, we propose a slightly different formula,

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-m})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}g_m\big)^n}}$$

with modular lambda function $\lambda(z)$. Values for $g_m$ have been given in Entry 183. Using some of them, we conjecture

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-6})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}(1+\sqrt2)^{1/6}\big)^n}}$$

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-22})\big) = \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac{(-1)^n}{\big(2^{1/4}(1+\sqrt2)^{1/2}\big)^n}}$$

and so on.

Entry 187

For the fourth McKay-Thompson series for the Monster $$j_{4}(\tau) = \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}$$ Since we have Ramanujan's $G_m$-function for $\tau = \sqrt{-m}$ as

$$2^{1/4}G_m = \frac{\eta^2(\tau)}{\eta(\tfrac12\tau)\,\eta(2\tau)}$$

then $j_4\big(\tfrac12\tau\big)$ is just the $24$th power of $2^{1/4}G_m$. Hence we propose 

$$K(k_m) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/4}G_m\big)^{24n}}}$$

Examples. The values for $G_5, G_{13}, G_{37}$ have already been given in Entry 181 and involve roots of quadratics, like $G_5 = \phi^{1/4}$ with golden ratio $\phi$. But for cubics, given the tribonacci constant $T$, the plastic ratio $P$ and the supergolden ratio $\psi$, the real root of

$$\begin{align}T^3-T^2-T-1 &= 0\\ P^3-P-1 &=0\\ \psi^3-\psi^2-1 &=0 \end{align}$$

then $G_{11}, G_{23}, G_{31}$ are

$$\begin{align}G_{11} &= 2^{-1/4}\left(\tfrac{T+1}T\right) \\ G_{23} &= 2^{1/4}P \\ \,G_{31} &=2^{1/4}\psi\end{align}$$

and we conjecture

$$\quad K(k_5) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/4}\phi^{1/4}\big)^{24n}}}$$

$$K(k_{11}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(\tfrac{T+1}T\big)^{24n}}}\quad$$

$$K(k_{23}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/2}P\big)^{24n}}}$$

$$K(k_{31}) = \frac{\pi}2 \sqrt{\sum_{n=0}^\infty \frac{(2n)!^3}{n!^6} \frac1{\big(2^{1/2}\psi\big)^{24n}}}$$

and so on.

Entry 186

Using the $j_3(\tau)$ from the previous entry, then the following quintics have a solvable Galois group

$$\begin{align}\qquad y^3(y-5)^2 &= -3\cdot 4^3\\ y^3(y-5)^2 &= -5\cdot 12^3\\ y^3(y-5)^2 &= -7\cdot 36^3\\ y^3(y-5)^2 &= -\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3\end{align}$$

as well as

$$\begin{align}y^3(y-5)^2 &= -(2\sqrt3)^6\\ y^3(y-5)^2 &= -(4\sqrt3)^6\\ y^3(y-5)^2 &= -(10\sqrt3)^6\qquad \qquad\end{align}$$

as discussed in Entry 142.

Entry 185

For the third McKay-Thompson series of the Monster, or $j_3(\tau)$ 

$$j_3(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3\left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2$$

I haven't yet found a general formula for the complete elliptic integral of the first kind $K(k_m)$ except for the special case $K(k_3)$. Note that,

$$\pi = \beta\big(\tfrac12,\tfrac12\big) = 2\int_0^1\frac1{(1-x^2)^{1/2}}dx = 3.14159$$

$$\pi_3 = \beta\big(\tfrac13,\tfrac13\big) = 3\int_0^1\frac1{(1-x^3)^{1/3}}dx = 5.29992$$

where $\pi_3 = 2^{4/3}\,3^{1/4}K(k_3) = \frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$, the real period of the Dixon elliptic functions, can be considered as a cubic analogue of $\pi$. Let 

$$\begin{align}j_3\big(\tfrac{1+3\sqrt{-1/3}}2\big) &= -3\cdot 4^3\\ j_3\big(\tfrac{1+5\sqrt{-1/3}}2\big) &= -5\cdot 12^3\\ j_3\big(\tfrac{1+7\sqrt{-1/3}}2\big) &= -7\cdot 36^3\\ j_3\big(\tfrac{1+11\sqrt{-1/3}}2\big) &= -\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3\end{align}$$

Hence we propose

$$2^{1/3}K(k_{3}) = 3^{1/4}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-3\cdot 4^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{5^{2/3}}{3^{3/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-5\cdot 12^3}\big)^n}}$$

$$2^{1/3}K(k_{3}) = \frac{7^{5/6}}{3^{5/4}}\,\frac{\pi}2 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-7\cdot 36^3}\big)^n}}$$

$$\frac{2^{1/3}K(k_{3})}{\sqrt{11}-\sqrt3\,} = \frac{11^{5/6}}{3^{3/4}}\,\frac{\pi}8 \,\sqrt{\sum_{n=0}^\infty \frac{(2n)!\,(3n)!}{n!^5} \frac1{\big({-\sqrt{11}\,(150\sqrt3+78\sqrt{11})^3}\big)^n}}$$

Entry 184

Using the $j_2(\tau)$ from the previous entries, then the following quintics have a solvable Galois group

$$\begin{align}y(y-5)^4 &= -(4\sqrt2)^4\\ y(y-5)^4 &=  -(12\sqrt2)^4\\ y(y-5)^4 &=  -(84\sqrt2)^4\quad\end{align}$$

as well as,

$$\begin{align}y(y-5)^4 &= (4\sqrt{3})^4\\ y(y-5)^4 &= 12^4\\ y(y-5)^4 &= (12\sqrt{11})^4\\ y(y-5)^4 &= 396^4\qquad\end{align}$$

and an example with class number $4$

$$\quad y(y-5)^4 = -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3$$

though for all radical $j_2(\tau)$ as also discussed in Entry 141.

Entry 183

The more famous value of $j_2(\tau)$ is

$$j_2\big(\tfrac12\sqrt{-58}\big) = 396^4$$

So for even $m$ and since $K(k) = \frac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12;1,k^2\big)$, we propose a slightly different formula,

$$\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-m})\big) = \frac{(2^{1/4}g_m)^3}{\;\big(j_2(\tau)\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

with modular lambda function $\lambda(z)$, Ramanujan g-function $g_m$ for integer $m\geq4$ and $\tau = \frac12{\sqrt{-m}}$,

$$2^{1/4}g_m = \frac{\eta\big(\frac12\sqrt{-m}\big)}{\eta(\sqrt{-m})}$$

For Ramanujan's $G_m$ and $g_m$-functions, he tabulated them for odd and even $m$, respectively and found,

$$\begin{align}g_6 &=  (1+\sqrt2)^{1/6}\\ g_{10} &=  \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\\ g_{22} &=  (1+\sqrt2)^{1/2}\\ g_{58} & = \left(\tfrac{5+\sqrt{29}}2\right)^{1/2}\end{align}$$ Therefore we conjecture

$$\begin{align}\,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-6})\big) &= \frac{(2^{1/4}g_{6})^3}{\;\sqrt{4\sqrt3}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(4\sqrt3)^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-10})\big) &= \frac{(2^{1/4}g_{10})^3}{\;\sqrt{12}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{12^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-22})\big) &= \frac{(2^{1/4}g_{22})^3}{\;\sqrt{12\sqrt{11}}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{(12\sqrt{11})^{4n}}}\\ \,_2F_1\big(\tfrac12,\tfrac12;1,\lambda(1+\sqrt{-58})\big) &= \frac{(2^{1/4}g_{58})^3}{\;\sqrt{396}}\; \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{396^{4n}}}\end{align}$$

Entry 182

Using class number $4$, in Entry 68 we calculated $j_{2}(\tau)$ at certain points and found,

$$\begin{align}j_{2}\Big(\tfrac{1+\sqrt{-17}}2\Big) &= -2^{11}\big(4+\sqrt{17}\big)^2 \big({-1}+\sqrt{17}\big)\\ j_{2}\Big(\tfrac{1+\sqrt{-73}}2\Big) &= -2^{9}\cdot3^4\big(111+13\sqrt{73}\big)^3\\ j_{2}\Big(\tfrac{1+\sqrt{-97}}2\Big) &= -2^{11}\cdot3^4\big(59+6\sqrt{97}\big)^4\big({-9}+\sqrt{97}\big)\\ j_{2}\Big(\tfrac{1+\sqrt{-193}}2\Big) &= -2^{11}\cdot3^4\big(208+15\sqrt{193}\big)^4\big(903+65\sqrt{193}\big)\end{align}$$ All the quadratic irrationals above with odd powers are odd fundamental solutions to Pell equations $x^2-my^2=-16$. For example, the initial solution to $x^2-193y^2=-16$ is $(x,y)=(903,\,65)$. And Ramanujan already found

$$\begin{align}G_{17} &=  \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &=  \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\\ G_{97} &=  \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\\ G_{193} &= \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$

These were used in Entry 162 to find formulas for $K(k_m)$. But we can combine these to find new formulas using 

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^3}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

where $\tau = \frac{1+\sqrt{-m}}2$. For example, let $m = 73$

$$K(k_{73}) = \frac{\pi}2  \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^3 \frac{2^{3/4}}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

where as given above

$$j_2(\tau) = j_2\Big(\tfrac{1+\sqrt{-73}}2\Big) = {-2^{9}}\cdot3^4\big(111+13\sqrt{73}\big)^3$$

and so on.

Entry 181

For the second McKay-Thompson series of the Monster, or $j_2(\tau)$ 

$$j_2(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6\left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2$$

we conjecture that the complete elliptic integral of the first kind $K(k_m)$

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^3}{\;\big({-j_2(\tau)}\big)^{1/8}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big(j_2(\tau)\big)^n}}$$

with Ramanujan G-function $G_m$ for integer $m\geq5$ and $\tau = \frac{1+\sqrt{-m}}2$, 

$$2^{1/4}G_m = \frac{\eta^2(\sqrt{-m})}{\eta(\tfrac12\sqrt{-m})\,\eta(2\sqrt{-m})} = \zeta_{48}\frac{\eta(\tau)}{\eta(2\tau)}$$ and $48$th root of unity $\zeta_{48} = e^{2\pi i/48}$. 

