## 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.