# Is $x^{2^{n+1}} - x^{2^n} + 1$ is irreducible over the integers for all $n$?

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?

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.
• 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. – lhf May 4 '17 at 10:56