Examples. For class number $h(-4m) = 2$, we've already come across 

$$\begin{align}G_{5} &= \left(\frac{1+\sqrt{5}}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ \,G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$

hence,

$$K(k_{5}) = \frac{\pi}2  \frac{\big(1+\sqrt{5}\big)^{3/4}}{\big(4\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(4\sqrt2)^4}\big)^n}}$$

$$K(k_{13}) = \frac{\pi}2  \frac{\big(3+\sqrt{13}\big)^{3/4}}{\big(12\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(12\sqrt2)^4}\big)^n}}$$

$$K(k_{37}) = \frac{\pi}2  \frac{2^{3/4}\big(6+\sqrt{37}\big)^{3/4}}{\big(84\sqrt2 \big)^{1/2}}\, \sqrt{\sum_{n=0}^\infty \frac{(4n)!}{n!^4} \frac1{\big({-(84\sqrt2)^4}\big)^n}}$$

and so on.

Entry 180

Using the $j$-functions from the previous entries, then the following quintics have a solvable galois group $$\begin{align}z^3(z^2+5z+40) &= -960^3\\ z^3(z^2+5z+40) &= -5280^3\\ z^3(z^2+5z+40) &= -640320^3\qquad\end{align}$$ and all other radical $j(\tau)$. Examples from class number $2$ $$\begin{align}z^3(z^2+5z+40) &= 2^3(25-13\sqrt{5})^3\\ z^3(z^2+5z+40) &= 30^3(31-9\sqrt{13})^3\\ z^3(z^2+5z+40) &= 60^3(2837-468\sqrt{37})^3\end{align}$$

a form related to the Brioschi quintic and discussed in Entry 140.

Entry 179

Continuing from Entry 178, we select some $d=4m$ with class number $h(-d) = 2$. It is known for $m = 5,13,17$ 

$$\begin{align}G_{5} &= \left(\frac{1+\sqrt{5}}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ \,G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$

and calculate the j-function at the points,

$$\begin{align}j\big(\tfrac{1+\sqrt{-5}}2\big) &= 2^3(25-13\sqrt{5})^3\\ j\big(\tfrac{1+\sqrt{-13}}2\big) &= 30^3(31-9\sqrt{13})^3\\ j\big(\tfrac{1+\sqrt{-37}}2\big) &= 60^3(2837-468\sqrt{37})^3\end{align}$$

which are all negative values just like in the previous entry. The proposed relation has a bound $m\geq7$, so the first value can't be used. But we conjecture 

$$K(k_{13}) = \frac{\pi}2  \frac{\big(3+\sqrt{13}\big)^{1/2}}{\big({-30^3}(31-9\sqrt{13})^3 \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(30^3(31-9\sqrt{13})^3\big)^n}}$$

$$K(k_{37}) = \frac{\pi}2  \frac{2^{3/4}\big(6+\sqrt{37}\big)^{3/4}}{\big({-60^3}(2837-468\sqrt{37} \big)^{1/12}}\, \sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({60^3(2837-468\sqrt{37}}\big)^n}}$$

and so on for other $j\big(\tfrac{1+\sqrt{-m}}2\big)$ with $m\geq7$.

Entry 178

We go back to the first four McKay-Thompson series of the Monster. We conjecture that the complete elliptic integral of the first kind $K(k_m)$ can be given by the first McKay-Thomson series $j_{1A}(\tau) = j(\tau)$ or simply the j-function as,

$$K(k_m) = \frac{\pi}2  \frac{(2^{1/4}G_m)^2}{\big({-j(\tau)}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big(j(\tau)\big)^n}}$$

with Ramanujan G-function $G_m$ for integer $m\geq7$ and $\tau = \frac{1+\sqrt{-m}}2$, 

$$2^{1/4}G_m = \frac{\eta^2(\sqrt{-m})}{\eta(\tfrac12\sqrt{-m})\,\eta(2\sqrt{-m})} = \zeta_{48}\frac{\eta(\tau)}{\eta(2\tau)}$$ and $48$th root of unity $\zeta_{48} = e^{2\pi i/48}$. Let odd discriminant $d=m$. Examples for class number $h(-d) = 1$,

$$K(k_{7}) = \frac{\pi}2 \frac{2}{\big({15^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-15^3}\big)^n}}$$

while for higher Heegner numbers, $x$ is a root of a simple cubic

$$K(k_{11}) = \frac{\pi}2 \left(\frac{1}{T}+1\right)^2\frac1{\big({32^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-32^3}\big)^n}}$$

with $T$ the tribonacci constant, the real root of $T^3-T^2-T-1=0$. For the largest

$$K(k_{163}) = \frac{\pi}2 \frac{x^2}{\big({640320^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-640320^3}\big)^n}}$$

where $x$ is the real root of $x^3-6x^2+4x-2 = 0$. 

P.S. The non-fundamental $d=27$ also has class number $1$ hence its j-function is an integer

$$K(k_{27}) = \frac{\pi}2 \frac{(2+2^{4/3}+2^{5/3})^{2/3}}{\big({{-3}\cdot160^3}\big)^{1/12}}\sqrt{\sum_{n=0}^\infty \frac{(6n)!}{(3n)!\,n!^3} \frac1{\big({-3\cdot160^3}\big)^n}}$$

Entry 177

From Entry 176, we gave the fundamental unit $U_d$ $$U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$ and from $e^{\pi\sqrt{163}}\approx 640320^3+744$, observed that $$640320=80\,(\color{blue}{8005}-1)$$ This may be just coincidence, but not when $U_{3d}$ for $d = 7,11,19,43,67,163$. This was also observed by H. H. Chan. Given the fundamental units

$$\begin{align}U_{21} &=\left(\tfrac{\sqrt3+\sqrt{7}}2\right)^2\\ U_{33} &=\big(2\sqrt3+\sqrt{11}\big)^2\\ U_{57} &=\big(5\sqrt3+2\sqrt{19}\big)^2\\ U_{129} &=\big(53\sqrt3+14\sqrt{43}\big)^2\\ U_{201} &=\big(293\sqrt3+62\sqrt{67}\big)^2\\ U_{489} &=\big(35573\sqrt3+4826\sqrt{163}\big)^2 \end{align}$$

Define the function

$$F_d = 3\sqrt3\big(\sqrt{U_{3d}}-1/\sqrt{U_{3d}}\big)+6$$

then we get the rather familiar $$\begin{align}F_{7} &= 15\\ F_{11} &= 6(1+\sqrt{33})\\ F_{19} &= 96\\ F_{43} &= 960\\ F_{67} &= 5280\\ F_{163} &= 640320\end{align}$$

which (except for $d=11$) are the cube roots of the j-function (negated). 

P.S. I don't know why $d=11$ does not obey the pattern, but it does yield the integer $42$ if the positive sign of the second square root $\pm 1/\sqrt{U_{3d}}$ is used, though the correct value should be $32$.

Entry 176

Ramanujan found the exact value of the Ramanujan $G$-function $$G_{69} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}$$ Note the fundamental units $U_n$ $$\begin{align}U_{23} &= 24+5\sqrt{23} = \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^2\\ U_{69} &= \frac{25+3\sqrt{69}}2 = \left(\frac{3\sqrt3+\sqrt{23}}2\right)^2 \end{align}$$ and how he uses the squared version. As a second example

$$G_{77} = \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\quad$$

and fundamental units $$\begin{align}U_7 &= 8+3\sqrt7 =  \left(\frac{3+\sqrt{7}}{\sqrt2}\right)^2\\ U_{77} &= \frac{9+\sqrt{77}}2 = \left(\frac{\sqrt7+\sqrt{11}}2\right)^2 \end{align}$$

Ramanujan mostly uses the squared version of the $U_n$ to get "simpler" expressions with smaller integers. For prime $p \equiv 3\,\text{mod}\,4$, one can always do since 

$$x^2-py^2=-2\\ x^2-py^2=+2$$ are solvable by $p \equiv 3\,\text{mod}\,8\,$ and $p \equiv 7\,\text{mod}\,8$, respectively. Checking  $U_{67}$ and $U_{163}$, yields the reductions $$U_{67}= 48842 + 5967\sqrt{67} = \left(\frac{\color{blue}{221} + 27\sqrt{67}}{\sqrt2}\right)^2\\ U_{163}=64080026 + 5019135\sqrt{163}= \left(\frac{\color{blue}{8005} + 627\sqrt{163}}{\sqrt2}\right)^2$$ And from $e^{\pi\sqrt{67}}\approx 5280^3+744$ and $e^{\pi\sqrt{163}}\approx 640320^3+744$, we find the relations $$5280=24\,(\color{blue}{221}-1)\\ 640320=80\,(\color{blue}{8005}-1)$$ though it may be just coincidence.

Entry 175

For discriminant $d=4m$ with class number $8$ and semiprime $m$: 

If $m =5p$, then distinguish between $p \equiv 1\,\text{mod}\, 8$ and $p \equiv 5\,\text{mod}\, 8$. 

If $m =7p$, then distinguish between $p \equiv 3\,\text{mod}\, 8$ and $p \equiv 7\,\text{mod}\, 8$. 

The semiprime $m =5p$ was in the previous entry. For $m =7p$, there are only four and Ramanujan found the radicals below. 

