This is the sextic overview. Using the McKay-Thompson series jn=jn(τ) for the Monster defined in Entry 145, if τ are complex quadratics such that jn(τ) is a radical, then the following sextics have a solvable Galois group, hence solvable in radicals j1=(x2+10x+5)3xj2=(x+1)4(x2+6x+25)xj32=j3(2x6+29x5+85x4+50x3)+54x6(2x−1)j4=(x+1)5(x+5)x
with the ones in red by Joachim König. (It would be nice if the one for j3 can be simplified.) They have discriminants,
D1=55(j1−1728)2j14D2=55(j2−256)3j23D3=55(j3−108)3j311D4=55(j4−64)2j44
Examples. Let τ=1+√−1632 so j1(τ)=−6403203. Let τ=√−582 so j2(τ)=3964. Then (x2+10x+5)3x=−6403203(x+1)4(x2+6x+25)x=3964
are sextics both solvable in radicals. There are infinitely many τ but special ones such that jn(τ) are integers can be found in Entry 145.
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