Solvable quintics

Here is a nifty sufficient but not necessary condition on whether a quintic is solvable in radicals or not.  Given,

[1]  x^5+10cx^3+10dx^2+5ex+f = 0

If there is an ordering of its roots such that,

[2]  x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - \\(x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0

or alternatively, its coefficients are related by,

[3]  -25c^6-40c^3d^2-16d^4+35c^4e+\\28cd^2 e -11c^2e^2+e^3-2c^2df-2def+cf^2 = 0

then [1] is solvable as,

x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}

where the z_i are the roots of the simple quartic,

z^4+fz^3+(2c^5-5c^3e-4d^2e+ce^2+2cdf)z^2-c^5fz+c^{10} = 0

Note that [3] in fact is the constant term of Cayley’s resolvent sextic and is only quadratic in f.  Using another relation among the x_i, Dummit’s resolvent has a constant term that is already a quartic in f, hence the choice of relation matters.

Example 1: This family of quintics by this author satisfies [3],

x^5+10x^3+5(n^2+3n+18)x^2-5(n^3+n^2+15n-14)x+\\(n^4-n^3+37n^2+441) = 0

Let n = 1, and we have,

x^5+10x^3+110x^2-15x+478 = 0

Quartic is,

z^4+478z^3+11994z^2-478z+1 = 0

such that,

x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5} = -4.50991\dots

Example 2: Another good example of [3] is Emma Lehmer’s quintic,

y^5 +n^2y^4-(2n^3+6n^2+10n+10)y^3+\\(n^4+5n^3+11n^2+15n+5)y^2+(n^3+4n^2+10n+10)y+1

The linear transformation,

y = x-n^2/5

will reduce it into the form of [1], and it will then be seen its coefficients obey [3]. As a particular example, let n = 5 and we have the reduced form,

x^5-710x^3+11005x^2-59640x+108701 = 0

Let its roots be,

(x_1, x_2, x_3, x_4, x_5) \approx (-33.15,\; 4.83,\; 12.16,\; 4.99,\; 11.16)

and we find that indeed it obeys [2].

Unfortunately, no similar simple relation between the coefficients of a solvable septic, or 7th degree equation, is yet known.

2 responses to this post.

  1. Posted by kadir saim uyanık (male, 49) on October 25, 2013 at 7:02 am

    Thank you very much for your valuable contributions for solvable quintics. I wish you will reach new findings and share them with us. Yours sincerely.


  2. Thanks Tito
    Interesting to also know that….


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