18
$\begingroup$

Motivated by this problem, and KCd's comment on my answer, I am left with the following question:

Question: Suppose that $n\not \equiv 2\pmod{3}$. Is $$x^n+x+1$$ irreducible over $\mathbb{Q}$?

I am not sure how to solve this, any thoughts are appreciated.

  • 0
    This is clearly false, we have $x^{4}+x+1=(x+1) \left(x^3-x^2+x+1\right)$2012-12-25
  • 2
    Your RHS is $x^4+2x+1$. I verified this for $n\leq 100$ in sage.2012-12-25
  • 4
    @user32240 What you have is incorrect. $(-1)^4 + (-1) + 1 = 1$2012-12-25
  • 0
    I see. But I believe now I found one.2012-12-25
  • 3
    Sorry ignored the modulo condition given.2012-12-25

1 Answers 1

13

Yes, it is true. See the second claim of Theorem 1 on page 289.

  • 2
    I found this by Googling "x^n+x+1 irreducible" which led me to [this mathoverflow link](http://mathoverflow.net/questions/56506/irreducibility-of-some-trinomials-modulo-p).2012-12-25
  • 0
    Very nice, thank you.2012-12-25
  • 7
    +1. Improvement suggestion: Add the word "Yes" to the answer.2012-12-25