For the 1st case, $p = 11, 43$, thus $m = 7p = 77,\, 301$  

$$\begin{align}G_{77} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{\sqrt7+\sqrt{11}}2\right)^{1/8} \left(\sqrt{\frac{2+\sqrt{11}}4}+\sqrt{\frac{6+\sqrt{11}}4}\right)^{1/2}\\ G_{301} &= \big(8+3\sqrt7\big)^{1/8}\left(\frac{57\sqrt7+23\sqrt{43}}2\right)^{1/8} \left(\sqrt{\frac{42+7\sqrt{43}}4}+\sqrt{\frac{46+7\sqrt{43}}4}\right)^{1/2}\end{align}$$

For the 2nd case, $p = 31, 79$, thus $m = 7p = 217,\, 553$ 

$$\begin{align}G_{217} &= \left(\sqrt{x_1-\tfrac12}+\sqrt{x_1+\tfrac12}\right)^{1/2} \left(\sqrt{y_1-\tfrac12}+\sqrt{y_1+\tfrac12}\right)^{1/2}\\ G_{553} &= \left(\sqrt{x_2-\tfrac12}+\sqrt{x_2+\tfrac12}\right)^{1/2} \left(\sqrt{y_2-\tfrac12}+\sqrt{y_2+\tfrac12}\right)^{1/2} \quad\end{align}$$ where $$x_1=\frac{10+4\sqrt{7}}2,\quad y_1=\frac{14+5\sqrt{7}}4$$ $$x_2=\frac{142+16\sqrt{79}}2,\quad y_2=\frac{98+11\sqrt{79}}4$$

But it seems not noticed that a fundamental unit is imbedded in these radicals as

$$x_1+2y_1 = \frac32(8+3\sqrt{7}) = \frac32\left(\frac{3+\sqrt7}{\sqrt2}\right)^2 =\frac32\,U_7$$

$$x_2+2y_2 = \frac32(80+9\sqrt{79}) = \frac32\left(\frac{9+\sqrt{79}}{\sqrt2}\right)^2 = \frac32\,U_{79}$$

Entry 174

Continuing from the previous entry, for $d=4m$ with class number $8$, the semiprime $m =5p$ with $p \equiv 5\,\text{mod}\, 8$ is also well-behaved. And it involves the golden ratio. There are only four, namely $p = 13, 29, 53, 101$, thus $m = 5p = 65, 145, 265, 505$. Ramanujan found the radicals below and the $G$-function have a common form

$$G_{5p} = \phi^{k}\,U_{p}^{1/4}\,x_{p}^{1/2}$$

with powers of the golden ratio $\phi$, fundamental unit $U_n$, and $x_{p}^2$ a root of a unit quartic

$$\begin{align}\frac{G_{65}}{\phi} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4} \left(\sqrt{\frac{1+\sqrt{65}}8}+\sqrt{\frac{9+\sqrt{65}}8}\right)^{1/2}\\ \frac{G_{145}}{\phi^3} &= \left(\frac{5+\sqrt{29}}2\right)^{1/4} \left(\sqrt{\frac{9+\sqrt{145}}8}+\sqrt{\frac{17+\sqrt{145}}8}\right)^{1/2}\\ \frac{G_{265}}{\phi^3} &= \left(\frac{7+\sqrt{53}}2\right)^{1/4} \left(\sqrt{\frac{81+5\sqrt{265}}8}+\sqrt{\frac{89+5\sqrt{265}}8}\right)^{1/2}\\ \frac{G_{505}}{\phi^7} &= \big(10+\sqrt{101}\big)^{1/4} \left(\sqrt{\frac{105+5\sqrt{505}}8}+\sqrt{\frac{113+5\sqrt{505}}8}\right)^{1/2}\end{align}$$

The case $p \equiv 1\,\text{mod}\, 8$ or $p = 41, 89$, thus $m = 5p = 205, 445$ behaves slightly differently though

$$\begin{align}\frac{G_{205}}{\phi} &= \left(\frac{43+3\sqrt{205}}2\right)^{1/8} \left(\sqrt{\frac{-1+\sqrt{41}}8}+\sqrt{\frac{7+\sqrt{41}}8}\right)\\  \frac{G_{445}}{\phi^{3/2}} &= \,\left(\frac{21+\sqrt{445}}2\right)^{1/4}\, \left(\sqrt{\frac{5+\sqrt{89}}8}+\sqrt{\frac{13+\sqrt{89}}8}\right)\end{align}$$

How Ramanujan found these is a mystery.

Entry 173

There are many in the set $d=4m$ with class number $8$. When $m$ is a prime or a semiprime (a product of two primes) like $m =3p$, then it may be well-behaved. For this set, there are only three, namely $p = 23, 47, 71$, thus $m = 3p = 69, 141, 213$. Ramanujan found the radicals below and the $G$-function have a common form

$$G_{3p} = U_{p}^{1/24}\,U_{3p}^{1/16}\,x_{p}^{1/2}$$

with fundamental unit $U_n$ and where $x_{p}^2$ is a root of a unit quartic

$$\begin{align}G_{69} &= \left(\frac{5+\sqrt{23}}{\sqrt2}\right)^{1/12} \left(\frac{3\sqrt3+\sqrt{23}}2\right)^{1/8} \left(\sqrt{\frac{2+3\sqrt3}4}+\sqrt{\frac{6+3\sqrt3}4}\right)^{1/2}\\ G_{141} &= \left(\frac{7+\sqrt{47}}{\sqrt2}\right)^{1/12}\; \big(4\sqrt3+\sqrt{47}\big)^{1/8}\; \left(\sqrt{\frac{14+9\sqrt3}4}+\sqrt{\frac{18+9\sqrt3}4}\right)^{1/2}\\ G_{213} &= \left(\frac{59+7\sqrt{71}}{\sqrt2}\right)^{1/12} \left(\frac{5\sqrt3+\sqrt{71}}2\right)^{1/8} \left(\sqrt{\frac{19+12\sqrt3}2}+\sqrt{\frac{21+12\sqrt3}2}\right)^{1/2} \end{align}$$ How Ramanujan found these is unknown as it is uncertain if he was aware of class field theory. Note also that without the factor $3$, then $d=23,47,71$ are the smallest $d$ with class number $h(-d) = 3,5,7$, respectively, while $h(-3d) = 8$.

P.S. Checking $h(-3d) = 16$, one finds the only primes are $d=167,191,239,383,311$ which are the smallest $d$ with class number $h(-d) = 11,13,15,17,19$, respectively. Makes me wonder if their $G_{3d}$ would be analogous.

Entry 172

There are many fundamental $d=4m$ with class number $8$, though only seven are prime namely $p = 41, 113, 137, 313, 337, 457, 577$. Their Ramanujan $G$-functions are

$$\begin{align}G_{41} &= \left(\frac{x+\sqrt{x^2-4}}2\right)^{1/2} =\sqrt{\frac{x-2}4}+\sqrt{\frac{x+2}4} \\ G_{113} &=\left(\frac{y+\sqrt{y^2-4}}2\right)^{1/2} =\sqrt{\frac{y-2}4}+\sqrt{\frac{y+2}4} \end{align}$$ where $$x = \left(\frac{5+\sqrt{41}}4\right) \left(1+\sqrt{\frac{-5+\sqrt{41}}8}\right)\\ y = \left(\frac{9+\sqrt{113}}4\right) \left(1+\sqrt{\frac{-7+\sqrt{113}}2}\right)$$ and similarly for the other $p$, though the quartic roots get more complicated.

Entry 171

Continuing from Entry 170, there is another way to express the Ramanujan $G_n$-function where $d=4n$ for prime $n$ has class number $6$ by using fundamental units $U_n$. Borrowing a trick from Ramanujan, he found

$$G_{169}=\frac13\Big(2+\sqrt{13}+\sqrt[3]{U_{13}(v+3\sqrt{3})\sqrt{13}}+\sqrt[3]{U_{13}(v-3\sqrt{3})\sqrt{13}}\Big)$$

where $$U_{13} = \frac{3+\sqrt{13}}2, \quad v = \frac{11+\sqrt{13}}2$$

Using a similar form, we propose that

$$\begin{align}G_{29} &=\frac1{3^{1/4}}\Big(\tfrac{9+\sqrt{29}}2+\sqrt[3]{U_{29}(x+24\sqrt{3})}+\sqrt[3]{U_{29}(x-24\sqrt{3})}\Big)^{1/4}\\ G_{53} &=\frac1{3^{1/4}}\Big(\tfrac{23+3\sqrt{53}}2+\sqrt[3]{U_{53}(y+120\sqrt{3})}+\sqrt[3]{U_{53}(y-120\sqrt{3})}\Big)^{1/4}\\ G_{61} &=\frac1{3^{1/4}}\Big(15+2\sqrt{61}+\sqrt[3]{U_{61}(z+72\sqrt{3})}+\sqrt[3]{U_{61}(z-72\sqrt{3})}\Big)^{1/4}\end{align}$$ where $$U_{29} = \frac{5+\sqrt{29}}2, \quad x = \frac{185+19\sqrt{29}}2$$ $$\quad U_{53} = \frac{7+\sqrt{53}}2, \quad y = \frac{1721+217\sqrt{53}}2$$ $$U_{61} = \frac{39+5\sqrt{61}}2, \quad z = \frac{601+93\sqrt{61}}2\;$$

and similarly for all seven prime $p = 29, 53, 61, 109, 157, 277, 397$. Note they have form $p \equiv 5\,\text{mod}\,8$. And all their fundamental units have form $U_p=\frac{a+b\sqrt{p}}2$, hence have odd solutions to the Pell equation $x^2-py^2=-4$.

Entry 170

There are only seven fundamental $d=4p$ with class number $6$, namely $p = 29, 53, 61, 109, 157, 277, 397$. Hence their Ramanujan $G$-functions are

$$\begin{align}G_{29} &= \left(\frac{y_1^2+\sqrt{y_1^4+4}}2\right)^{1/4} \\ G_{53} &= \left(\frac{y_2^2+\sqrt{y_2^4+4}}2\right)^{1/4} \\ G_{61} &=  \left(\frac{y_3^2+\sqrt{y_3^4+36}}6\right)^{1/4} \end{align}$$ where $$y^3 - y^2 - 4y - 4 = 0\\ y^3 - 7y^2 + 13y - 11 = 0\\ y^3 - 6y^2 - 27y - 54 = 0$$ and similarly for the other $p$. 

Entry 169

Recall Ramanujan's $G$-function $$2^{1/4}G_n=\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(\tau)}$$ where $\tau=\sqrt{-n}$. There are only three fundamental $d=4p$ with class number $2$ for prime $p$, namely $p = 5,13,37$. Hence $$\begin{align}G_5 &= \left(\frac{1+\sqrt5}2\right)^{1/4}\\ G_{13} &= \left(\frac{3+\sqrt{13}}2\right)^{1/4}\\ G_{37} &=\, \big(6+\sqrt{37}\big)^{1/4}\end{align}$$ Going higher, there are only four fundamental $d=4p$ with class number $4$ for prime $p$, namely $p = 17,73,97,193$.

$$\begin{align}G_{17} &= \sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8} \\ G_{73} &= \sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8} \\ G_{97} &= \sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8} \\ G_{193} &=  \sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8} \end{align}$$ all of which were already known to Ramanujan. But as was shown in Entry 161 and Entry 162, it turns out these also appear in the closed-form of the complete elliptic integral of the first kind $K(k_n)$. The next entry will be for class number $6$ where one had to extract $4$th roots again.

