Entry 133

Given fundamental discriminants $d=4m$ with class number $h(-d)=2$, there are exactly four even $m = 6, 10, 22, 58$. The most well-known is $m=58$ because of the near-integer

$$\quad e^{\pi\sqrt{58}} = 396^4-104.00000017\dots$$ and the appearance of $396^4$ in the denominator of Ramanujan's famous $1/\pi$ formula. These $m=2p$ for prime $p$ have other interesting properties. Recall the modular lambda function $\lambda(\tau)$ also discussed in Entry 112 $$\lambda(\tau) = \left(\frac{\sqrt2\,\eta\big(\tfrac12\tau\big)\,\eta^2\big(2\tau\big)}{\eta^3(\tau)}\right)^8$$ We focus on $m=2p$ for prime $p = 3\,\text{mod}\,4$ hence $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= U_3\sqrt{U_6} = (2+\sqrt3)(\sqrt2+\sqrt3)\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &= U_{11}\sqrt{U_{22}}= (10+3\sqrt{11})(7\sqrt2+3\sqrt{11}) \end{align}$$ with fundamental units $U_n$. However, special $d$ with class number $h(-d)=2^k$ surprisingly can be expressed by nested radicals using only the square root of 2. So, $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-6})}} &= (1+\sqrt2)^2+\sqrt{1+ (1+\sqrt2)^4}\\ \frac1{\sqrt{\lambda(\sqrt{-22})}} &=  (1+\sqrt2)^6+\sqrt{1+ (1+\sqrt2)^{12}} \end{align}\quad$$ Similar behavior can also be observed for $2p$ for $p=7,23,71$ which now have class number 4.