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I followed the exact steps from this forum post to solve quintic polynomials of the form:

$x^5 - x + d$

But I got a different answer in number form from Mathematica than the original quintic polynomial root.

Am I doing something wrong?

Also, where it says, $(k^2)^{1/8}$, isn't that supposed to be $k^{1/4}$?

machinegun3455
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  • Please show us your calculations first. – Somos Oct 10 '19 at 00:32
  • NOTE: $k^2$ has eight $8$th roots, whereas $k$ has only four $4$th roots. Example, $k=1$. Then $(k^2)^{1/8}$ has $8$ values,the $8$th roots of unity, whereas $k^{1/4}$ has only $4$ values, the $4$th roots of unity. – GEdgar Oct 10 '19 at 00:52
  • I want to solve for $x^5 - x + 1 = 0$, so we know that d = 1, here is what I put in for Mathematica: k = Tan[ArcSin[16/(25Sqrt[5]d^2)]/4] p = IEllipticK[1-k^2]/EllipticK[k^2] T[J_] = (E^(2PiIJ)/8)Sqrt[2]DedekindEta[(p+2J)/10](DedekindEta[4(p+2J)/10])^2/(DedekindEta[2*(p+2J)/10])^3 T5 = Sqrt[2]DedekindEta[5p/2](DedekindEta[10p])^2/(DedekindEta[5*p])^3 u = (1/(25^(3/4)))((k^2)^(1/8)/Sqrt[k*(1-k^2)])(T[0] + T5)(T[1] + IT[4])(I*T[2]+T[3]) – machinegun3455 Oct 10 '19 at 02:05

1 Answers1

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d = RandomInteger[{-100, 100}];
Print["\nEquation: x^5 - x ", If[d < 0, " - ", " + "], Abs[d], " = 0\n"];
Print["Ordinary solution:"];
Print[x /. (x^5 - x + d // NSolve) // Sort, "\n"];
Print["Solution with Dedekind Eta:"];
k = Tan[ArcSin[16/(25 Sqrt[5] d^2)]/4];
p = I EllipticK[1 - k^2]/EllipticK[k^2];
S[J_] = Exp[2 Pi I/8]^J Sqrt[2] DedekindEta[(p + 2 J)/10] DedekindEta[4 (p + 2 J)/10]^2/DedekindEta[2 (p + 2 J)/10]^3 ;
S5 = Sqrt[2] DedekindEta[5 p/2] DedekindEta[10 p]^2/DedekindEta[5 p]^3 ;
Sign[d] (1/(2 5^(3/4))) ((k^2)^(1/8)/Sqrt[k (1 - k^2)]) (S[0] + S5) (S[1] + I S[4]) (I S[2] + S[3]) // N
Dmitry Ezhov
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