Entry 144

Define the McKay-Thompson series of Class 6A for the Monster $$j_6 = j_{6}(\tau) =\left( \left(\frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)}\right)^3 + 2^3\left(\frac{\eta(2\tau)\,\eta(6\tau)}{\eta(\tau)\,\eta(3\tau)}\right)^3\right)^2$$ and the quintic $$x^5-5\alpha x^3+10\alpha^2x-\alpha^2=0$$ where \(\small\alpha = -\dfrac1{j_6-32}.\) Alternatively $$(y^2+15)^2(y-5) = 32\big(j_6-32\big) $$ $$z^5 + 5z^3 - 10z^2 = j_6 $$

Conjecture: "If \(\tau\) is a complex quadratic such that \(j_6=j_{6}(\tau)\) is an algebraic number with \(j_6\neq 32\), then the quintics above have a solvable Galois group."

Example: Let \(j_6\big(\tfrac{1\sqrt{-59/3}}2\big)=-1060^2\), then $$(y^2+15)^2(y-5) = 32\big({-1060^2}-32\big)$$ $$z^5 + 5z^3 - 10z^2=-1060^2 $$ are solvable in radicals.