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$$1^2-2\cdot4^2=-31\\ 9^2-2\cdot8^2=-47$$