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This problem is inspired by this work, which is related to musical applications.

Briefly, given $A(X)=1+X+X^{2^k}$ in $\mathbb{F}_2[X]$, we would like to find the smallest $L$ (and optionally the polynomial $E(X)$ of $\mathbb{F}_2[X]$) such that we have $1+\ldots+X^{L-1}=A(X)E(X)$. The above paper proves, in a rather complex way, that $L=4^k-1$.

Looking at a simpler approach, I observed that in $\mathbb{F}_2[X]/A(X)$, we have $X^{4^k}=1+X^{2^k}=1+1+X=X$, so that we can deduce easily that $L$ divides $4^k-1$.

The question is: how do we prove from here that $L$ is exactly $4^k-1$ ?

OliverX1
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  • Possibly related/useful: Assuming I made no mistakes here, the irreducible factors of $A(X)$ are of the form $p(X^2+X)$, where $p(X)$ is an irreducible polynomial of degree $m\mid k$ such that A) the coefficient of degree $m-1$ term of $p(x)$ is equal to $1$, and B) $k/m$ is odd. – Jyrki Lahtonen May 12 '17 at 20:17
  • A straightforward approach would be to find the prime factors $p_i$ of $4^k-1$ and then check that $L=(4^k-1)/p_i$ doesn't work for any of those primes. Doesn't sound easy, does it? – Jyrki Lahtonen May 12 '17 at 20:19
  • $L=(4^k-1)$ is not necessarily minimal for all the irreducible factors of $A(X)$. When $k=4$ we have $m(x):=(1 + x + x^3 + x^4 + x^5 + x^6 + x^8)\mid x^{16}+x+1$, and $x^{85}\equiv1\pmod{m(x)}$. I think this question is a bit delicate. Of course, it suffices to prove that to each prime $p_i\mid 4^k-1$ the polynomial $A(x)$ has at least one factor such that $L=(4^k-1)/p_i$ doesn't work. I have to stop for now, but I'm not giving up. An interesting question. – Jyrki Lahtonen May 12 '17 at 20:44

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