Entry 81

Level 10. Define $d_k = \eta(k\tau)$ with Dedekind eta function $\eta(k\tau)$ and the McKay-Thompson series of class 10D for the Monster (A132130). $$j_{10D}(\tau) = \left(\frac{d_2\,d_5}{d_1\,d_{10}}\right)^6$$

Examples. We select $d=20m$ with class number $h(-d)=4$ and find odd $m=17$ as well as $$m=6, 14, 26, 38$$ such that the following are special quadratic irrationals $$\begin{align}j_{10D}\Big(\tfrac{1+\sqrt{-17/5}}{2}\Big) &= -U_{5}^{12} = -\left(\tfrac{1+\sqrt5}2\right)^{12}\\ j_{10D}\big(\tfrac12\sqrt{-6/5}\big) &=\, U_{10}^{2}\, = \left(3+\sqrt{10}\right)^{2}\\ j_{10D}\big(\tfrac12\sqrt{-14/5}\big) &=\, U_{2}^{6}\, = \left(1+\sqrt2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-26/5}\big) &=\, U_{13}^{6}\, = \left(\tfrac{3+\sqrt{13}}2\right)^{6}\\ j_{10D}\big(\tfrac12\sqrt{-38/5}\big) &=\, U_{5}^{18}\, = \left(\tfrac{1+\sqrt5}2\right)^{18} \end{align}$$

as they are fundamental units $U_n$. Since $U_5=\phi$ is the golden ratio, then the first and last implies the integers $$\left(\sqrt{-\phi^{12}}-1/\sqrt{-\phi^{12}}\right)^2 = -18^2\\ \left(\sqrt{\phi^{18}}-1/\sqrt{\phi^{18}}\right)^2 = 76^2$$ and similarly for the others as discussed in the previous entry.