On solvable quintics and septics Here is a nice sufficient (but not necessary) condition on whether a quintic is solvable in radicals or not.  Given,
$x^5+10cx^3+10dx^2+5ex+f = 0\tag{1}$ 
If there is an ordering of its roots such that,
$x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_5 + x_5 x_1 - (x_1 x_3 + x_3 x_5 + x_5 x_2 + x_2 x_4 + x_4 x_1) = 0\tag{2}$
or alternatively, its coefficients are related by the quadratic in f,
$(c^3 + d^2 - c e) \big((5 c^2 - e)^2 + 16 c d^2\big) = (c^2 d + d e - c f)^2
\tag{3}$
then (1) is solvable.  This also implies that if $c\neq0$, then it has a solvable twin,
$x^5+10cx^3+10dx^2+5ex+f' = 0\tag{4}$
where $f'$ is the other root of (3). The Lagrange resolvent are the roots of,
$z^4+fz^3+(2c^5-5c^3e-4d^2e+ce^2+2cdf)z^2-c^5fz+c^{10} = 0\tag{5}$
so,
$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag{6}$
Two questions though:
I. Does the septic (7th deg) analogue,
$x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 – (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0\tag{7}$
imply such a septic is solvable?
II. The septic has a $5! = 120$-deg resolvent. While this is next to impossible to explicitly construct, is it feasible to construct just the constant term? Equating it to zero would then imply a family of solvable septics, just like (3) above.
More details and examples for (2) like the Emma Lehmer quintic in my blog.

It turns out there is an infinite number of such septics, such as the Hashimoto-Hoshi septic,
$$\small x^7 - (a^3 + a^2 + 5a + 6)x^6 + 
  3(3a^3 + 3a^2 + 8a + 4)x^5 + (a^7 + a^6 + 9a^5 - 5a^4 - 15a^3 - 22a^2 - 
        36a - 8)x^4 - a(a^7 + 5a^6 + 12a^5 + 24a^4 - 6a^3 + 2a^2 - 20a - 16)x^3 + a^2(2a^6 + 7a^5 + 19a^4 + 14a^3 + 2a^2 + 8a - 8)x^2 - a^4(a^4 + 4a^3 + 8a^2 + 4)x + a^7=0$$
For example, let $a=1$ so,
$$1 - 17 x + 44 x^2 - 2 x^3 - 75 x^4 + 54 x^5 - 13 x^6 + x^7=0$$
which is the equation involved in $\cos\frac{\pi k}{43}$ and order its roots as,
$$x_1,\,x_2,\,x_3,\,x_4,\,x_5,\,x_6,\,x_7 =\\
r_1,\,r_2,\,r_5,\,r_6,\,r_3,\,r_7,\,r_4 =
\\ -0.752399,\; 0.0721331,\; 2.63744,\; 3.62599,\; 0.480671,\; 6.29991,\; 0.636246$$
where the $r_i$ is the root numbering in Mathematica. Then,
$$
x_1 x_2 + x_2 x_3 + \dots + x_7 x_1 - (x_1 x_3 + x_3 x_5 + \dots + x_6 x_1) = 0$$