Entry 168

For class number $5$, there are only four $d$ of the kind $d \equiv 7\,\text{mod}\,8$, namely $d = 47, 79, 103, 127$. We illustrate only the first two and propose, $$\begin{align}K(k_{47}) &=\frac{\sqrt{2\pi}}{2\sqrt{47}}\,x^{2/3} \left(\prod_{m=1}^{47}\Big[\Gamma\big(\tfrac{m}{47}\big)\Big]^{\big(\tfrac{-47}{m}\big)}\right)^{\color{red}{1/10}}\\ K(k_{79}) &=\frac{\sqrt{2\pi}}{2\sqrt{79}}\,y^{2/3} \left(\prod_{m=1}^{79}\Big[\Gamma\big(\tfrac{m}{79}\big)\Big]^{\big(\tfrac{-79}{m}\big)}\right)^{\color{red}{1/10}}\end{align}$$ where $(x,y)$ are the real roots of the quintics solvable in radicals $$x^5 - 2x^4 - 10x^3 - 13x^2 - 6x - 1=0\\ y^5 - 11y^4 + 17y^3 - 2y^2 - 5y - 1=0$$ And similarly for $d= 103,127$. But for the next discriminant or $d=131$ which is of kind $d \equiv 3\,\text{mod}\,8$, one now has to deal with an algebraic number of degree $3\times5 = 15$. And it seems difficult to find the eta quotients responsible for $(x,y)$.

Entry 167

For odd class number $h(-d)$, the kind that is $d \equiv 7\,\text{mod}\,8$ seems more well-behaved than $d \equiv 3\,\text{mod}\,8$. For class number $3$, there are only two of the first kind: $d = 23,31$. Hence, $$\begin{align}K(k_{23}) &=\frac{\sqrt{2\pi}}{2\sqrt{23}}\,x^{4/3} \left(\prod_{m=1}^{23}\Big[\Gamma\big(\tfrac{m}{23}\big)\Big]^{\big(\tfrac{-23}{m}\big)}\right)^{\color{red}{1/6}}\\ K(k_{31}) &=\frac{\sqrt{2\pi}}{2\sqrt{31}}\,y^{4/3} \left(\prod_{m=1}^{31}\Big[\Gamma\big(\tfrac{m}{31}\big)\Big]^{\big(\tfrac{-31}{m}\big)}\right)^{\color{red}{1/6}}\end{align}$$ where $(x,y)$ are the real roots of the cubics $$x^3-x-1=0\\ y^3-y^2-1=0$$ or the plastic ratio and supergolden ratio, respectively. But for $d=59$ which is of the second kind, then the radical involved will be an algebraic number of degree $3\times3 =9$.

Entry 166

The case $d=8$ is special since it is even but has odd class number $h(-d)=1$. Given the Kronecker symbol $\big(\tfrac{-d}{m}\big)$, we propose a symmetrical closed-form 

$$\begin{align}K(k_8) &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\prod_{m=1}^{8}\Big[\Gamma\big(\tfrac{m}{8}\big)\Big]^{\big(\tfrac{-8}{m}\big)}\right)^{1/2}\\ &=\frac{\sqrt{2\pi}}{16}\frac1{\sqrt{U_2}}\left(\sqrt{\frac{U_2-1}2}+\sqrt{\frac{U_2+1}2}\right)^2 \left(\frac{\Gamma\big(\frac18\big)\,\Gamma\big(\frac38\big)}{\Gamma\big(\frac58\big)\,\Gamma\big(\frac78\big)}\right)^{1/2}\end{align}$$ with fundamental unit $U_2=1+\sqrt2$ and form reminiscent of the odd $d$ with class number $1$.

Entry 165

The previous entries dealt with even class numbers. For odd class number $h(-d)=1$, there are the nine Heegner numbers $d=1,2,3,7,11,19,43,67,163$. Given the Kronecker symbol $\big(\tfrac{-d}{m}\big)$, we propose 

Conjecture. Let $d>3$ with class number $h(-d)=1$. Then $x$ below is an algebraic number $$\frac1{x_d^2}  = \frac1{K(k_d)}\frac{\sqrt{2\pi}}{\color{blue}4\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/2}}$$ specifically $$x_d=2^{1/4}G_n=\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(\tau)}$$ with Ramanujan's $G$-function. This function $x_d$ has been discussed in Entry 159. For $d=7$, then $x_7=\sqrt2$, but for the five $d\geq11$, then $x_d$ are the real roots of the following five simple cubics, 

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$ also discussed in Entry 160. 

Entry 164

To summarize, given fundamental discriminant $d$, class number $h(-d)=n$, complete elliptic integral of the first kind $K(k_p)$, and Kronecker symbol $\big(\tfrac{-d}{m}\big)$. 

Conjecture 1. Let even $d=4p$ for prime $p\equiv 1\,\text{mod}\,4$ with even class number $h(-d)=n$ and $$x = \frac1{K(k_{\color{blue}{d/4}})}\frac{\sqrt{2\pi}}{2\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/(2n)}}$$ Conjecture 2. Let odd $\,d=p\,$ for prime $p\,\equiv\, 3\,\text{mod}\,4\,$ with odd class number $h(-d)=n$ and  $$y = \frac1{K(k_{d})}\frac{\sqrt{2\pi}}{2\sqrt{d}} \left(\prod_{m=1}^{d}\Big[\Gamma\big(\tfrac{m}{d}\big)\Big]^{\big(\tfrac{-d}{m}\big)}\right)^{\color{red}{1/(2n)}}$$ then $(x,y)$ are algebraic numbers as seen in entries $160-163$ and $165-168$. They seem to have a closed-form in term of eta quotients but I haven't found it yet.

Entry 163

There are only seven fundamental $d = 4p$ with class number $6$, namely $p=29, 53, 61, 109, 157, 277, 397$.  To prevent clutter, only the first three will be stated. Given the Kronecker symbol $\big(\frac{d}{m}\big)$, we propose

$$\begin{align}K(k_{29}) &= \frac{\sqrt{2\pi}}{2\sqrt{116}} \frac1{\;(x_1)^{1/3}} \left(\frac{y_1^2+\sqrt{y_1^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{116}\Big[\Gamma\big(\tfrac{m}{116}\big)\Big]^{\big(\tfrac{-116}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{53}) &= \frac{\sqrt{2\pi}}{2\sqrt{212}} \frac1{\;(x_2)^{1/3}} \left(\frac{y_2^2+\sqrt{y_2^4+4}}2\right)^{3/4} \left(\prod_{m=1}^{212}\Big[\Gamma\big(\tfrac{m}{212}\big)\Big]^{\big(\tfrac{-212}{m}\big)}\right)^{\color{red}{1/12}}\\ K(k_{61}) &= \frac{\sqrt{2\pi}}{2\sqrt{244}} \frac1{\;(x_3)^{1/3}} \left(\frac{y_3^2+\sqrt{y_3^4+36}}6\right)^{3/4} \left(\prod_{m=1}^{244}\Big[\Gamma\big(\tfrac{m}{244}\big)\Big]^{\big(\tfrac{-244}{m}\big)}\right)^{\color{red}{1/12}}\end{align}$$ where the $x_n$ are the real roots of the following cubics  $$x^3 - 5x^2 - 3x - 1 = 0\\ x^3 - 15x^2 - x - 1= 0\\ x^3 - 27x^2 - 5x - 1= 0$$ while the $y_n$ are also the real roots of cubics  $$y^3 - y^2 - 4y - 4 = 0\\ y^3 - 7y^2 + 13y - 11 = 0\\ y^3 - 6y^2 - 27y - 54 = 0$$ and similarly for the other $p$.

Entry 162

There are only four fundamental $d = 4p$ with class number $4$, namely $p=17,73,97,193$.  Given the Kronecker symbol $\big(\frac{d}{m}\big)$, we propose

$$\begin{align}K(k_{17}) &= \frac{\sqrt{2\pi}}{2\sqrt{68}} \frac1{(U_{17})^{1/8}} \left(\sqrt{\frac{-3+\sqrt{17}}8} +\sqrt{\frac{5+\sqrt{17}}8}\right)^{3} \left(\prod_{m=1}^{68}\Big[\Gamma\big(\tfrac{m}{68}\big)\Big]^{\big(\tfrac{-68}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{73}) &= \frac{\sqrt{2\pi}}{2\sqrt{292}}  \frac1{(U_{73})^{1/8}} \left(\sqrt{\frac{1+\sqrt{73}}8} +\sqrt{\frac{9+\sqrt{73}}8}\right)^{3} \left(\prod_{m=1}^{292}\Big[\Gamma\big(\tfrac{m}{292}\big)\Big]^{\big(\tfrac{-292}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{97}) &= \frac{\sqrt{2\pi}}{2\sqrt{388}} \frac1{(U_{97})^{1/8}} \left(\sqrt{\frac{5+\sqrt{97}}8} +\sqrt{\frac{13+\sqrt{97}}8}\right)^{3}  \left(\prod_{m=1}^{388}\Big[\Gamma\big(\tfrac{m}{388}\big)\Big]^{\big(\tfrac{-388}{m}\big)}\right)^{\color{red}{1/8}}\\ K(k_{193}) &= \frac{\sqrt{2\pi}}{2\sqrt{772}} \frac1{(U_{193})^{1/8}} \left(\sqrt{\frac{22+2\sqrt{193}}8} +\sqrt{\frac{30+2\sqrt{193}}8}\right)^{3} \left(\prod_{m=1}^{772}\Big[\Gamma\big(\tfrac{m}{772}\big)\Big]^{\big(\tfrac{-772}{m}\big)}\right)^{\color{red}{1/8}}\end{align}$$ where the $U_n$ are fundamental units $$U_{17} = 4+\sqrt{17}$$ $$U_{73} = 1068+125\sqrt{73}$$ $$U_{97} = 5604 + 569\sqrt{97}$$ $$U_{193} = {1764132} + 126985\sqrt{193}$$ Going to class number $6$, the red exponent will now be $1/12$.

