## Solvable Quintics, Part 2

In the 1850’s, Jerrard showed that a Tschirnhausen transformation could reduce in radicals the general quintic into one missing three terms,

$\text{(1)}\;\; x^5+5ax+4b = 0$

In 1864, Bring independently would do the same. It is now known as the Bring-Jerrard quintic.  Such a reduction is important because it proved that a formula for the general quintic does exist, albeit it went beyond radicals and used elliptic functions, as was first done by Hermite.

In 1885, Runge et al showed that all solvable quintics [1] with rational coefficients have the form,

$x^5+5\,\frac{4u+3}{u^2+1}xz^4+4\,\frac{(2u+1)(4u+3)}{u^2+1}z^5 = 0$

A century later, Spearman and Williams gave their version,

$x^5+5\,\frac{4v+3}{v^2+1}xz^4+4\,\frac{-2v+11}{\,v^2+1}z^5 = 0$

So which is it? It turns out they are two sides of the same coin. Using the Spearman-Williams parametrization, let,

$a = \frac{4v+3}{v^2+1}z^4$

$b = \frac{-2v+11}{\,v^2+1}z^5$

and eliminating v between them using resultants, one gets,

$\text{(2)}\;\; 4az^6+8bz^5-5a^2z^2+2abz-b^2 = 0$

This is one of the simplest sextic resolvents for the quintic:  given {a, b}, if one can solve for z, then [1] is solvable. Since [2] is only a quadratic for b, we can easily solve for it,

$b = az+4z^5+2z\sqrt{(a+z^4)(-a+4z^4)}$

Still let,

$a= \frac{4v+3}{v^2+1}z^4$

and substituting it into the positive case of the square root yields the b of the 1885 version, while the negative case gives the b of the 1994 one, proving that they are indeed two sides of the same coin.