Galois Group of $x^5+ 5x^3 + 5x + 1$. I've been asked to determine the Galois Group of $x^5+  5x^3 + 5x + 1$. This is what I know so far.
1) The polynomial is irreducible.
2) Its discriminant is $78125=5^7$
Since the discriminant is not a square, the Galois group should be $S_5$ or $F_{20}$. How can I know? In case it was $F_{20}$, is it solvable?
Thanks in advance
 A: The paper Solving Solvable Quintics - D. S. Dummit shows the construction of a resolvent $f_{20}$ for the quintic such that $f_{20}$ has a rational root iff the Galois group is contained in $F_{20}$.
Let $x_i$ be roots of the general quintic.
Let $$\theta = x_{0} x_{1}^{2} x_{2} + x_{0}^{2} x_{2} x_{3} + x_{1} x_{2}^{2} x_{3} + x_{0} x_{1} x_{3}^{2} + x_{0}^{2} x_{1} x_{4} + x_{0} x_{2}^{2} x_{4} + x_{1}^{2} x_{3} x_{4} + x_{2} x_{3}^{2} x_{4} + x_{1} x_{2} x_{4}^{2} + x_{0} x_{3} x_{4}^{2}$$
Then we define $f_{20}(x) = \prod_{\sigma} (x - \sigma \theta)$ where the permutations $\sigma \in S_5$ act by permuting the variables $x_{i}$ of $\theta$. The product is taken over the whole orbit. This orbit has size 6. Note that $5!/6 = 20$.
Now $f_{20}$ is invariant under conjugation therefore it lies in the base field. Using symmetic polynomial calculations you may recover (the very long) formula (2) from the paper. The resolvent for our polynomial in particular is:
$$f_{20}(x) = x^{6} + 40 \, x^{5} + 250 \, x^{4} - 4375 \, x^{3} - 21875 \, x^{2} + 25000 \, x - 1187500$$
and $(x-10)|f_{20}(x)$ so combining this with your observation that the discriminant is nonsquare we find that the Galois group of this polynomial is $F_{20}$.
A: There is no doubt that this is solvable.  Render $x=r-1/r$ and show that $r^5-1/r^5=-1$.  Thus $r^5=(-1\pm\sqrt5)/2$, etc, and the Galois group can't be larger than $F_{20}$.
A: We can use the Magma online calculator to find the Galois group of the equation $f(x)=  x^5 + 5 x^3 + 5 x + 1$ in the field of rational numbers.
> P<x> := PolynomialRing(IntegerRing()); 
> f :=  x^5 + 5 * x^3 + 5 * x + 1; 
>G:=GaloisGroup (f);
>G;
>IsCyclic (G);
>IsSolvable (G);

We can get that the Galois group of this equation in the rational number field has order 20 and is a solvable group:

This result is for reference only, and cannot replace theoretical analysis.
