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 \(G_{127}\).