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I am currently investigating whether the polynomial $x^6 + x^3 + 1 \in \mathbb{Z}_2[x]$ divides $(x^4 + 1)^n + 1$ for some $n \geq 1$.

I think it is impossible, and so far, I've checked in the case where $n$ is a power of 2:

Suppose $n = 2^m$ for some $m \in \mathbb{N}$. In this case, $(x^4 + 1)^n + 1 = x^{4n} + 1 + 1 = x^{4n}$. So if we assume $x^6 + x^3 + 1 \mid x^{4n}$, then $x \in \sqrt{\langle x^6 + x^3 + 1 \rangle} = \langle x^6 + x^3 + 1 \rangle$, which is not possible. (Here, $\sqrt{\cdot}$ denotes the radical of an ideal.)

However, I'm struggling seeing an argument to prove this is impossible when $n$ is not a power of $2$.

Arturo Magidin
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beanshadow
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1 Answers1

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This $n$ exists. Let $F$ be the splitting field of $x^6+x^3+1$ over $\Bbb F_2$. Let $n=|F|-1$. Then for every $y\ne 0$ in $F$ we have $y^n=1=-1$, so $y^n+1=0$. The polynomial $x^6+x^3+1$ has $6$ roots in $F$ and no double roots. Since every root of $x^6+x^3+1$ is a root of $(x^4+1)^n+1$, $x^6+x^3+1$ divides $(x^4+1)^n+1$.

markvs
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