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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
Your RHS is $x^4+2x+1$. I verified this for $n\leq 100$ in sage. – JSchlather Dec 25 '12 at 6:54
@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
Sorry ignored the modulo condition given. – Bombyx mori Dec 25 '12 at 6:58
up vote 13 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
+1. Improvement suggestion: Add the word "Yes" to the answer. – user3533 Dec 25 '12 at 9:21

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