Entry 136

Given fundamental discriminants $d = 4m$ with class number $h(-d)=8$. Then there are exactly ten $m = 2p$ for prime $p$, namely $p \equiv 1\,\text{mod}\,4 = 89, 113, 233, 281$ and $p \equiv 3\,\text{mod}\,4 = 31, 47, 79, 191, 239, 431$. The modular lambda function $\lambda(\sqrt{-2p})$ for both is a root of a deg-$16$ equation, but the latter is easier to factor into two deg-$8$ equations. Define the two simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$ It seems for the $p \equiv 3\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n)$$ where $\alpha(n),\beta(n)$ are octic units and $n$ is just a quartic root given by $$n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with $\lambda=\lambda(\tau)$ for simplicity. Examples,

Let $p=31$ and $n= 2(1+\sqrt2)^2\Big(1+3\sqrt2+2\sqrt{1+4\sqrt2}\Big)$ then $$\frac1{\lambda(\sqrt{-62})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ Let $p=47$ and $n= 2(1+\sqrt2)^3\Big(9+2\sqrt{9+8\sqrt2}\Big)$ then $$\frac1{\lambda(\sqrt{-94})} = \alpha(n)\,\beta(n) = \Big(n+\sqrt{n^2-1}\Big)^2 \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$$ and so on for $p = 31, 47, 79, 191, 239, 431$. P.S. Note that solutions to the Pell-like equation $x^2-2y^2 = -p$ appear in the nested radicals $\sqrt{x+y\sqrt2\,}$ above, namely $$1^2-2\cdot4^2=-31\\ 9^2-2\cdot8^2=-47$$