For what positive integers $n$ does $x^n - 1$ factor as $(1+x)(1 + x +\cdots + x^{n-1})$ over $\mathbb{F}_2[x]$

Question

For what positive integers $n$ does $x^n - 1$ factor as $(1+x)(1 + x + \cdots + x^{n-1})$ over $\mathbb{F}_2[x]$

I think the first few terms are $1,3,5,11,13,\dots$

Answer 1

In $\mathbf Z[x]$ we have $x^n - 1 = (x-1)(x^{n-1} + \cdots + x + 1)$ for all $n \geq 1$, as you surely know, so in $\mathbf F_2[x]$ we have $x^n - 1 = (x+1)(x^{n-1} + \cdots + x + 1)$. Your question is asking for which $n \geq 2$ the irreducible factorization of $x^n-1$ in $\mathbf F_2$ is $(x-1)(x^{n-1} + \cdots + x + 1)$ (since $x-1 = x+1$ in $\mathbf F_2[x]$). The answer is: when $n = 2$ or when $n$ is an odd prime such that $2 \bmod n$ has order $n-1$. I don't know why you left out $n = 2$ in your list: $x^2 - 1 = (x-1)(x+1) = (x+1)(x+1)$ in $\mathbf F_2[x]$.

Let $n = 2^km$ where $k \geq 0$ and $m$ is odd. Then in $\mathbf F_2[x]$, $x^n - 1 = x^{2^km}-1 = (x^m - 1)^{2^k}$. If $m = 1$ this is $(x-1)^{2^k}$, which has $2$ irreducible factors only when $k = 1$ (corresponding to $n = 2$). If $m \geq 3$ and $k \geq 1$ there are more than $2$ irreducible factors of $(x^m - 1)^{2^k}$, so if $m \geq 3$ we need $k = 0$: $n = m$ is odd.

For odd $n$, Galois theory for finite fields tells us that the $n$-th roots of unity in characteristic $2$ generate a field extension of $\mathbf F_2$ with degree over $\mathbf F_2$ equal to the multiplicative order of $2 \bmod n$ since that order is the degree over $\mathbf F_2$ of the primitive $n$-th roots of unity. (Primitive $n$-th roots of unity exist in characteristic $2$ when $n$ is not divisible by $2$, i.e., when $n$ is odd.) Therefore you want to know when $2 \bmod n$ has order $n-1$. The order of $2 \bmod n$ divides $\varphi(n)$, which in turn is at most $n-1$, and if $\varphi(n) = n-1$ then $n$ must be prime. Thus if $2 \bmod n$ has order $n-1$, $n$ has to be prime.

There is no genuinely simpler description of the primes $n$ such that $2 \bmod n$ has order $n-1$ other than that description itself. Well, you could rephrase this in terms of something about repeating binary decimals, but that's not really going to help.

There is not much special here about using $\mathbf F_2$: for $n > 1$ and primes $p>2$, the irreducible factorization of $x^n - 1$ in $\mathbf F_p[x]$ is $(x-1)(x^{n-1}+\cdots + x + 1)$ exactly when $n$ is a prime other than $p$ such that $p \bmod n$ has order $n-1$. For $n = 2$ the second factor has degree $1$ and thus is irreducible mod $2$, but this isn't the case in $\mathbf F_p[x]$ if $n = p$ is an odd prime.

Answer 2

$\forall n \in \mathbb{N}:\phi(X):=X^n-1=(X-1)(X^{n-1}+X^{n-2}+\dots+X+1) \in \mathbb{Z}[X]$

Since $\phi$ does only have coefficients $1$, the equation is identical considering $\phi$ as an element in $\mathbb{F}_2[X]$. Moreover, $X-1=X+1$ in $\mathbb{F}_2$ thus one can write $(X+1)(X^{n-1}+X^{n-2}+\dots+X+1)$ in this case.