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It seems $x^2-x+1$ or $x^4-x^2+1$ are irreducible over the integers.

Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible for all non-negative integer $n$? If so, how to prove?

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Multiplying by $ (X^{2^n} + 1) $, we have that

$$ (X^{2^n} + 1)(X^{2^{n+1}} - X^{2^n} + 1) = X^{3 \cdot 2^n} + 1 $$

It is clear from this that the roots of $ X^{2^{n+1}} - X^{2^n} + 1 $ are the primitive $ 3 \cdot 2^{n+1} $th roots of unity, and the degree of the corresponding minimal polynomial is $ \varphi(3 \cdot 2^{n+1}) = 2^{n+1} $. Thus, $ X^{2^{n+1}} - X^{2^n} + 1 $ is the $ 3 \cdot 2^{n+1} $th cyclotomic polynomial, and it follows that it is irreducible.

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  • $\begingroup$ A simpler argument: $y=x^{2^n}$ transforms $x^{2^{n+1}} - x^{2^n} + 1$ to $y^2-y+1$, whose roots are the primitive sixth roots of unity. $\endgroup$ – lhf May 4 '17 at 10:56
  • $\begingroup$ What primitive are you reffering? $\endgroup$ – user2860452 May 5 '17 at 11:24

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