Entry 134

Given fundamental discriminants $d = 4m$ with class number $h(-d)=4$, there are exactly five $m = 2p$ for prime $p$, namely $p=7,17,23,41,71$. The cases $p = 17, 41$ were in Entry 115 while $p = 7,23,71$ which are $p \equiv 3\,\text{mod}\,4$ will be tackled here. Define the two simple functions $$\begin{align}\alpha(n) &= \Big(n+\sqrt{n^2-1}\Big)^2\\ \beta(n) &= \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2 \end{align}$$ and let $$\begin{align}n_1 &=2+2\sqrt2 \\ n_2 &= 26+18\sqrt2 \\ n_3 &=1450+1026\sqrt2\end{align}$$ Then the first quartic unit can be an eighth power 

$$\begin{align}\alpha(n_1) &=\frac1{2^8}\Big(\sqrt{0+\sqrt2}+\sqrt{4+\sqrt2}\Big)^8\\ \alpha(n_2)  &=\frac1{2^8}\Big(\sqrt{4+3\sqrt2}+\sqrt{8+3\sqrt2}\Big)^8\\ \alpha(n_3)  &=\frac1{2^8}\Big(\sqrt{36+27\sqrt2}+\sqrt{40+27\sqrt2}\Big)^8 \end{align}$$ while the second is a product of quadratic units 

$$\begin{align}\beta(n_1) &= U_{7}\,\sqrt{U_{14}}=\big(8+3\sqrt{7}\big) \big(2\sqrt2+\sqrt{7}\big) \\ \beta(n_2)  &= U_{23}\,\sqrt{U_{46}}=\big(25+5\sqrt{23}\big) \big(78\sqrt2+23\sqrt{23}\big) \\ \beta(n_3)  &= U_{71}\,\sqrt{U_{142}}=\big(3480+413\sqrt{71}\big) \big(1710\sqrt2+287\sqrt{71}\big) \\ \end{align}$$ which gives the radical expressions of  $$\begin{align}\frac1{\lambda(\sqrt{-14})} &= \alpha(n_1)\,\beta(n_1),\quad n_1=2+2\sqrt2\\ \frac1{\lambda(\sqrt{-46})} &= \alpha(n_2)\,\beta(n_2),\quad n_2=26+18\sqrt2 \\ \frac1{\lambda(\sqrt{-142})} &= \alpha(n_3)\,\beta(n_3),\quad n_3=1450+1026\sqrt2 \end{align}$$ And since $\beta(n) = \Big(\sqrt{n^2-2}+\sqrt{n^2-1}\Big)^2$, then they are also nested radicals that use only $\sqrt2$.