Entry 106

Let $q = e^{2\pi i \tau}$. In Entry 99, we gave the cubic continued fraction

$$\begin{align}\text{K}_6(q)  &= \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}\\ & = \cfrac{q^{1/3}}{1+\cfrac{q+q^2}{1+\cfrac{q^2+q^4}{1+\cfrac{q^3+q^6}{1+\ddots}}}}\end{align}$$ Define the McKay-Thompson series of Class 6E for Monster (A128633) $$\begin{align}j_{6E}(\tau) &=\frac1{\big(\text{K}_6(q)\big)^3}+1 \\ &= \left(\frac{\eta(2\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(6\tau)}\right)^3 + 1 = \left(\frac{\eta^2(2\tau)\,\eta(3\tau)}{\eta(\tau)\,\eta^2(6\tau)}\right)^4\end{align}$$ While it is possible to evaluate $K_6(q)$, doing it for $j_{6E}(\tau)$ seems more "easy" since it is a $4$th power. Let $d=3m$ with class number $h(-d)=2$ for $m=17,41,89$. Then for $$\tau_1=\tfrac{1+\sqrt{-17/3}}2,\quad \tau_2=\tfrac{1+\sqrt{-41/3}}2,\quad \tau_3=\tfrac{1+\sqrt{-89/3}}2$$ we get the evaluations $$\begin{align} -\sqrt{\frac{-3\;}{j_{6E}(\tau_1)}}+\sqrt{-\frac13\, j_{6E}(\tau_1)} &= 2^3+2(-2+2\sqrt{17})^{2/3}+2(2+2\sqrt{17})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_2)}}+\sqrt{-\frac13\, j_{6E}(\tau_2)} &= 4^3+4(-2+10\sqrt{41})^{2/3}+4(2+10\sqrt{41})^{2/3}\\ -\sqrt{\frac{-3\;}{j_{6E}(\tau_3)}}+\sqrt{-\frac13\, j_{6E}(\tau_3)} &= 10^3+10(-2+106\sqrt{89})^{2/3}+10(2+106\sqrt{89})^{2/3}\end{align}$$ where the RHS are roots of cubics and all quadratic irrationals, for example $72+8\sqrt{17} = (2+2\sqrt{17})^2$, are squares.