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

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