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V. Level 5. The McKay-Thompson series of class 5B for the Monster.

$$j_{5B}(\tau) =\left(\frac{\eta(\tau)}{\eta{(5\tau)}}\right)^{6}$$

Examples. Surprisingly, the following "cubes" are powers of the golden ratio \(\phi=\frac{1+\sqrt5}2\)  $$\begin{align}j_{5B}\Big(\tfrac{1+\sqrt{-7/5}}{2}\Big) &= -(\sqrt5)^3\,\phi^3\\ j_{5B}\Big(\tfrac{1+\sqrt{-23/5}}{2}\Big) &=  -(\sqrt5)^3\,\phi^9\\ j_{5B}\Big(\tfrac{1+\sqrt{-47/5}}{2}\Big) &=  -(\sqrt5)^3\,\phi^{15}\end{align}$$ These discriminants \(d=5m\) have class number \(h(-d)=2\).

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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.

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IV. Level 4. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\). Then the McKay-Thompson series of class 4C for the Monster (A007248) $$j_{4C}(\tau) = \left(\frac{d_1}{d_4}\right)^8$$ Examples. $$\begin{align}j_{4C}(\tfrac12\sqrt{-3}) &= 2^4\,U_3^2 = 2^4(2+\sqrt3)^2\\ j_{4C}(\tfrac12\sqrt{-7}) &= 2^4\,U_7^2 = 2^4(8+3\sqrt7)^2\end{align}$$ where \(U_n\) are fundamental units. Since $$\left(\frac{d_1}{d_4}\right)^8+16 = \left(\frac{d_2^3}{d_1\,d_4^2}\right)^8$$ where the RHS is related to the modular lambda function, then we can use \( \left(\frac{d_1}{d_4}\right)^8\) for the case \(\tau = \tfrac12\sqrt{-n}\) to solve $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)} = \sqrt{n}$$ where \(x=\dfrac{16}{\left(\frac{d_1}{d_4}\right)^8+16}\). For example, $$\frac{_2F_1\big(\tfrac12,\tfrac12,1,1-x\big)}{_2F_1\big(\tfrac12,\tfrac12,1,x\big)} = \sqrt{3}$$ has solution $$x=\frac{16}{16(2+\sqrt3)^2+16}$$

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 IV. Level 4. The McKay-Thompson series of class 4A for the Monster (A097340)

$$\begin{align}j_{4A}(\tau) &= \left(\left(\frac{d_1}{d_4}\right)^4+4^2\left(\frac{d_4}{d_1}\right)^4\right)^2\\ &=\left(\frac{d_2^2}{d_1\,d_4}\right)^{24}\end{align}$$ The second form shows they can be \(12\)th powers. Examples: 

$$j_{4A}\big(\tfrac12\sqrt{-7}\big)=2^{12}$$

which has class number 1 (but non-fundamental \(d\)). For class number 2,

$$\begin{align}j_{4A}\Big(\tfrac{1+\sqrt{-6}}{2}\Big) &= -2^6\left(1+\sqrt2\right)^{4}\\ j_{4A}\Big(\tfrac{1+\sqrt{-10}}{2}\Big) &= -2^6\left(\frac{1+\sqrt5}2\right)^{12}\\ j_{4A}\Big(\tfrac{1+\sqrt{-22}}{2}\Big) &= -2^6\left(1+\sqrt2\right)^{12}\\ j_{4A}\Big(\tfrac{1+\sqrt{-58}}{2}\Big) &= -2^6\left(\frac{5+\sqrt{29}}2\right)^{12}\end{align}$$

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III. Level 3. The McKay-Thompson series of class 3B for the Monster.

$$j_{3B}(\tau) =\left(\frac{d_1}{d_3}\right)^{12}$$ Examples: $$\begin{align}j_{3B}\Big(\tfrac{1+\sqrt{-5/3}}{2}\Big) &= -3^3U_{5}^2 =-3^3\left(\tfrac{1+\sqrt{5}}2\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-17/3}}{2}\Big) &= -3^3U_{17}^2 =-3^3\left(4+\sqrt{17}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-41/3}}{2}\Big) &=-3^3U_{41}^2 =-3^3\left(32+5\sqrt{41}\right)^2\\ j_{3B}\Big(\tfrac{1+\sqrt{-89/3}}{2}\Big) &= -3^3U_{89}^2=-3^3\left(500+53\sqrt{89}\right)^2\end{align}$$ These \(d=3m\) have class number \(h(-d)=2\). The quadratic irrationals have already appeared in Level 1 and are fundamental units \(U_n\), solutions to Pell equations \(x^2-ny^2=\pm1\). See also class 7B.