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?

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.

You can test its Galois group using this online Magma calculator. For example, to test the solvable but irreducible quintic $x^5-5x+12=0$, use the command,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-5*x+12; G, R := GaloisGroup(f); G;

Copy and paste. One then finds the order is $10$, hence that quintic is solvable. (All groups with order $<60$ is solvable, but there are solvable groups with order $>60$.)

For the pentanacci,

Z := Integers(); P < x > := PolynomialRing(Z); f := x^5-x^4-x^3-x^2-x-1; G, R := GaloisGroup(f); G;

It says it is the symmetric group $S_5$. And for higher $n$-nacci (I tried up to $n=12$), we get $S_n$ which are not solvable for $n\geq5$.

P.S. If you are testing other equations, don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x. (I learned that after a while.)

According to Maple's galois function, the polynomial $x^n - (x^{n-1} + \ldots + 1)$ has Galois group $S_n$ for $n$ from $3$ to $9$ (it can't handle polynomials of degree greater than $9$). In particular, for $n = 5$ to $9$ these are not solvable by radicals. I suspect that none of the polynomials for $n \ge 5$ are solvable by radicals.

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: $$\begin{array}{c|l} n & p \\ \hline 15 & 11 \, (\sigma^5 \text{ a $2$-cycle}), \, 199 \, (\sigma \text{ a $14$-cycle}) \\ 16 & 5 \, (\sigma^6 \text{ a $2$-cycle}), \, 59 \, (\sigma \text{ a $15$-cycle}) \\ 17 & 3 \, (\sigma^{36} \text{ a $2$-cycle}), \, 5 \, (\sigma \text{ a $17$-cycle}) \\ 18 & 5 \, (\sigma^{26} \text{ a $2$-cycle}), \, 17 \, (\sigma \text{ a $17$-cycle}) \\ 19 & 3 \, (\sigma \text{ a $18$-cycle}), \, 13 \, (\sigma^{33} \text{ a $2$-cycle}) \\ 20 & 5 \, (\sigma^{66} \text{ a $2$-cycle}), \, 23 \, (\sigma \text{ a $19$-cycle}) \end{array}$$