Saturday, June 14, 2025

Entry 171

Continuing from Entry 170, there is another way to express the Ramanujan \(G_n\)-function where \(d=4n\) for prime \(n\) has class number \(6\) by using fundamental units \(U_n\). Borrowing a trick from Ramanujan, he found

$$G_{169}=\frac13\Big(2+\sqrt{13}+\sqrt[3]{U_{13}(v+3\sqrt{3})\sqrt{13}}+\sqrt[3]{U_{13}(v-3\sqrt{3})\sqrt{13}}\Big)$$

where $$U_{13} = \frac{3+\sqrt{13}}2, \quad v = \frac{11+\sqrt{13}}2$$

Using a similar form, we propose that

$$\begin{align}G_{29} &=\frac1{3^{1/4}}\Big(\tfrac{9+\sqrt{29}}2+\sqrt[3]{U_{29}(x+24\sqrt{3})}+\sqrt[3]{U_{29}(x-24\sqrt{3})}\Big)^{1/4}\\ G_{53} &=\frac1{3^{1/4}}\Big(\tfrac{23+3\sqrt{53}}2+\sqrt[3]{U_{53}(y+120\sqrt{3})}+\sqrt[3]{U_{53}(y-120\sqrt{3})}\Big)^{1/4}\\ G_{61} &=\frac1{3^{1/4}}\Big(15+2\sqrt{61}+\sqrt[3]{U_{61}(z+72\sqrt{3})}+\sqrt[3]{U_{61}(z-72\sqrt{3})}\Big)^{1/4}\end{align}$$ where$$U_{29} = \frac{5+\sqrt{29}}2, \quad x = \frac{185+19\sqrt{29}}2$$ $$\quad U_{53} = \frac{7+\sqrt{53}}2, \quad y = \frac{1721+217\sqrt{53}}2$$ $$U_{61} = \frac{39+5\sqrt{61}}2, \quad z = \frac{601+93\sqrt{61}}2\;$$

and similarly for all seven prime \(p = 29, 53, 61, 109, 157, 277, 397\). Note they have form \(p \equiv 5\,\text{mod}\,8\). And all their fundamental units have form \(U_p=\frac{a+b\sqrt{p}}2\), hence have odd solutions to the Pell equation \(x^2-py^2=-4\).

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