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Sunday, June 8, 2025

Entry 147

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(2x1)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(j11728)2j14D2=55(j2256)3j23D3=55(j3108)3j311D4=55(j464)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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