Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?

While considering this post, it made me wonder about its generalization in another direction and from the perspective of Galois theory.

Question: Is it true that, given four constants ($\alpha,\beta,\gamma,\delta$), then the system, \begin{aligned} x_1^a+x_2^a+x_3^a+x_4^a &= \alpha\\ x_1^b+x_2^b+x_3^b+x_4^b &= \beta\\ x_1^c+x_2^c+x_3^c+x_4^c &= \gamma\\ x_1^d+x_2^d+x_3^d+x_4^d &= \delta \end{aligned}\tag1 can be solved in radicals $x_i$ for any positive integer power ($a,b,c,d$) and $a<b<c<d$?

Example: Let's choose the year Back to the Future came out,

\begin{aligned} x_1+x_2+x_3+x_4 &= \color{brown}1\\ x_1^2+x_2^2+x_3^2+x_4^2 &= \color{brown}9\\ x_1^4+x_2^4+x_3^4+x_4^4 &= \color{brown}8\\ x_1^8+x_2^8+x_3^8+x_4^8 &= \color{brown}5 \end{aligned}

Two solutions are given by the roots of the octic,

$$4707 + 9036 x - 7495 x^2 - 5096 x^3 + 3024 x^4 + 1848 x^5 - 896 x^6 - 256 x^7 + 128 x^8 = 0$$

Magma says this is $8T47$, has order $1152=2^7\cdot3^2$ and is the largest solvable order for deg $8$. (In fact, it factors over $\sqrt{1441}$). Using the root $r_i$ numbering system of Mathematica, we find that,

$$x_1,\, x_2,\, x_3,\, x_4 = r_1,\, r_2,\, r_5,\, r_6\quad \text{(Solution 1)}\\ \text{or}\quad\quad\\ x_1,\, x_2,\, x_3,\, x_4 = r_3,\, r_4,\, r_7,\, r_8\quad \text{(Solution 2)}$$

Remarks: Using generic ($\alpha,\beta,\gamma,\delta$)

1. For exponents $a,b,c,d = 1,2,c,8$ and $c=3,4,5$, one ends up with a $8T47$ octic with order $1152=2^7\cdot3^2$.
2. For $a,b,c,d = 1,2,6,7$, one gets a $12T294$ with order $82944 = 2^{10} \cdot 3^4$.
3. For $a,b,c,d = 1,2,c,8$ and $c=6,7$, one now gets a $16T1947$ with order $7962624 = 2^{15}\cdot3^5$. These three are the largest solvable orders for their respective degrees.
4. Using others yields deg $20,24,$ etc.

So, is the system $(1)$ always solvable in radicals?

• The equations are symmetric in the $x_i$ of course, so they are actually 4 equations in the four unknowns given by the first four elementary symmetric polynomials in the $x_i$. This divides the degree of everything by 24 and puts an $S_4$ on the top (which is solvable) which makes me worry that what you have seen computationally is a low-dimensional coincidence. I think also that this is a question about geometry rather than arithmetic; probably this is a question about the extension $C(x_1,x_2,x_3,x_4)/C(\alpha,\beta,\gamma,\delta)$ with $\alpha$ etc independent transcendental vars.
– eric
Jan 6 '16 at 19:47
• (the point being that by standard Hilbert irreducibility methods coming from arithmetic, if you can compute the geometric Galois group then you'll in general be able to find rational values of $\alpha$ etc such that eveything specialises but the Galois group doesn't get any smaller). What I'm saying is that this question is tagged number theory but is actually a question in complex algebraic geometry.
– eric
Jan 6 '16 at 19:51

For $(a,b,c,d)=(1,2,3,20)$ and $(\alpha,\beta,\gamma,\delta)=(1,1,1,0)$ one computes that $x_1$ is a root of the irreducible polynomial $4 x^{20} - 20 x^{19} + 40 x^{18} - 40 x^{17} + 195 x^{16} - 704 x^{15} + 1050 x^{14} - 700 x^{13} + 475 x^{12} - 900 x^{11} + 900 x^{10} - 300 x^{9} + 130 x^{8} - 260 x^{7} + 130 x^{6} + 20 x^{4} - 20 x^{3} + 1$, which according to magma, has a non-solvable Galois group, namely the full wreath product $S_4^5\rtimes S_5$.
Without Magma: With the same values of $a,b,c,d,\alpha,\beta,\gamma,\delta$ as above, one computes (e.g. using Sage) that $x_1x_2x_3x_4$ is a root of $h(x)=4 x^{5} - 175 x^{4} + 300 x^{3} - 130 x^{2} + 20 x - 1$. Now $h(x-1)$ is an Eisenstein polynomial with respect to $5$, and modulo $3$ $h(x)=(x + 1) \cdot (x + 2) \cdot (x^{3} + 2 x^{2} + x + 1)$ is a factorization into irreducibles. Thus the Galois group of $h$ contains the alternating group $A_5$.