Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
    
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. –  Ian Coley Mar 18 '13 at 19:27
    
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 $n=2^k-1$. –  Jean-Claude Arbaut Mar 18 '13 at 19:30
    
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. –  Ian Coley Mar 18 '13 at 19:36
    
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. –  Jean-Claude Arbaut Mar 18 '13 at 19:39

1 Answer 1

up vote 6 down vote accepted

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 $\pm1 \ \Rightarrow\Leftarrow$.

share|improve this answer
1  
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. –  Jean-Claude Arbaut Mar 19 '13 at 8:18
    
By the way, is there some reference about Perron's criterion ? I'd like to know more about it. –  Jean-Claude Arbaut Mar 19 '13 at 8:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.