# Galois group of the polynomial $x^n+x^{n-1} +⋯+x^2+x−1$

As we already know, the following polynomial is irreducible over $$\mathbb Q[x]$$:

$$x^n + x^{n-1} + \cdots + x^2 + x - 1 = \frac{x^{n+1} - 2x + 1 }{ x-1}.$$

By Descartes' rule of signs, it has only 1 positive real root. It seems, though I do not know how to prove it yet, that there is only 1 real root if $$n$$ is odd and 2 real roots of opposite signs if $$n$$ is even.

Computational tests, conducted with GAP, suggest that the Galois group of this polynomial is $$S_n$$. How we can actually prove it?

Any help is greatly appreciated.

• It is perhaps possible to argue like here, where the polynomial is $x^n-x^{n-1}-\ldots -1$. May 3 at 15:00
• It may be that the article referred to here will help. I haven't looked it up. And in this case the trinomial is not irreducible. May 3 at 17:21
• It is really interesting, @DietrichBurde, the polynomial $x^n - x^{n-1} - ... - 1$ is a negative reciprocal $- x^n f(1/x)$, where $f(x) = x^n + x^{n-1} + ... x - 1$. There should be some results about Galois group of reciprocal polynomials... May 3 at 19:17
• Yes, the Galois groups then are isomorphic. May 3 at 19:50
• @DietrichBurde could you please elaborate a little bit more why it is true or give a link to this result about reciprocal? I'm only starting to learn Galois theory and the help of specialists is priceless. May 4 at 15:09

The polynomial $$f(x)=x^n+x^{n-1}+\cdots +x^2+x-1$$ is the negative reciprocal of the generalised Fibonacci polynomial $$x^n-x^{n-1}-\cdots -x-1$$. Its Galois group is indeed isomorphic to $$S_n$$, because both $$f(x)$$ and $$g(x)=-x^nf(1/x)$$ do have the same Galois group. The Galois group of $$g(x)$$ was computed in the article The Galois group of $$x^n-x^{n-1}-\cdots -x-1$$ by P. A Martin in $$2004$$, for even $$n$$ and odd prime $$n$$. The result is the symmetric group $$S_n$$. He conjectures that it is true for all $$n$$.