TITLE: Why does the polynomial equation $1 + x + x^2 + \cdots + x^n = S$ have at most two solutions in $x$?
QUESTION [11 upvotes]: Américo Tavares pointed out in his answer to this question that finding the ratio of a geometric progression only from knowledge of the sum of its first $n+1$ terms $S = 1+x+x^2+\cdots+x^n$ amounts to solving a polynomial of degree $n$. This suggested to me that there might be up to $n$ real solutions of $x$ for a given sum, but I could not find any. In fact, it turned out that the following fact is true:
For $n \ge 1$ and $S \in \mathbb{R}$, the polynomial equation $x^n + x^{n-1} + \cdots + x + 1 = S$ has at most two real solutions.
A corollary is that if $n$ is odd, there is exactly one real solution. I was only able to prove this using a rather contrived geometric argument based on the shape of the graph of $y = x^{n+1}$. Is there a simple, direct (and ideally, intuitive) proof of this fact?

REPLY [13 votes]: In between the previous two answers - one very specific and one very general, there is the following one:
Rolle's theorem states that between any two zeros of the function lies a zero of the derivative. So each time you differentiate the number of real roots can not decrease by more than 1. This actually holds with multiplicity as well - if a root is of multiplicity k, then it is of multiplicity (k-1) for the derivative, and there are still roots of the derivative strictly between consecutive distinct roots of the original function. So $x^{n+1}−Sx+S−1$ has at most one root more than $(n+1)x^n-S$ and at most two more than $n(n+1)x^{n-1}$, so not more than three total (counted with multiplicity). One of those is $x=1$, so there are no more than two for the $1+x+\ldots+x^n=S$. QED.