While considering [this post](https://math.stackexchange.com/questions/1598329/can-we-find-x-1-x-2-x-n), it made me wonder about its generalization in another direction and from the perspective of Galois theory. 




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**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 $a<b<c<d$?




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**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?



Is this system always solvable in radicals by quartics, octics, $12$-ics, etc?
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 $A\_5$. 