Entry 161

Entry 160 dealt with discriminants $d$ with class number $1$. Going higher, there are only three fundamental $d = 4p$ with class number $2$, namely $p=5,13,17$.  Given the Kronecker symbol $\big(\frac{d}{m}\big)$, we propose

$$\begin{align}K(k_5) &= \frac{\sqrt{2\pi}}{2\sqrt{20}}\left(\frac{1+\sqrt5}2\right)^{3/4}\left(\prod_{m=1}^{20}\Big[\Gamma\big(\tfrac{m}{20}\big)\Big]^{\big(\tfrac{-20}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{13}) &= \frac{\sqrt{2\pi}}{2\sqrt{52}}\left(\frac{3+\sqrt{13}}2\right)^{3/4}\left(\prod_{m=1}^{52}\Big[\Gamma\big(\tfrac{m}{52}\big)\Big]^{\big(\tfrac{-52}{m}\big)}\right)^{\color{red}{1/4}}\\ K(k_{37}) &= \frac{\sqrt{2\pi}}{2\sqrt{148}}\left(6+\sqrt{37}\right)^{3/4}\left(\prod_{m=1}^{148}\Big[\Gamma\big(\tfrac{m}{148}\big)\Big]^{\big(\tfrac{-148}{m}\big)}\right)^{\color{red}{1/4}}\end{align}$$ Going to class number $4$, the red exponent will now be $1/8$.

Entry 160

We continue with closed-forms for the Dedekind eta function $\eta^2(\sqrt{-d})$ for $d=11,19,43,67,163$. As discussed in Entry 159, the closed-form for the complete elliptic integral of the first kind $K(k_d)$ then necessarily follows. 

$$\eta^2(\sqrt{-11}) = \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}$$ $$\eta^2(\sqrt{-19}) = \frac1{x_{19}^2}\frac{\Gamma\big(\tfrac1{19}\big)\,\Gamma\big(\tfrac4{19}\big)\,\Gamma\big(\tfrac5{19}\big)\,\Gamma\big(\tfrac6{19}\big)\dots\Gamma\big(\tfrac{17}{19}\big)}{19^{1/4}\,(2\pi)^5}$$

$$\eta^2(\sqrt{-43}) = \frac1{x_{43}^2}\frac{\Gamma\big(\tfrac1{43}\big)\,\Gamma\big(\tfrac4{43}\big)\,\Gamma\big(\tfrac6{43}\big)\,\Gamma\big(\tfrac9{43}\big)\dots\Gamma\big(\tfrac{41}{43}\big)}{43^{1/4}\,(2\pi)^{11}}$$

$$\eta^2(\sqrt{-67}) = \frac1{x_{67}^2}\frac{\Gamma\big(\tfrac1{67}\big)\,\Gamma\big(\tfrac4{67}\big)\,\Gamma\big(\tfrac6{67}\big)\,\Gamma\big(\tfrac9{67}\big)\dots\Gamma\big(\tfrac{65}{67}\big)}{67^{1/4}\,(2\pi)^{17}}$$

$$\eta^2(\sqrt{-163}) = \frac1{x_{163}^2}\frac{\Gamma\big(\tfrac1{163}\big)\,\Gamma\big(\tfrac4{163}\big)\,\Gamma\big(\tfrac6{163}\big)\,\Gamma\big(\tfrac9{163}\big)\dots\Gamma\big(\tfrac{161}{163}\big)}{163^{1/4}\,(2\pi)^{41}}$$

where the $x_d$ are the real roots of the following simple cubics, respectively

$$\begin{align}& x^3-2x^2+2x-2 = 0\\ & x^3-2x-2 = 0 \\ & x^3-2x^2-2 = 0 \\ & x^3-2x^2-2x-2 = 0 \\ & x^3-6x^2+4x-2=0\end{align}$$

The numerators, for example, of $\Gamma\big(\frac{n}{19}\big)$ can be found using the Kronecker symbol and this Wolfram command which yields the $\frac{19-1}2 = 9$ numerators as $n = 1, 4, 5, 6, 7, 9, 11, 16, 17$.

Entry 159

Given the McKay-Thompson series of Class 4A for the Monster (A097340)

$$j_{4A}(\tau)=\left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)}\right)^{24} = \frac1q+24 + 276q + 2048q^2 +\dots$$ We take the $6$th root, scale $\tau \to \frac{\tau}2$, and use the version, $$(x_d)^4=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4=\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda(\tau)\big)}{\eta^2(\tau)}=\frac{\frac2{\pi}\times K(k_d)}{\eta^2(\tau)}$$ where $K(k_d)$ is a complete elliptic integral of the first kind. Compared also to Ramanujan's G-function, $$2^{1/4}G_n = \frac{\eta^2(\tau)}{\eta\big(\tfrac{\tau}{2}\big)\eta(2\tau)} = q^{-\frac{1}{24}}\prod_{n>0}(1+q^{2n-1})\qquad\qquad$$ where Ramanujan used odd $n$ for $G_n$ and calculated a lot of $n$. Mathworld also has a partial list of closed-forms for $K(k_d)$. Knowing $x_d$ or $G_n$ immediately leads to a closed-form for $\eta^2(\tau)$ as well. Examples, 

$$\begin{align}\frac{2}{\pi}\times K(k_7) &= x_7^2\,\frac{\Gamma\big(\tfrac17\big)\,\Gamma\big(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\\ \eta^2(\sqrt{-7}) &= \frac1{x_7^2}\frac{\Gamma\big(\tfrac17\big)\,\Gamma(\tfrac27\big)\,\Gamma\big(\tfrac47\big)}{7^{1/4}\,(2\pi)^2}\end{align}$$

where $x_7 =\sqrt2$. And

$$\begin{align}\frac{2}{\pi}\times K(k_{11}) &= x_{11}^2\,\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\\ \eta^2(\sqrt{-11}) &= \frac1{x_{11}^2}\frac{\Gamma\big(\tfrac1{11}\big)\,\Gamma\big(\tfrac3{11}\big)\,\Gamma\big(\tfrac4{11}\big)\,\Gamma\big(\tfrac5{11}\big)\,\Gamma\big(\tfrac9{11}\big)}{11^{1/4}\,(2\pi)^3}\end{align}$$

where $x_{11} =\dfrac1T+1$ and $T$ is the tribonacci constant, the real root of $x^3-x^2-x-1=0$. For $d=11,19,43,67,163$, then $x_d$ is just the real root of a cubic.

Entry 158

In the previous entry, given the Jacobi theta functions $\vartheta_n(0,q)$ with nome $q = e^{\pi i\tau}$, modular lambda function $\lambda(\tau)$, and complete elliptic integral of the first kind $K(k) = \tfrac{\pi}2\, {_2F_1}\big(\tfrac12,\tfrac12,1,k^2)$ we proposed three identities, one as,

$$\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 \overset{\color{red}?}=\lambda$$ where $\lambda = \lambda(\tau)$. Using the known definitions of $\vartheta_n(0,q)$ and $\lambda(\tau)$, the proposed identity implies 

$$\frac{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}{\eta^2(\tau)}=\left(\frac{\eta^2(\tau)}{\eta(\tfrac{\tau}2)\,\eta(2\tau)}\right)^4$$

For appropriate complex quadratics such as $\tau=\sqrt{-n}$, then the RHS is a radical. But if the numerator of the LHS has a closed-form (implied in this list), then this also gives a closed-form for the denominator $\eta^2(\tau)$ such as

$$\begin{align}\eta^2(\sqrt{-1}) &=\frac{\Gamma^2(\tfrac14)}{(2\pi)^{3/2}}\frac1{2^{1/2}}\\ \eta^2(\sqrt{-2}) &=\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)}{(2\pi)^{3/2}}\frac1{2^{5/4}}\\ \eta^2(\sqrt{-3}) &=\frac{\Gamma^3(\tfrac13)}{(2\pi)^{2}}\frac{3^{1/4}}{2^{2/3}}\end{align}$$

and so on. Since an eta quotient is involved, then $\eta(\sqrt{-d})$ for $d$ with class number $1$ will behave quite orderly and will be discussed in the next entry.

Entry 157

Given the Jacobi theta functions $\vartheta_n(0,q)$ which traditionally uses the nome $q = e^{\pi i\tau}$. Define the modular lambda function $\lambda(\tau)$,

$$\lambda(\tau) = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$

Then we propose the three identities below and for appropriate $\tau$ such as $\tau = \sqrt{-d}$ and $\lambda = \lambda(\tau)$ that the ratios below are radicals,

$$\begin{align}\left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=\lambda\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\lambda\big)}}\right)^4 &\overset{\color{red}?}=1\end{align}$$

Adding the first two implies the third. Hence, after removing the common denominator

$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$

which is known to be true. As eta quotients in the same order above, 

$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4$$

Also, the equalities $$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} = \frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}$$ which has a cubic version given in Entry 156.

Entry 156

Given the square of the nome, so $q = e^{2\pi i\tau}$ and the Borwein cubic theta functions $a(q),b(q),c(q)$. Define,

$$\beta = \left(\frac{3}{\left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3=\left(\frac{c(q)}{a(q)}\right)^3$$

Then we conjecture,

$$\begin{align}\left(\frac{c(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=\beta\\ \left(\frac{b(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1-\beta\\ \left(\frac{a(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1\end{align}$$

Adding the first two implies the third,

$$\big(c(q)\big)^3+\big(b(q)\big)^3=\big(a(q)\big)^3$$

which is a known relationship of the Borwein cubic theta functions. As eta quotients,

$$\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3+\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$

Like in the previous entry, the RHS is also a sum,

$$a(q) = \sum_{m,n\,=-\infty}^\infty q^{m^2+mn+n^2} = \frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}$$

Entry 155

It turns out we can use the previous entries to parameterize the equation $$x^n+y^n = z^n$$ for $n=2,3,4$. Given the square of the nome $q = e^{2\pi i\tau}$ and assume the functions $A(q), B(q), C(q)$. Define

$$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$

Then we conjecture,

$$\begin{align}\left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1\end{align}$$

where $C(q), B(q), A(q)$ are defined by the relation and eta quotients below,

$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$

$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$ Note also that $$\begin{align}A(q) &= 1+24\sum_{n=1}^\infty\frac{n q^n}{1+q^n}\\ &=\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_2(0,q)\big)^4\\ &=\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\end{align}$$ where, in this particular instance, we let the Jacobi theta functions $\vartheta_n(0,q)$ use $q = e^{2\pi i\tau}$. 

