Define,

$$f_n = \frac{x(x^n - 1)}{x - 1}m^{p - n}$$

then for $p=5,7,13$, respectively,

$$f_5 + 5(6f_4 + 63 f_3 + 260 f_2 + 315 f_1) - 5^3m^6 = 0\tag{1}$$

$$f_7 + 7 (4 f_6 + 46 f_5 + 272 f_4 + 845 f_3 + 1232 f_2 + 574 f_1) - 7^2m^8 = 0\tag{2}$$

$$f_{13}  + 13 (2 f_{12} + 25 f_{11} + 196 f_{10} + 1064 f_9 + 4180 f_8 + 12086 f_7 + 25660 f_6 + 39182 f_5 + 41140 f_4 + 27272 f_3 + 9604 f_2 + 1165 f_1) - 13m^{14} = 0\tag{3} $$

***Questions***: 

1. ***As equations of deg $p$ in $x$, are these solvable in radicals for any $m$***? (After testing various $m$, I'm 99.99% sure they are, but it would be nice to know their Galois group.)

2. Given the Dedekind eta function $\eta(\tau)$, these were found by determining integer relations between $m = \left(\frac{\eta\,(p\tau)}{\eta\,(\tau)}\right)^k$ and $x = \left(\frac{\sqrt{p}\,\eta\,(p^2\tau)}{\eta\,(\tau)}\right)^k$ where $k=\frac{24}{p-1}$. Does the relation $F(m,x)$, as a polynomial in $x$, always have a solvable Galois group for *any prime p*?