Entry 115

For fundamental discriminants $d=4m$ with class number $h(-d)=4$, there are exactly twelve $m$ that are even. In Entry 113, the seven with 8 divisors were discussed. The remaining five are $m = 14, 34, 46, 82, 142$ which are of form $m=2p$ for prime $p=7, 17, 23, 41, 71$. From experience, $p\equiv 1\, (\text{mod}\, 4)$ are more well-behaved, hence for $m=34,82$ $$\begin{align}\frac1{\sqrt{\lambda(\sqrt{-34})}} &= (1+\sqrt2)^2\sqrt{35+6\sqrt{34}} \left(\sqrt{\frac{5+\sqrt{17}}4}+\sqrt{\frac{1+\sqrt{17}}4}\right)^4 \\ \frac1{\sqrt{\lambda(\sqrt{-82})}} &\,=\, (1+\sqrt2)^4\, (9+\sqrt{82})\,\left(\sqrt{\frac{7+\sqrt{41}}2}+\sqrt{\frac{5+\sqrt{41}}2}\right)^4 \end{align}\quad$$ For $m = 14, 46,142$, presumably they may be products of two quartic units $$\frac1{\sqrt{\lambda(\sqrt{-m})}} = \big(a+\sqrt{a^2\pm1}\big)\big(b+\sqrt{b^2\pm1}\big)$$ where $(a,b)$ are roots of quadratics, but I haven't figured out the correct values yet.