Entry 154

As an overview of the previous four entries, Ramanujan's theory of elliptic functions to alternative bases uses the hypergeometric function $_2F_1(a,b;c;z)$ with $a+b=c=1$ for the cases $a=\frac16, \frac14, \frac13, \frac12$. This can be related to the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145 for $n = 1,2,3,4$. Consider the equations, 

$$\begin{align}\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)} &=-\tau\sqrt{-1}\\ \frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)} &=-\tau\sqrt{-2}\\ \frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)} &=-\tau\sqrt{-3}\\ \frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)} &=-\tau\sqrt{-4}\end{align}$$ Then the $\alpha_n$ can be solved and expressed in terms of the $j_n(\tau)$ as discussed in the said entries.

Entry 153

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145. Define,

$$\alpha_4(\tau) = \frac{16}{\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{8}+16} = \left(\frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\right)^8$$

Let $\alpha_4 = \alpha_4(\tau)$. Then we conjecture that,

$$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}=-\tau\sqrt{-4}$$ as well as

$$\begin{align}j_{4}(\tau) &=  \frac{16}{\alpha_4\,(1-\alpha_4)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4}+4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}\\ &= \left(\frac{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(4\tau)}\right)^{24/5}\end{align}$$

Example. Let $\tau =\sqrt{-3}$. Then $\alpha_4=(\sqrt3-\sqrt2)^4(\sqrt2-1)^4$ solves $$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\alpha_4\big)}{_2F_1\big(\frac12,\frac12,1,\,\alpha_4\big)}=\sqrt4\times\sqrt3$$ Alternatively and more familiar

$$\frac{_2F_1\big(\frac12,\frac12,1,\,1-\lambda(\sqrt{-r})\big)}{_2F_1\big(\frac12,\frac12,1,\,\lambda(\sqrt{-r})\big)}=\sqrt{r}$$ where $\lambda(\tau)$ is the modular lambda function.

Entry 152

We relate Ramanujan's theory of elliptic functions to alternative bases to the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145. Define,

$$\alpha_3(\tau) = \frac{27}{\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{12}+27} = \left(\frac{3}{\left(\frac{\eta(\tau/3)}{\eta(3\tau)}\right)^{3}+3}\right)^3$$

Let $\alpha_3 = \alpha_3(\tau)$. Then we conjecture that,

$$\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}=-\tau\sqrt{-3}$$ as well as

$$\begin{align}j_{3}(\tau) &=  \frac{27}{\alpha_3\,(1-\alpha_3)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2\\ &= \left(\frac{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(3\tau)}\right)^{24/4}\end{align}$$

Example. Let $\tau =\sqrt{-3}$. Then $\alpha_3=\frac1{250}(187-171\cdot2^{1/3}+18\cdot2^{2/3})$ solves $$\frac{_2F_1\big(\frac13,\frac23,1,\,1-\alpha_3\big)}{_2F_1\big(\frac13,\frac23,1,\,\alpha_3\big)}=\sqrt3\times\sqrt3$$

Entry 151

Ramanujan's theory of elliptic functions to alternative bases can be related to the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145. Define,

$$\alpha_2(\tau) = \frac{64}{\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24}+64} = \left(\frac{8}{\left(\frac{\eta(\tau/2)}{\eta(2\tau)}\right)^{8}+8}\right)^2$$

Let $\alpha_2 = \alpha_2(\tau)$. Then we conjecture that,

$$\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}=-\tau\sqrt{-2}$$ as well as

$$\begin{align}j_{2}(\tau) &=  \frac{64}{\alpha_2\,(1-\alpha_2)}\\ &= \left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2\\ &= \left(\frac{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}{\eta^2(\tau)} \times\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24/3}\end{align}$$

Example. Let $\tau =\sqrt{-3}$. Then $\alpha_2=\frac1{3(2+\sqrt3)^2(13+4\sqrt3)}$ solves $$\frac{_2F_1\big(\frac14,\frac34,1,\,1-\alpha_2\big)}{_2F_1\big(\frac14,\frac34,1,\,\alpha_2\big)}=\sqrt2\times\sqrt3$$

Entry 150

Ramanujan's theory of elliptic functions to alternative bases considers the hypergeometric function $_2F_1(a,b;c;z)$ with $a+b=c=1$ for the cases $a=\frac16, \frac14, \frac13, \frac12$. We can relate this to the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145 for $n = 1,2,3,4$. Define,

$$\alpha_1(\tau) = \frac12\left(1-\sqrt{1-\frac{1728}{j_{1}(\tau)}}\right)$$

where $j = j_1(\tau)$ is the j-function. Let $\alpha_1 = \alpha_1(\tau)$. Then we conjecture that,

$$\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}=-\tau\sqrt{-1}$$ as well as

$$\begin{align}j_{1}(\tau)  &=\frac{432}{\alpha_1\,(1-\alpha_1)}\\ &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3\\  &=\left(\frac{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}{\eta^2(\tau)} \right)^{24/2}\end{align}$$

Example. Let $\tau =\sqrt{-3}$. Then $\alpha_1=\frac1{5\sqrt5}\left(\frac{-1+\sqrt5}2\right)^5$ solves $$\frac{_2F_1\big(\frac16,\frac56,1,\,1-\alpha_1\big)}{_2F_1\big(\frac16,\frac56,1,\,\alpha_1\big)}=\sqrt3$$ since $-\tau\sqrt{-1}=-\sqrt3\, i\times i=\sqrt3$.

Entry 149

This is the octic overview. Using the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145, if $\tau$ are complex quadratics such that $j_n(\tau)$ is a radical, then the following octics have a solvable Galois group, hence solvable in radicals $$\begin{align}\qquad{j_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\ {j_2}^2 &=\frac{j_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\ {j_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\ {j_4}^2 &=\frac{7 j_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x \end{align}$$

It would be good if the one for $j_2$ can be simplified. They have discriminants (with another factor of $D_2$ suppressed),

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

Examples. Let $\tau = \tfrac{1+\sqrt{-41/3}}2$ so $j_3(\tau) = -(4\sqrt3)^6$. Let $\tau = \tfrac{1+\sqrt{-89/3}}2$ so $j_3(\tau) = -(10\sqrt3)^6$. Then $$\frac{(x + 1)^6(x^2 + x + 7)}x = -(4\sqrt3)^6\; \\ \frac{(x + 1)^6(x^2 + x + 7)}x =  -(10\sqrt3)^6$$ are octics both solvable in radicals. And also $$e^{\pi\sqrt{41/3}} =  (4\sqrt3)^6+41.993\dots\quad\\ e^{\pi\sqrt{89/3}} =  (10\sqrt3)^6+41.99997\dots$$ There are infinitely many $\tau$ but special ones such that $j_n(\tau)$ are integers can be found in Entry 145.

Entry 148

This is the 7th-deg overview though only results by Klein for the j-function $j = j_1(\tau)$ are known. In Klein's "On the Order-Seven Transformations of Elliptic Functions", he gave two elegant resolvents of degrees 7 and 8 in pages 306 and 313. Translated to more understandable notation, we have,

$$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$$

$$y^8+14y^6+63y^4+70y^2-7 = y\sqrt{j-1728}$$

If $\tau$ are complex quadratics such that $j= j_1(\tau)$ is a radical, then the two resolvents have a solvable Galois group, hence solvable in radicals

Example. Let $\tau = \tfrac{1+\sqrt{-163}}2$, then $j = -640320^3$ and $$x\left(x^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3  = -640320^3$$ is solvable in radicals. There are infinitely many such $\tau$ and some can be found in Entry 145. Note also that $$\frac{(y^4 + 14y^3 + 63y^2 + 70y - 7)^2}y + 1728 = \frac{(y^2 + 5y + 1)^3 (y^2 + 13y + 49)}y$$ where the octic on the RHS will appear in the next entry.

Entry 147

This is the sextic overview. Using the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145, if $\tau$ are complex quadratics such that $j_n(\tau)$ is a radical, then the following sextics have a solvable Galois group, hence solvable in radicals $$\begin{align}\qquad j_1 &=\frac{(x^2 + 10x + 5)^3}x\\ \color{red}{j_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\ \color{red}{{j_3}^2} &= \frac{j_3\,(2 x^6 + 29x^5 + 85x^4 + 50x^3) + 5^4 x^6}{(2x - 1)}\\ j_4 &=\frac{(x + 1)^5 (x + 5)}x \end{align}$$

with the ones in red by Joachim König. (It would be nice if the one for $j_3$ can be simplified.) They have discriminants,

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

Examples. Let $\tau = \tfrac{1+\sqrt{-163}}2$ so $j_1(\tau) = -640320^3$. Let $\tau = \tfrac{\sqrt{-58}}2$ so $j_2(\tau) = 396^4$. Then $$\frac{(x^2 + 10x + 5)^3}x = -640320^3\\ \frac{(x+1)^4(x^2+6x+25)}x=396^4$$ are sextics both solvable in radicals. There are infinitely many $\tau$ but special ones such that $j_n(\tau)$ are integers can be found in Entry 145.

Entry 146

This is the quintic overview of Entries 140-144. Using the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145, if $\tau$ are complex quadratics such that $j_n(\tau)$ is a radical, then the following simple quintics have a solvable Galois group hence solvable in radicals $$\begin{align}x^5 + 5x^4 + 40x^3 &= j_1\\ x(x - 5)^4 &= j_2\\ x^3(x - 5)^2 &= j_3\\ x^5+5x &= \sqrt{\frac{64}{\sqrt{j_4}}-\sqrt{j_4}}\\ x^5 + 5x^3 - 10x^2 &= j_6\end{align}$$ Example. Let $\tau = \tfrac{1+\sqrt{-163}}2$ so $j_1(\tau) = -640320^3$. Let $\tau = \tfrac{\sqrt{-58}}2$ so $j_2(\tau) = 396^4$. Then $$x^5 + 5x^4 + 40x^3 = -640320^3\\ x(x - 5)^4 = 396^4$$ are quintics both solvable in radicals. Of course, it is well-known that $$\qquad e^{\pi\sqrt{163}} = 640320^3+743.99999999999925\dots\\ e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ There are infinitely many $\tau$ but special ones such that $j_n(\tau)$ are integers can be found in Entry 145.

