For class number 5, there are only four d of the kind d≡7mod8, namely d=47,79,103,127. We illustrate only the first two and propose, K(k47)=√2π2√47x2/3(47∏m=1[Γ(m47)](−47m))1/10K(k79)=√2π2√79y2/3(79∏m=1[Γ(m79)](−79m))1/10 where (x,y) are the real roots of the quintics solvable in radicals x5−2x4−10x3−13x2−6x−1=0y5−11y4+17y3−2y2−5y−1=0 And similarly for d=103,127. But for the next discriminant or d=131 which is of kind d≡3mod8, 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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