Entry 140

Given the Dedekind eta function \(\eta(\tau)\) and define the McKay-Thompson series of Class 1A for the Monster, or better known as the j-function \(j\) $$j=j_1(\tau) =\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3$$ and the Brioschi quintic $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ where \(\small\alpha = -\dfrac1{j-1728}\). Alternatively $$(y^2+20)^2(y-5) = j-1728$$ $$z^5 + 5z^4 + 40z^3=j$$Conjecture: "If \(\tau\) is a complex quadratic such that \(j\) is an algebraic number \(j\neq1728\), then the quintics above have a solvable Galois group." 

Example. Let \(j\big(\tfrac{1+\sqrt{-163}}2\big) = -640320^3\) and \(\alpha = \tfrac1{640320^3+1728}\), then $$x^5-10\alpha x^3+45\alpha^2x-\alpha^2=0$$ $$(y^2+20)^2(y-5) = -640320^3-1728$$ $$z^5 + 5z^4 + 40z^3=-640320^3$$ are quintics solvable in radicals. (In fact, they factor into a quadratic and a cubic.)