Assume it is reducible, then $g(x)h(x)=x^n-x-1$ Note $g(0),h(0),g(1),h(1)=\pm1$ Then I am stuck.
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@Sil This doesn't appear to be exactly the same question. – Math1000 Oct 23 '19 at 21:59
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1@Math1000 Yes, it is. It is a standard fact that a monic polynomial with integer coefficient is irreducible over $\Bbb{Z}$ if and only if it is irreducible over $\Bbb{Q}$. – Jyrki Lahtonen Oct 24 '19 at 04:23
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Very well, I am not so knowledgeable about algebra. I will vote to close this question then. – Math1000 Oct 24 '19 at 04:25