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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.

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This is clearly false, we have $x^{4}+x+1=(x+1) \left(x^3-x^2+x+1\right)$ –  Bombyx mori Dec 25 '12 at 6:53
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Your RHS is $x^4+2x+1$. I verified this for $n\leq 100$ in sage. –  JSchlather Dec 25 '12 at 6:54
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@user32240 What you have is incorrect. $(-1)^4 + (-1) + 1 = 1$ –  user17762 Dec 25 '12 at 6:55
    
I see. But I believe now I found one. –  Bombyx mori Dec 25 '12 at 6:56
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Sorry ignored the modulo condition given. –  Bombyx mori Dec 25 '12 at 6:58
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1 Answer

up vote 12 down vote accepted

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

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I found this by Googling "x^n+x+1 irreducible" which led me to this mathoverflow link. –  alex.jordan Dec 25 '12 at 7:41
    
Very nice, thank you. –  Eric Naslund Dec 25 '12 at 7:50
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+1. Improvement suggestion: Add the word "Yes" to the answer. –  user3533 Dec 25 '12 at 9:21
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