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How to prove that $x^n+x^{n-1} +⋯+x^2+x−1$ is irreducible over $\mathbb Q$?

Below I show you my (unsuccessful) attempt to prove it using Eisenstein's criterion when $n+1$ is a prime. First, observe

$$\frac{x^{n+1} - 2x + 1 }{ x-1} = x^n + x^{n-1} + \cdots + x^2 + x - 1.$$

Next, substitute $x+1$ for $x$, and obtain

$$ {\frac {(x+1)^{n+1}-2x - 1}{x}}=x^n + {\binom {n+1}{n}}x^{n-1}+\cdots +{\binom {n+1}{2}}x+ (n-1).$$

We need that all coefficients, except the coefficient of $x_n$, be divisible by $n+1$ and $(n+1)^2$ must not divide the constant coefficient $n-1$. But $n+1$ must divide the constant coefficient! And this is not the case...

How to proceed further? What to do in general case?

kerzol
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  • See solution by Nikolslife here. – Jean Marie Apr 28 '21 at 22:10
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    @JeanMarie Did you notice the negative sign? Besides, OP already knows how to do it for $n$ prime. – Ted Shifrin Apr 28 '21 at 22:11
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    Are you sure Eisenstein applies when $n$ is prime? That works when the constant coefficient is $+1$, not here. And it works when $n+1$ is prime, no? – Ted Shifrin Apr 28 '21 at 22:19
  • probably I was in a hurry, let me check it out... – kerzol Apr 28 '21 at 22:25
  • Good remark, @TedShifrin my attempt does not work. – kerzol Apr 28 '21 at 22:56
  • I have never seen this exercise before. Mathematica is working on verifying that it is correct for $n\le 1000$, so I'm inclined to believe it's correct. But I have no idea how to do it. Where did you find this? (Yes, yours is the standard trick with the $+1$ there.) – Ted Shifrin Apr 28 '21 at 23:00
  • So, maybe it is more suitable for Mathoverflow... I do not know, because my knowledge in algebra is not perfect. – kerzol Apr 28 '21 at 23:18
  • It is not a classic textbook exercise. It actually arises when I try study the asymptotic number of 1s in all n-Fibonacci binary words of length k. A binary word of length k is called n-Fibonacci when it avoids a consecutive subword $\underbrace{1..1}_{n \text{ times}}$. For instance, for n=2 and k=2 there is three such words containing in total two 1s {00,10,01}; and for n=2 and k=3 there is five such words containing in total five 1s {000,010,101,001,100} (see Knuth's TAoCP Vol 3, p 286) – kerzol Apr 28 '21 at 23:41

2 Answers2

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You can also use the following criterion by Brauer on the (negative) reciprocal $$-x^nf(1/x)=x^n-x^{n-1}-\dots-1.$$

Let $a_1 \geq a_2 \geq \dots \geq a_n$ be positive integers and $n \geq 2$. Then the polynomial $p(x)=x^n-a_1x^{n-1}-a_2x^{n-2}-\dots-a_n$ is irreducible over $\mathbb{Z}$.

For the proof see for example Prasolov's Polynomials (Theorem 2.2.6).

Sil
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    It works also, thanks Sil. If only I could mark with a green checkmark the two answers... – kerzol Apr 29 '21 at 12:43
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This is result of Perron and Selmer, after your $x+1$ trick. See Theorem 2 here:

Selmer, Ernst S., On the irreducibility of certain trinomials, Math. Scand. 4, 287-302 (1956). ZBL0077.24602.

Igor Rivin
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  • If you don't have access: https://documentcloud.adobe.com/link/track?uri=urn:aaid:scds:US:7a86ef36-6140-4cf3-80ac-96d64b33edad Eat this comment after using. – Igor Rivin Apr 28 '21 at 23:24
  • Thanks Igor. It's exactly what I'm needed. Now I'll read it :) – kerzol Apr 28 '21 at 23:30
  • @kerzol unfortunately the body of the argument is in Perron's paper referenced in German, which may or may not bother you. – Igor Rivin Apr 28 '21 at 23:36
  • Don't worry, I'll use some art of googling. For example, one of the Perron's results is also proven in Victor v. Prasolov, Polynomials (Thm 2.2.5). – kerzol Apr 29 '21 at 00:12