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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.

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  • $\begingroup$ Check Perron's criteria math.stackexchange.com/questions/270599/… $\endgroup$ – Emre May 11 '16 at 6:58
  • $\begingroup$ Perron’s criterion doesn't apply here. $\endgroup$ – user261263 May 11 '16 at 7:03
  • $\begingroup$ Yes, but one can change the proof accordingly for the equality case. $\endgroup$ – Emre May 11 '16 at 7:05
  • $\begingroup$ @E.Girgin Might not be that easy ... $\endgroup$ – user261263 May 11 '16 at 7:15
  • $\begingroup$ He's not actually wrong... $\endgroup$ – Shubhodip Mondal May 11 '16 at 7:22
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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.

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    $\begingroup$ It's very easy to prove $z^n - 2 z^{n-1} - 1$ has no zero for $|z| = 1$ so +1 from me $\endgroup$ – user261263 May 11 '16 at 7:47
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    $\begingroup$ Yes. Let me just write it down anyway. $z^n - 2z^{n-1} = 1$ for $|z| = 1$ implies $|z-2| = 1$. So, the only possibility is $z=1$, which is not a root. $\endgroup$ – Shubhodip Mondal May 11 '16 at 7:49
  • $\begingroup$ Thank you very much, this is an excellent argument. Do you know if there is something similar for $x^n - 3x^{n-2} - 1$ whenever $n > 3$? I think I can follow through most of the argument in this case but I finish with this polynomial has $n-2$ roots in the unit disk (and ones close to $\sqrt{3}$ and $-\sqrt{3}$) and so I can't conclude that it's irreducible. I should probably open a second question for this. $\endgroup$ – Mark Bell May 11 '16 at 8:32
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    $\begingroup$ For even $n$ I have an easy fix. Let $u$ be a root outside the unit circle. Then the only other root outside the unit circle has to be $-u$. Note that like before if you write $p(z) := z^{n} - 3 z^{n-2} - 1 = f(z) g(z)$, then both $f(z), g(z)$ has to be irreducible polynomials, one will have $u$ as a root and other will have $-u$. Then by uniqueness of minimal polynomial, we will have $f(z) = g(-z)$, implying that $f(0) = g(0)$. Contradiction since $p(0) = -1$. $\endgroup$ – Shubhodip Mondal May 11 '16 at 11:15

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