Entry 138

This summarizes the last several entries. Let fundamental discriminant $d = 4m$ with class number $h(-d)=2^k$ and even $m = 2p$ for prime $p$. Previously, $p \equiv 1\,\text{mod}\,4$ and $p \equiv 3\,\text{mod}\,4$ were distinguished, but we can have a more unified approach. Given the modular lambda function $\lambda(\tau)$ and define the three simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2\\ \gamma(n) &= \Big(n+\sqrt{n^2+1}\Big)^2 \end{align}$$ If $p \equiv 3\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\beta(n),\quad \text{where}\, n = \frac{2\sqrt{\lambda}+1-\lambda}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ If $p \equiv 1\,\text{mod}\,4$, then $$\frac1{\lambda(\sqrt{-2p})} = \alpha(n)\,\gamma(n),\quad \text{where}\, n = \frac{\lambda+1}{2\lambda^{1/4}\sqrt{1-\lambda}}$$ with $\lambda=\lambda(\tau)$ for simplicity and $\alpha(n),\,\beta(n),\,\gamma(n)$ are units. Examples. Let $m = 2p$ with class number $4$. 

For $p = 7,23$ $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n)\,\beta(n),\quad n=2(1+\sqrt2)\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n)\,\beta(n),\quad n=2(13+9\sqrt2) \end{align}$$ For $p=17,41$ $$\begin{align}\frac1{\lambda(\sqrt{-34})} &= \alpha(n)\,\gamma(n),\quad n=3(4+\sqrt{17})\\ \frac1{\lambda(\sqrt{-82})} &= \alpha(n)\,\gamma(n),\quad n=3(51+8\sqrt{41})\end{align}$$ One can observe that $n$ is an algebraic number of degree half that of the class number. For $m = 2p$ with class number $8$, examples for $p \equiv 3\,\text{mod}\,4$ were given in Entry 136. For $p \equiv 1\,\text{mod}\,4$ like $p=89$, then $$n^4 - 8886 n^3 + 648 n^2 - 10314 n + 5751=0$$ and so on for other $p$.