Show an infinite family of polynomials is irreducible

Question

I would like to show that the polynomials \begin{equation} x^n - 2x^{n-1} - 1 \end{equation} are irreducible over $\mathbf{Z}$, whenever $n > 2$. I've used some computer algebra systems to check that this is true whenever $n < 1000$ but I couldn't find, for example, a substitution to allow me to apply Eisenstein's criterion.

Answer 1

Hint:

First prove that $z^n - 2 z^{n-1} - 1$ has no zero for $|z| = 1$.

Then apply extended Rouche. We have $|z^n - 1| \le 2 < |z^n - 2z^{n-1} - 1| + |2 z^{n-1} |$ on the unit circle. So $z^n - 2z^{n-1} - 1$ has same number of zeros as $z^{n-1}$ inside the unit circle. i.e., $(n-1)$.

Hence if $z^n - 2z^{n-1} - 1 = f(z)g(z)$ at least one of $f$ and $g$ has all roots of modulus less than one. But then the constant coefficient has absolute value less than $1$. Which means that the constant term of that polynomial has to be zero, as it is an integer polynomial. Which gives a contradiction as zero is not a root of $z^n - 2z^{n-1} - 1$ either.

Edit: (In reference to E.Girgin's comment) Even in the equality case, one can still use Perron, as long as the polynomial has no zero on the unit circle. Just follow the above proof.