Entry 145

Summarizing the McKay-Thompson series of the Monster discussed in Entries 140-144, 

$$\begin{align}\quad j_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\ \quad j_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\ \quad j_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\ \quad j_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}\\ j_{6} &=\left( \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^3 + 2^3\left(\frac{\eta(2\tau)\,\eta(6\tau)}{\eta(\tau)\,\eta(3\tau)}\right)^3\right)^2  \end{align}$$

where $j_1(\tau)$ is the j-function. Let $\tau$ be complex quadratics $\tau = \tfrac12\sqrt{-d}$ or $\tau =\frac12+ \sqrt{-d}$ such that the $j_n(\tau)$ are radicals. For the following special $\tau$, then $j_n(\tau)$ are integers 

$$j_1(\tau)\; \text{where}\; \tau=\sqrt{-d}\;\text{for}\; d = 1, 2, 3, 4, 7,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =1, 3, 7, 11, 19, 127, 43, 67, 163.$$

$$j_2(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d}}2\;\text{for}\; d = 4, 6, 10, 18, 22, 58,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =5, 7, 9, 13, 25, 37.$$

$$j_3(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d/3}}2\;\text{for}\; d = 4,8,16,20,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d/3}}2\,\text{for}\; d =5, 9, 17, 25, 41, 49, 89.$$

$$j_4(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d}}2\;\text{for}\; d = 3, 7,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d}}2\,\text{for}\; d =1, 2, 4.$$

$$j_6(\tau)\; \text{where}\; \tau=\tfrac{\sqrt{-d/3}}2\;\text{for}\; d = 10, 14, 26, 34,\; \text{and}\, \tau =\tfrac{1+\sqrt{-d/3}}2\,\text{for}\; d = 7, 11, 19, 31, 59.$$

Entry 144

Define the McKay-Thompson series of Class 6A for the Monster $$j_6 = j_{6}(\tau) =\left( \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^3 + 2^3\left(\frac{\eta(2\tau)\,\eta(6\tau)}{\eta(\tau)\,\eta(3\tau)}\right)^3\right)^2$$ and the quintic $$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0$$ where $\small\alpha = -\dfrac1{j_6-32}.$ Alternatively $$(y^2+15)^2(y-5) = 32\big(j_6-32\big)$$ $$z^5 + 5z^3 - 10z^2 = j_6$$

Conjecture: "If $\tau$ is a complex quadratic such that $j_6=j_{6}(\tau)$ is an algebraic number with $j_6\neq 32$, then the quintics above have a solvable Galois group."

Example: Let $j_6\big(\tfrac{1\sqrt{-59/3}}2\big)=-1060^2$, then $$(y^2+15)^2(y-5) = 32\big({-1060^2}-32\big)$$ $$z^5 + 5z^3 - 10z^2=-1060^2$$ are solvable in radicals.

Entry 143

Define the McKay-Thompson series of Class 4A for the Monster $$j_4 = j_{4}(\tau) = \left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}$$ and the Bring-Jerrard quintic $$y^5+5y=\left(\frac{64}{\sqrt{j_4}}-\sqrt{j_4}\right)^{1/2}$$

Conjecture: "If $\tau$ is a complex quadratic such that $j_4=j_{4}(\tau)$ is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let $j_4\big(\tfrac12\sqrt{-7}\big)=2^{12}$, then $$y^5+5y=\sqrt{-63}$$ is solvable in radicals.

Entry 142

Define the McKay-Thompson series of Class 3A for the Monster $$j_3 = j_{3}(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2$$ and the Euler-Jerrard quintic $$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0$$ Alternatively $$y^3(y-5)^2 =j_3$$

Conjecture: "If $\tau$ is a complex quadratic such that $j_3=j_{3}(\tau)$ is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let $j_3\big(\tfrac{1+\sqrt{-89/3}}2\big)=-300^3$, then $$y^3(y-5)^2 = -300^3$$ is solvable in radicals.

Entry 141

Define the McKay-Thompson series of Class 2A for the Monster $$j_2 = j_{2}(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2$$ and the Bring-Jerrard quintic $$x^5-5\alpha x -\alpha=0$$ Alternatively $$y(y-5)^4 =j_2$$

Conjecture: "If $\tau$ is a complex quadratic such that $j_2=j_{2}(\tau)$ is an algebraic number, then the quintic above has a solvable Galois group."

Example: Let $j_2\big(\tfrac12\sqrt{-10}\big)=12^4$, then $$\quad y(y-5)^4=12^4\\ z^5-5z-12=0$$ are solvable in radicals.

Entry 140

Given the Dedekind eta function $\eta(\tau)$ and define the McKay-Thompson series of Class 1A for the Monster, or better known as the j-function $j$ $$j=j_1(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3$$ and the Brioschi quintic $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ where $\small\alpha = -\dfrac1{j-1728}$. Alternatively $$(y^2+20)^2(y-5) = j-1728$$ $$z^5 + 5z^4 + 40z^3=j$$ Conjecture: "If $\tau$ is a complex quadratic such that $j$ is an algebraic number $j\neq1728$, then the quintics above have a solvable Galois group." 

Example. Let $j\big(\tfrac{1+\sqrt{-163}}2\big) = -640320^3$ and $\alpha = \tfrac1{640320^3+1728}$, then $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ $$(y^2+20)^2(y-5) = -640320^3-1728$$ $$z^5 + 5z^4 + 40z^3=-640320^3$$ are quintics solvable in radicals. (In fact, they factor into a quadratic and a cubic.)

Entry 139

The general quintic can be reduced to the following one-parameter forms

$$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0\tag1$$

$$x^5-5\alpha x -\alpha = 0\tag2$$

$$x^5+5\sqrt{\alpha}\, x^2 -\sqrt{\alpha} = 0\tag3$$

$$\; x^5+5x+\left(\frac1{\sqrt{\alpha}}-64\sqrt{\alpha}\right)^{1/2} = 0\tag4$$

$$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0\tag5$$ with the last found by yours truly. They have neat discriminants

$$\begin{align}D_1 &= 5^5\,(1-1728\alpha)^2\,\alpha^8\\ D_2 &= 5^5\,(1-256\alpha)\,\alpha^4\\ D_3 &= 5^5\,(1-108\alpha)\,\alpha^2\\ D_4 &= 5^5\,(1+64\alpha)^2\,\alpha^{-1}\\ D_5 &= 5^5\,(1-36\alpha)(1-32\alpha)\,\alpha^8\end{align}$$ The integers $(1728, 256, 108, 64)$ appear in Ramanujan's theory of elliptic functions to alternative bases and we will connect these quintics to the McKay-Thompson series of class 1A, 2A, 3A, 4A, 6A for the Monster in subsequent entries. 

Entry 138

This summarizes the last several entries. Let fundamental discriminant $d = 4m$ with class number $h(-d)=2^k$ and even $m = 2p$ for prime $p$. Previously, $p \equiv 1\,\text{mod}\,4$ and $p \equiv 3\,\text{mod}\,4$ were distinguished, but we can have a more unified approach. Given the modular lambda function $\lambda(\tau)$ and define the three simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2\\ \gamma(n) &= \Big(n+\sqrt{n^2+1}\Big)^2 \end{align}$$ If $p \equiv 3\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n),\quad \text{where}\, n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ If $p \equiv 1\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\gamma(n),\quad \text{where}\, n = \frac{\lambda+1}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with $\lambda=\lambda(\tau)$ for simplicity and $\alpha(n),\,\beta(n),\,\gamma(n)$ are units. Examples. Let $m = 2p$ with class number $4$. 

For $p = 7,23$ $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n)\,\beta(n),\quad n=2(1+\sqrt2)\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n)\,\beta(n),\quad n=2(13+9\sqrt2) \end{align}$$ For $p=17,41$ $$\begin{align}\frac1{\lambda(\sqrt{-34})} &= \alpha(n)\,\gamma(n),\quad n=3(4+\sqrt{17})\\ \frac1{\lambda(\sqrt{-82})} &= \alpha(n)\,\gamma(n),\quad n=3(51+8\sqrt{41})\end{align}$$ One can observe that $n$ is an algebraic number of degree half that of the class number. For $m = 2p$ with class number $8$, examples for $p \equiv 3\,\text{mod}\,4$ were given in Entry 136. For $p \equiv 1\,\text{mod}\,4$ like $p=89$, then $$n^4 - 8886 n^3 + 648 n^2 - 10314 n + 5751=0$$ and so on for other $p$.

Entry 137

This continues Entry 136. Recall the function $$\beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ and the examples of $n$ which were quartic roots. It turns out these $n$ have additional properties which yield fundamental units $U_k$ though I don't know why.

For $p=31$, let $n_{\color{red}\pm} = 2(1+\sqrt2)^2\Big(1+3\sqrt2\color{red}{\pm}2\sqrt{1+4\sqrt2}\Big)$ or the two real roots of the quartic. Then $$\frac{\beta(n_{+})}{\beta(n_{-})} = U_{31}\sqrt{U_{62}} = (1520+273\sqrt{31})\,(4\sqrt2+\sqrt{31})$$ For $p=47$, let $n_{\color{red}\pm} = 2(1+\sqrt2)^3\Big(9\color{red}{\pm}2\sqrt{9+8\sqrt2}\Big)$ or again the two real roots. Then $$\frac{\beta(n_{+})}{\beta(n_{-})} = U_{47}\sqrt{U_{94}} = (48+7\sqrt{47})\,(732\sqrt2+151\sqrt{47})$$ and so on for $p = 31, 47, 79, 191, 239, 431$.

