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

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