Entry 79

Level 7. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 7B for the Monster $$j_{7B}(\tau) = \left(\frac{d_1}{d_7}\right)^4$$ Examples. We select \(d=7m\) with class number \(h(-d)=2\) and find \(m=5,13,61\) such that the following are special quadratic irrationals $$\begin{align}j_{7B}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\,U_{5}^{2} = -7\left(\tfrac{1+\sqrt5}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\,U_{13}^{2} = -7\left(\tfrac{3+\sqrt{13}}2\right)^{2}\\ j_{7B}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\,U_{61}^{2} = -7\left(\tfrac{39+5\sqrt{61}}2\right)^{2}\end{align}$$

as they involve fundamental units \(U_n\). These are analogous to the examples in Level 3B which have the form \(-3^3\,U_n^2\). The last implies the integer $$\left(\sqrt{-7\,U_{61}}+7/\sqrt{-7\,U_{61}}\right)^2 = -7\times39^2=-22^3+1$$ and solutions to the curve \(x^3-1 = 7y^2\) as discussed in the previous entry.

Entry 78

Level 7. This level is special since we have to consider the curve \(x^3-1 = 7y^2\). Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and the McKay-Thompson series of class 7A for the Monster.  $$j_{7A}(\tau) = \left(\left(\frac{d_1}{d_7}\right)^2+7\left(\frac{d_7}{d_1}\right)^2\right)^2$$ Examples. We select \(d=7m\) with class number \(h(-d)=2\) and find \(m=5,13,61\) such that the following are well-behaved integers $$\begin{align}j_{7A}\Big(\tfrac{1+\sqrt{-5/7}}{2}\Big) &= -7\times1^2 = -2^3+1\\ j_{7A}\Big(\tfrac{1+\sqrt{-13/7}}{2}\Big) &= -7\times3^2 = -4^3+1\\ j_{7A}\Big(\tfrac{1+\sqrt{-61/7}}{2}\Big) &= -7\times39^2 = -22^3+1 \end{align}$$ A WolframAlpha search for positive integer solutions to \(x^3-1 = 7y^2\) reveals only these three. However, if we allow \((x,y)\) to be higher algebraic integers with the same odd degree and a solvable Galois group, then it seems there are infinitely many. For example, we select \(d=7m\) with class number \(h(-d)=6\) and find \(m=101\) and others so $$j_{7A}\Big(\tfrac{1+\sqrt{-101/7}}{2}\Big) =-x^3+1=-7y^2$$ where \((x,y)\) are the real roots of cubics $$x^3 - 46x^2 - 380x - 800 = 0\\ y^3 - 145y^2 - 357y - 1235=0$$ For class number \(h(-d)=10\), we find \(m=17\) and others so $$j_{7A}\Big(\tfrac{1+\sqrt{-17/7}}{2}\Big) =-x^3+1=-7y^2$$ where \((x,y)\) are the real roots of solvable quintics $$x^5 - 5x^4 + 4x^3 - 15x^2 - 23x - 11 = 0\\ y^5 - 5y^4 + 5y^3 - 7y^2 - 1=0$$ and so on.

Entry 77

Level 6. Define \(d_k = \eta(k\tau)\) with Dedekind eta function \(\eta(k\tau)\) and the McKay-Thompson series of class 6B for the Monster.

$$j_{6B}(\tau) = \left(\frac{d_2\,d_3}{d_1\,d_6}\right)^{12}$$

Example. We select \(d=12m\) with class number \(h(-d)=4\) and find  

$$m=10, 14, 26, 34\\ m=7, 11, 19, 31, 59$$ such that the following are special quadratic irrationals $$\begin{align}j_{6B}\big(\tfrac12\sqrt{-10/3}\big) &= U_5^{12}=\big(\tfrac{1+\sqrt5}2\big)^{12}\\ j_{6B}\big(\tfrac12\sqrt{-14/3}\big) &= U_{14}^2 =\big(15+4\sqrt{14}\big)^2 \\ j_{6B}\big(\tfrac12\sqrt{-26/3}\big) &= U_{26}^4 =\big(5+\sqrt{26}\big)^4\\ j_{6B}\big(\tfrac12\sqrt{-34/3}\big) &= U_2^{12}=\big(1+\sqrt{2}\big)^{12}\end{align}$$

$$\quad \begin{align}j_{6B}\Big(\tfrac{1+\sqrt{-7/3}}{2}\Big) &= -U_{21}^{3}= -\big(\tfrac{5+\sqrt{21}}2\big)^{3}\\ j_{6B}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -U_{11}^2 = -\big(10+3\sqrt{11}\big)^2\\ j_{6B}\Big(\tfrac{1+\sqrt{-19/3}}{2}\Big) &= -U_{3}^6 = -\big(2+\sqrt{3}\big)^6\\ j_{6B}\Big(\tfrac{1+\sqrt{-31/3}}{2}\Big) &= -U_{93}^{3}= -\big(\tfrac{29+3\sqrt{93}}2\big)^{3}\\ j_{6B}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -U_{59}^2 = -\big(530+69\sqrt{59}\big)^2 \end{align}$$

since they are fundamental units \(U_n\). Note the integer $$\left(\sqrt{-\big(530+69\sqrt{59}\big)^2}-1/\sqrt{-\big(530+69\sqrt{59}\big)^2}\right)^2 = -1060^2$$ and similarly for the others as discussed in the previous entry.

Entry 76

Level 6. Define \(d_k=\eta(k\tau)\) with the Dedekind eta function \(\eta(\tau)\) and the McKay-Thompson series of class 6A for the Monster. 

$$j_{6A}(\tau) = \left(\left(\frac{d_2\,d_3}{d_1\,d_6}\right)^6-\left(\frac{d_1\,d_6}{d_2\,d_3}\right)^6\right)^2$$ Examples. We select \(d=12m\) with class number \(h(-d)=4\) and find  

$$m=10, 14, 26, 34\\ m=7, 11, 19, 31, 59$$ such that the following are well-behaved integers $$\begin{align}j_{6A}\big(\tfrac12\sqrt{-10/3}\big) &= (8\sqrt5)^2\\ j_{6A}\big(\tfrac12\sqrt{-14/3}\big) &= (8\sqrt{14})^2\\ j_{6A}\big(\tfrac12\sqrt{-26/3}\big) &= (20\sqrt{26})^2\\ j_{6A}\big(\tfrac12\sqrt{-34/3}\big) &= (140\sqrt2)^2 \end{align}$$ as well as $$\begin{align}j_{6A}\Big(\tfrac{1+\sqrt{-7/3}}{2}\Big) &= -(4\sqrt7)^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-11/3}}{2}\Big) &= -20^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-19/3}}{2}\Big) &= -52^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-31/3}}{2}\Big) &= -(28\sqrt{31})^2\\ j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}{2}\Big) &= -1060^2\end{align}$$ Note that the prime-generating polynomials $$F(n)=n^2-n+41\\ F(n)=6n^2-6n+31$$ where the latter is prime for \(30\) consecutive values \(n=0 - 29\). Solving \(F(n)=0\) yields \(n=\frac{1+\sqrt{-163}}{2}\) and \(n=\frac{1+\sqrt{-59/3}}{2}\), respectively hence $$j_{1A}\Big(\tfrac{1+\sqrt{-163}}2\Big) = -640320^3\\ j_{6A}\Big(\tfrac{1+\sqrt{-59/3}}2\Big) = -1060^2\quad$$

Entry 75

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 \(h(-d)=2\).