Consider the following family of polynomials over $\mathbb{Q}$:

$$f_n = x^n - x^{n-1} - \dots - 1$$

Notice that these polynomials satisfy the recurrence

$$ f_{n+1} = x f_n - 1 $$

I would like to show that every polynomial in this family is irreducible over $\mathbb{Q}$. My strategy for showing this is to show that for each of these polynomials, all but one of the roots are in the unit circle (this would imply irreducibility since if $f_n(x) = g(x) h(x)$, then since $f_n$ has only one root of size greater than one, only one of $g, h$ can have non-zero constant term. This implies that 0 is a root of either $g$ or $h$, but 0 is not a root of $f_n$ for any $n$).

I am attempting to proceed by induction using Rouche's theorem. The idea is that if $f_n$ has $n-1$ roots in the unit circle, then $x f_n$ has $n$ roots in the unit circle. Therefore, if it's true that

$$ |f_{n+1} - x f_n | = 1 < |f_{n+1}| + |x f_{n}| $$

for all $x$ on the unit circle, then $f_{n+1}$ has $n$ roots inside the unit circle. However, I am not able to prove that we always have strict inequality above.

I suspect that this family $f_n$ has been studied before, but I was not able to find it in the literature. Any help would be appreciated.


This family of polynomials is studied in The Galois group of $x^n - x^{n-1} - \cdots - x - 1$ by Paulo Martin. It is shown that each $f_n$ is irreducible in Corollary 2.2.

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    $\begingroup$ To summarize the argument, which itself refers to a previous result: the reason is that non-real roots have absolute values less than 1, it immediately yields that each factor over $\mathbb{Z}$ has to have a real root, but it is unique. $\endgroup$ – Fedor Petrov Jul 14 '16 at 22:47

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