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Saturday, June 14, 2025

Entry 168

For class number 5, there are only four d of the kind d7mod8, namely d=47,79,103,127. We illustrate only the first two and propose, K(k47)=2π247x2/3(47m=1[Γ(m47)](47m))1/10K(k79)=2π279y2/3(79m=1[Γ(m79)](79m))1/10 where (x,y) are the real roots of the quintics solvable in radicals x52x410x313x26x1=0y511y4+17y32y25y1=0 And similarly for d=103,127. But for the next discriminant or d=131 which is of kind d3mod8, one now has to deal with an algebraic number of degree 3×5=15. And it seems difficult to find the eta quotients responsible for (x,y).

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