Entry 74

In the previous entries, Levels \(1, 2, 3, 4\) were discussed. Level \(5\) is intimately connected to the famous Rogers-Ramanujan continued fraction \(R(q)\). Let \(q = e^{2\pi i \tau}\) then 

$$x=\frac1{R(q)}-R(q) =  \frac{\eta(\tau/5)}{\eta(5\tau)}+1\\ \quad y=\frac1{R^5(q)}-R^5(q) = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11$$ Eliminating \(R(q)\) between the two and we get the relationship between \((x,y)\) as the solvable DeMoivre quintic,

$$x^5+5x^3+5x=y$$ which is solvable in radicals for any \(y\). As usual, define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and $$j_{5A}(\tau) = \left(\frac{d_1}{d_5}\right)^6+5^3\left(\frac{d_5}{d_1}\right)^6+22$$ Examples

$$\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-7/5}}{2}\Big) &= -(2\sqrt7)^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &= -(6\sqrt{23})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &= -(18\sqrt{47})^2\\ \end{align}$$ Compare to a similar phenomenon for Level 5. These have discriminant \(d=5p\) for prime \(p=7,23,47\) and have class number 2. For class number 6,

$$\begin{align}j_{5A}\Big(\tfrac{1+\sqrt{-103/5}}{2}\Big) &= -(x\sqrt{103})^2\\ j_{5A}\Big(\tfrac{1+\sqrt{-167/5}}{2}\Big) &= -(y\sqrt{167})^2\\ \end{align}$$

where \((x,y)\) are the real roots of the cubics

$$103 x^3 + 12566 x^2 - 12316 x + 15272 = 0\\ 167 y^3 + 113226 y^2 + 6372 y + 216= 0$$ and so on.