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Prove that $P_n(X) = X^n - X^{n-1} - X^{n-2} - ... - X - 1$ is irreducible over $\mathbb{Z}$ for all $n$.

I was able to prove the result for $n=2^k-1$ by applying Eisenstein's criterion to $P_n(X+1)$. But for other values of $n$, I'm stuck. Has anyone an idea on this ?

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  • $\begingroup$ Try substituting $X\mapsto X-a$ and check the irreducibility of $P_n(X-a)$. My gut tells me you should try $X-1$ but I'm not totally sure about that. $\endgroup$ – Ian Coley Mar 18 '13 at 19:27
  • $\begingroup$ Actually, I've written a program in Maxima to check values of $a$ automatically (using Eisenstein), but it seems the only usable value is $a=-1$ for $n=2^k-1$. $\endgroup$ – Jean-Claude Arbaut Mar 18 '13 at 19:30
  • $\begingroup$ Ah, okay. How about we view $P_n(X)=X^n-(X^{n-1}+\ldots+1)$? Maybe you can show that we can never have $x^{n-1}+\ldots+1=x^n$ for any $x\in\mathbb Z$? Edit: sorry I'm not doing more directly and just offering maybe-helpful suggestions; I'm strapped for time at the moment. $\endgroup$ – Ian Coley Mar 18 '13 at 19:36
  • $\begingroup$ It's not enough. One can easily prove $P_n$ has no root in $\mathbb{Z}$ (an integer root would be $+1$ or $-1$, given leading coefficient and constant term), but that does not prove it's irreducible. $\endgroup$ – Jean-Claude Arbaut Mar 18 '13 at 19:39
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Use the same idea used in the proof of Perron's irreducibility criterion:

  • Prove that $P_n(x)$ has one and only one root $a$ with $|a|>1$ and none with $|a|=1$ (this is the difficult part. To prove this use that $(x-1)P_n(x)=x^{n+1}-2x^n+1$).

  • If $P_n(x)$ was reducible with $P_n(x)=f(x)g(x)$ ($1\leq\deg(f)<n$) then one of the polynomials $f,g$ has all of its roots inside the unit circle and constant term $\pm1 \ \Rightarrow\Leftarrow$.

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    $\begingroup$ I see the idea. Since product of roots equals constant term, if $P=f \ g$ there must be at least two roots with absolute value > 1. And for the first part, it's easy to prove there is only one real root with absolute value > 1. For complex roots, it looks more difficult, but at least, numerically it seems to be true. I'll try Rouché's theorem, it's the only one I know to find number of roots in a region. $\endgroup$ – Jean-Claude Arbaut Mar 19 '13 at 8:18
  • $\begingroup$ By the way, is there some reference about Perron's criterion ? I'd like to know more about it. $\endgroup$ – Jean-Claude Arbaut Mar 19 '13 at 8:19

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