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.