TITLE: How can I tell if $x^5 - (x^4 + x^3 + x^2 + x^1 + 1)$ is/is not part of the solvable group of polynomials?
QUESTION [9 upvotes]: I have developed an interest in generalisations of the fibonacci sequence, from tribonacci sequence up to what I'll coin the 'infinibonacci' sequence.
I'm aware that these nth-bonacci sequences require roots of matched nth-degree polynomials in the form $x^n - (x^{n-1} + x^{n-2} + \dots +x^0)=0$ and that 5th-degree polynomials are not guaranteed to be solvable, but I'm wondering if there is a systematic way of showing whether these particular polynomials are solvable by root extraction?
I am prepared to learn anything required to understand the result but am currently starting from smart 2nd year undergraduate level.
Thanks.

REPLY [6 votes]: A result of Dedekind says that for any polynomial $p \in \mathbb{Z}[x]$ and any prime $q$ not dividing the discriminant of $p$, then if $p$ factors modulo $q$ into a product of irreducible polynomials with degrees $d_1, \ldots, d_r$, then the Galois group $\text{Gal}(p)$ contains a permutation with cycle structure $(d_1, \ldots, d_r)$.
The discriminant of the Pentanacci polynomial $p_5(x) := x^5 - (x^4 + x^3 + x^2 + x + 1)$ is $9584 = 2^4 \cdot 599$. It is irreducible modulo $5$ and so $\text{Gal}(p_5)$ contains a $5$-cycle. Modulo $3$, we have
$$p_5(x) \equiv (x^3 + x^2 + 2 x + 1) (x^2 + x + 2),$$ so $\text{Gal}(p_5)$ contains a product $\sigma$ of a $2$-cycle and a $3$-cycles and thus also the $2$-cycle $\sigma^3$. Now, if $r$ is prime, then a $2$-cycle and an $r$-cycle in $S_r$ together generate all of $S_r$, and in particular the Galois group of $p_5$ is $S_5$. Now, $S_n$ is only solvable iff $n \leq 4$, so the roots of $p_5$ cannot be extracted with radicals.
Tito and Robert's answers used a dedicated CAS command to show that $\text{Gal}(p_n) \cong S_n$ for $n \leq 12$, and we can readily use Dedekind's result to extend this to, say, $n \leq 20$.
For the $13$-nacci polynomial $p_{13}$, applying an argument similar to the $n = 5$ case, and now instead considering the primes $p = 5$ (which gives a $13$-cycle) and $17$ (which gives an element $\sigma$ with $\sigma^{11}$ a transposition), leads to the conclusion that $\text{Gal}(p_{13}) \cong S_{13}$.
For the $14$-nacci polynomial $p_{14}$, factoring modulo $5$ gives a product $\sigma$ of a $4$-cycle and a $9$-cycle, so $\sigma^{18}$ is a transposition, and factoring modulo $19$ gives a $13$-cycle. But $p_{14}$ is irreducible (as it is irreducible modulo $5$), and so its Galois group is transitive, and being a transitive subgroup of $S_n$ that contains a $2$-cycle and an $(n - 1)$-cycle, $\text{Gal}(p_{14})$ is necessarily $S_{14}$ itself. 
We can handle the remaining cases similarly. As above, $\sigma$ is an element with cycle structure given by Dedekind's result for the corresponding prime:
$$\begin{array}{c|l}
n & p \\
\hline
15 & 11 \, (\sigma^5    \text{ a  $2$-cycle}), \, 199 \, (\sigma \text{ a $14$-cycle}) \\
16 &  5 \, (\sigma^6    \text{ a  $2$-cycle}), \,  59 \, (\sigma \text{ a $15$-cycle}) \\
17 &  3 \, (\sigma^{36} \text{ a  $2$-cycle}), \,   5 \, (\sigma \text{ a $17$-cycle}) \\
18 &  5 \, (\sigma^{26} \text{ a  $2$-cycle}), \,  17 \, (\sigma \text{ a $17$-cycle}) \\
19 &  3 \, (\sigma      \text{ a $18$-cycle}), \,  13 \, (\sigma^{33} \text{ a $2$-cycle}) \\
20 &  5 \, (\sigma^{66} \text{ a  $2$-cycle}), \,  23 \, (\sigma \text{ a $19$-cycle})
\end{array}$$