Entry 149

This is the octic overview. Using the McKay-Thompson series $j_n = j_n(\tau)$ for the Monster defined in Entry 145, if $\tau$ are complex quadratics such that $j_n(\tau)$ is a radical, then the following octics have a solvable Galois group, hence solvable in radicals

$$\begin{align}\qquad{j_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\ {j_2}^2 &=\frac{j_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\ {j_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\ {j_4}^2 &=\frac{7 j_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x \end{align}$$

It would be good if the one for $j_2$ can be simplified. They have discriminants (with another factor of $D_2$ suppressed),

$$\begin{align}D_1 &= -7^7(j_1-1728)^4\,{j_1}^4\\ D_2 &= -7^7(j_2-256)^4\,{j_2}^6 \\ D_3 &= -7^7(j_3-108)^3\,{j_3}^5\\ D_4 &= -7^7(j_4-64)^4\,{j_4}^{12}\quad \end{align}$$

Examples. Let $\tau = \tfrac{1+\sqrt{-41/3}}2$ so $j_3(\tau) = -(4\sqrt3)^6$. Let $\tau = \tfrac{1+\sqrt{-89/3}}2$ so $j_3(\tau) = -(10\sqrt3)^6$. Then

$$\frac{(x + 1)^6(x^2 + x + 7)}x = -(4\sqrt3)^6\; \\ \frac{(x + 1)^6(x^2 + x + 7)}x =  -(10\sqrt3)^6$$ are octics both solvable in radicals. And also $$e^{\pi\sqrt{41/3}} =  (4\sqrt3)^6+41.993\dots\quad\\ e^{\pi\sqrt{89/3}} =  (10\sqrt3)^6+41.99997\dots$$ There are infinitely many $\tau$ but special ones such that $j_n(\tau)$ are integers can be found in Entry 145.