Entry 136

Given fundamental discriminants $d = 4m$ with class number $h(-d)=8$. Then there are exactly ten $m = 2p$ for prime $p$, namely $p \equiv 1\,\text{mod}\,4 = 89, 113, 233, 281$ and $p \equiv 3\,\text{mod}\,4 = 31, 47, 79, 191, 239, 431$. The modular lambda function $\lambda(\sqrt{-2p})$ for both is a root of a deg-$16$ equation, but the latter is easier to factor into two deg-$8$ equations. Define the two simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$ It seems for the $p \equiv 3\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n)$$ where $\alpha(n),\beta(n)$ are octic units and $n$ is just a quartic root given by $$n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with $\lambda=\lambda(\tau)$ for simplicity. Examples,

Let $p=31$ and $n= 2(1+\sqrt2)^2\Big(1+3\sqrt2+2\sqrt{1+4\sqrt2}\Big)$ then $$\frac1{\lambda(\sqrt{-62})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ Let $p=47$ and $n= 2(1+\sqrt2)^3\Big(9+2\sqrt{9+8\sqrt2}\Big)$ then $$\frac1{\lambda(\sqrt{-94})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ and so on for $p = 31, 47, 79, 191, 239, 431$. P.S. Note that solutions to the Pell-like equation $x^2-2y^2 = -p$ appear in the nested radicals $\sqrt{x+y\sqrt2\,}$ above, namely $$1^2-2\cdot4^2=-31\\ 9^2-2\cdot8^2=-47$$

Entry 135

Mathworld has a list of the modular lambda function $\lambda(\tau)$ with the particular case $\tau=\sqrt{-14}$ as the rather complicated $$\sqrt{\lambda(\sqrt{-14})}=-11-8\sqrt2-2(2+\sqrt2)\sqrt{5+4\sqrt2}\qquad \\\ \qquad+\sqrt{11+8\sqrt2}\,\Big(2+2\sqrt2+\sqrt2\sqrt{5+4\sqrt2}\Big)\qquad$$ which is approximately $0.011208.$ It can be calculated in Mathematica or WolframAlpha as ModularLambda[tau]. However, we can simplify and factor that into two quartic units as $$\begin{align}\frac1{\lambda(\sqrt{-14})} & =\frac1{2^8}(8+3\sqrt7)\,(\sqrt7+\sqrt8)\,\Big(2^{1/4}+\sqrt{4+\sqrt2}\Big)^8\\ &=7960.423255\dots\end{align}$$ It is then just a matter of getting the reciprocal and square roots. One can do so similarly for discriminants $d=4m$ with class number $h(-d)=4$ and even $m=14,62,142$ as described in Entry 134.

Entry 134

Given fundamental discriminants $d = 4m$ with class number $h(-d)=4$, there are exactly five $m = 2p$ for prime $p$, namely $p=7,17,23,41,71$. The cases $p = 17, 41$ were in Entry 115 while $p = 7,23,71$ which are $p \equiv 3\,\text{mod}\,4$ will be tackled here. Define the two simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$ and let $$\begin{align}n_1 &=2+2\sqrt2 \\ n_2 &= 26+18\sqrt2 \\ n_3 &=1450+1026\sqrt2\end{align}$$ Then the first quartic unit can be an eighth power 

$$\begin{align}\alpha(n_1) &=\frac1{2^8}\Big(\sqrt{0+\sqrt2}+\sqrt{4+\sqrt2}\Big)^8\\ \alpha(n_2)  &=\frac1{2^8}\Big(\sqrt{4+3\sqrt2}+\sqrt{8+3\sqrt2}\Big)^8\\ \alpha(n_3)  &=\frac1{2^8}\Big(\sqrt{36+27\sqrt2}+\sqrt{40+27\sqrt2}\Big)^8 \end{align}$$ while the second is a product of quadratic units 

$$\begin{align}\beta(n_1) &= U_{7}\,\sqrt{U_{14}}=\big(8+3\sqrt{7}\big) \big(2\sqrt2+\sqrt{7}\big) \\ \beta(n_2)  &= U_{23}\,\sqrt{U_{46}}=\big(25+5\sqrt{23}\big) \big(78\sqrt2+23\sqrt{23}\big) \\ \beta(n_3)  &= U_{71}\,\sqrt{U_{142}}=\big(3480+413\sqrt{71}\big) \big(1710\sqrt2+287\sqrt{71}\big) \\ \end{align}$$ which gives the radical expressions of  $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n_1)\,\beta(n_1),\quad n_1=2+2\sqrt2\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n_2)\,\beta(n_2),\quad n_2=26+18\sqrt2 \\ \frac1{\lambda(\sqrt{-142})} &= \alpha(n_3)\,\beta(n_3),\quad n_3=1450+1026\sqrt2 \end{align}$$ And since $\beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$, then they are also nested radicals that use only $\sqrt2$. 

Entry 133

Given fundamental discriminants $d=4m$ with class number $h(-d)=2$, there are exactly four even $m = 6, 10, 22, 58$. The most well-known is $m=58$ because of the near-integer $$\quad e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ and the appearance of $396^4$ in the denominator of Ramanujan's famous $1/\pi$ formula. These $m=2p$ for prime $p$ have other interesting properties. Recall the modular lambda function $\lambda(\tau)$ also discussed in Entry 112 $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We focus on $m=2p$ for prime $p = 3\,\text{mod}\,4$ hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)(\sqrt2+\sqrt3)\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})(7\sqrt2+3\sqrt{11}) \end{align}$$ with fundamental units $U_n$. However, special $d$ with class number $h(-d)=2^k$ surprisingly can be expressed by nested radicals using only the square root of 2. So, $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= (1+\sqrt2)^2+\sqrt{1+ (1+\sqrt2)^4}\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &=  (1+\sqrt2)^6+\sqrt{1+ (1+\sqrt2)^{12}} \end{align}\quad$$ Similar behavior can also be observed for $2p$ for $p=7,23,71$ which now have class number 4.

Entry 132

Given $_2F_1(a,b;c;z)$ and Dedekind eta function $\eta(\tau)$ where $\tau = \frac{1+n\sqrt{-3}}2$ for positive integer $n$. Then for type $a+b=c=\color{blue}{\tfrac56}$ $$\begin{align}&\,_2F_1\big(\tfrac12,\tfrac13;\tfrac56;(1-2\delta_1)^2\big),\quad\;\delta_1 = \delta_2\\ &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\sqrt{\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}}\\ &\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1\,=\,\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both $(z_1, z_2)$ in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when $n$ is a positive integer. Note that $_2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big)$ so the first form is superfluous. Examples: Let $\tau = \frac{1+5\sqrt{-3}}2$, $$_2F_1\Big(\frac13,\frac12;\frac56;\frac45\Big)=\;\frac35\sqrt5$$ $$\quad _2F_1\Big(\frac16,\frac23;\frac56;\frac{80}{81}\Big)=\frac35\,(9\sqrt5)^{1/3}$$

Entry 131

Given $_2F_1(a,b;c;z)$ and Dedekind eta function $\eta(\tau)$ where $\tau = \frac{1+n\sqrt{-1}}2$ for positive integer $n$. Then for type $a+b=c=\color{blue}{\tfrac34}$ $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\sqrt{-\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}}\\ &\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\ &\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1\,=\, -\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}\\ &\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both $(z_1, z_2)$ in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when $n$ is a positive integer. Examples: Let $\tau = \frac{1+5\sqrt{-1}}2$, 

$$_2F_1\Big(\frac14,\frac12;\frac34;\frac{80}{81}\Big)=\frac95$$

$$\quad _2F_1\Big(\frac18,\frac58;\frac34;\frac{25920}{25921}\Big)=\frac35\,161^{1/4}$$

Entry 130

Given $_2F_1(a,b;c;z)$ and j-function $j = j(\tau)$ where $\tau = \frac{1+n\sqrt{-3}}2$ for positive integer $n$. Then for type $a+b=c=\color{blue}{\tfrac23}$ $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;(1-2\beta_1)^2\big),\qquad \color{red}{\beta_1 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;(1-2\beta_2)^2\big),\qquad \frac{1}{\beta_2}-1=\sqrt{\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}}\\ &\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;(1-2\beta_3)^2\big),\qquad \color{red}{\beta_3 =\,?} \\ &\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;(1-2\beta_4)^2\big),\qquad \frac{1}{\beta_4}-1=\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}\\\end{align}$$ For this type, there are infinitely many hypergeometrics such that both $(z_1, z_2)$ in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when $n$ is a positive integer. Examples: Let $\tau = \frac{1+3\sqrt{-3}}2$, $$_2F_1\Big(\frac16,\frac12;\frac23;\frac{125}{128}\Big) =\frac43\times2^{1/6}$$ $$_2F_1\Big(\frac1{12},\frac7{12};\frac23;\frac{64000}{64009}\Big) =\frac23\times253^{1/6}$$ Let $\tau = \frac{1+5\sqrt{-3}}2$, $$_2F_1\left(\frac16,\frac12;\frac23;\Big(\frac45\Big)^2\Big(\frac{15-\sqrt5}{11}\Big)^3\right) =\frac35(5+4\sqrt5)^{1/6}$$

Entry 129

Given $_2F_1(a,b;c;z)$ where $a+b=c$. The previous four (Entries 125-128) discuss closed-forms and can be neatly summarized as type $a+b=c=\color{blue}{\tfrac12}$  $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\qquad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\qquad \tau_1=n\sqrt{-4}\\ &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\qquad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\qquad \tau_2=n\sqrt{-3}\\ &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\qquad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\qquad \tau_3=n\sqrt{-2}\\ &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad\; \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\qquad\tau_4=n\sqrt{-1}\end{align}$$ For this type, there are infinitely many hypergeometrics such that both $(z_1, z_2)$ in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when $n$ is a positive integer. However, for the parameters $c=\frac12, \frac23,\frac34. \frac56$, it seems only the first is easy. For type $a+b=c=\color{blue}{\tfrac23}$, it's more challenging and will be discussed in the next entry.

Entry 128

Assume $\tau=n\sqrt{-\color{blue}{4}}$ for some positive integer $n$. Given $_2F_1(a,b;c;z)$ where $a+b=c=\frac12$ for the case $a=\tfrac1{4}$. Let $z_1 = (1-2w)^2$ where $w$ is $$w=\frac{16}{16+\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8}$$ Then $(z_1, z_2)$ are algebraic numbers in

$$_2F_1\left(\tfrac14,\tfrac14;\tfrac12;z_1\right) = z_2$$

Examples: 

If $n=2$ so $\tau=2\sqrt{-4}$, then,

$$_2F_1\left(\frac14,\frac14;\frac12;\,\frac{9\,(3+\sqrt2)^2}{(1+\sqrt2)^6}\right)=\frac38\big(2+\sqrt2\big)$$  

If $n=3$ so $\tau=3\sqrt{-4}$, then,

$$\quad _2F_1\left(\frac14,\frac14;\frac12;\,\frac{8\sqrt3}{(2+\sqrt3)^2}\right)=\frac23\big(3+2\sqrt3\big)^{1/2}$$