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.