http://en.wikipedia.org/wiki/Quintic "During the second half of 19th century, John Stuart Glashan<http://en.wikipedia.org/w/index.php?title=John_Stuart_Glashan&action=edit&redlink=1>, George Paxton Young, and Carl Runge<http://en.wikipedia.org/wiki/Carl_Runge>found that any irreducible <http://en.wikipedia.org/wiki/Irreducible_polynomial> quintic with rational coefficients in Bring<http://en.wikipedia.org/wiki/Erland_Samuel_Bring> -Jerrard <http://en.wikipedia.org/wiki/George_Jerrard> form, [image: x^5 + ax + b = 0\,] (x^5+ax+b=0) is solvable by radicals if and only if either *a* = 0 or it is of the following form: [image: x^5 + \frac{5\mu^4(4\nu + 3)}{\nu^2 + 1}x + \frac{4\mu^5(2\nu + 1)(4\nu + 3)}{\nu^2 + 1} = 0] (x^5 + 5*\[Mu]^4*(4*\[Nu] + 3)*x/(\[Nu]^2 + 1) + 4*\[Mu]^5*(2*\[Nu] + 1)*(4*\[Nu] + 3)/(\[Nu]^2 + 1)) where [image: \mu] and [image: \nu] are rational. In 1994, Blair Spearman and Kenneth S. Williams gave an alternative, [image: x^5 + \frac{5e^4( 4c + 3)}{c^2 + 1}x + \frac{-4e^5(2c-11)}{c^2 + 1} = 0.] (x^5 + 5*e^4*(4*c + 3)/(c^2 + 1)*x - 4*e^5*(2*c - 11)/(c^2 + 1)) "In at least two cases where the quintic is reducible, 1/8 (1 + x) (3 - 8 x + 8 x^2 - 8 x^3 + 8 x^4)==0, and 1/3 (1 + x) (8 - 3 x + 3 x^2 - 3 x^3 + 3 x^4)==0 these parameters come out Gaussian rationals: {\[Mu] -> -(1/4) + I/4, \[Nu] -> 1/4 + (3 I)/4} and {\[Mu] -> -(1/2) + I/2, \[Nu] -> -(3/2) - I}, i.e., {e -> -(1/4) + I/4, c -> 1/4 - (3 I)/4} and {e -> -(1/2) + I/2, c -> -(3/2) + I} The quintic solver goes ahead and produces such unlikelihoods as 1/10 ((-79 + 3 I) ((-7 - 24 I) + 3 Sqrt[-41 + 38 I]))^(1/5) + 1/10 ((3 + 79 I) ((7 + 24 I) + 3 I Sqrt[41 - 38 I]))^(1/5) + 1/10 ((3 + 79 I) ((7 + 24 I) - 3 I Sqrt[41 - 38 I]))^(1/5) E^(( 2 I \[Pi])/5) + 1/10 ((79 - 3 I) ((7 + 24 I) + 3 Sqrt[-41 + 38 I]))^(1/5) E^(-((4 I \[Pi])/5)) == 1/4 (1 - (-1)^(3/4) Sqrt[2] 5^(1/4) + I Sqrt[5]) , which embarrasses RootReduce. --rwg
-------- Original Message -------- Subject: [math-fun] Solvable, irreducible quintic trinomials Date: 2011-11-08 00:44 From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Reply-To:
The only ones of height<=5 are x^5 - 5 x^2 - 3, 2*x^5 - 5*x + 4, and 4*x^5 + 5*x + 4 (modulo x->1/x). The only real root of the 2nd one is (1/(2^(2/5) 5^( 3/5)))(-(150 - 100 Sqrt[2] + Sqrt[10 (4250 - 2999 Sqrt[2])])^( 1/5) + (-150 + 100 Sqrt[2] + Sqrt[10 (4250 - 2999 Sqrt[2])])^( 1/5) - (150 + 100 Sqrt[2] - Sqrt[10 (4250 + 2999 Sqrt[2])])^( 1/5) - (150 + 100 Sqrt[2] + Sqrt[10 (4250 + 2999 Sqrt[2])])^(1/5))
This seems almost a case of "*a stupid man asking such a question to which one hundred wise men would not be able to answer*."
Various sources, e.g., http://en.wikipedia.org/wiki/**Quintic_function<http://en.wikipedia.org/wiki/Quintic_function> give a pair of equations which have rational roots iff x^5+ax+b is solvable and irreducible, but sometimes reducible cases that fail this test, e.g., {3/8 - (5 x)/8 + x^5, 8/3 + (5 x)/3 + x^5}, yield to the quintic solver, producing obscure radical identities of the form biquadratic = sum of four fifth roots of biquadratics, which paralyze FullSimplify.
I don't know if it's proven or merely conjectured that there are no solvable irreducible septic trinomials. See http://www.math.harvard.edu/~**elkies/trinomial.html<http://www.math.harvard.edu/~elkies/trinomial.html> --rwg