Entry 73

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 $$x=\frac{16}{16(2+\sqrt3)^2+16}$$