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Assume F, G, M, and N are Laurent polynomials over $\mathbb{R}$, with coefficients either 0 or 1. Assume that $F \neq M$, and that we have $$F(X^{2^n})G(X^{-1})=M(X^{2^n})N(X^{-1}),$$ for $n \in \mathbb{N}, n\geq1$.

I want to show that we have $F(X)G(X^{-1}) \neq M(X)N(X^{-1})$. Unfortunately, I have no idea how this can be proved, except maybe in the case $n=1$ (I'm not even sure this conjecture is valid). In that case, my reasoning is as follows:

We have $F(X^2)G(X^{-1})=M(X^2)N(X^{-1})$, which is also valid over the finite field $\mathbb{F}_2$. In this field $F(X^2)=F(X)^2$, thus we get $F(X)^2G(X^{-1})=M(X)^2N(X^{-1})$. If we assume $F(X)G(X^{-1}) = M(X)N(X^{-1})$, then we have $F(X)=M(X)$ which is a contradiction.

My questions are: is this a valid proof ? Can it be extended to the general case ?

OliverX1
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  • Shouldn't they be if their coefficients is either 0 or 1 ? – OliverX1 Mar 19 '15 at 13:08
  • I think you are right, @OliverX1. As for extending it, simply observe that $F(X^{2^n})=F(X)^{2^n}$ in $\mathbb{F}_2$. – A.P. Mar 19 '15 at 13:54
  • I actually thought of that, but then, following the same reasoning, I would get $F(X)^{2^{n-1}}=M(X)^{2^{n-1}}$, and I don't know if I am allowed to deduce $F(X)=M(X)$ (and thus the contradiction) from that... – OliverX1 Mar 19 '15 at 14:29
  • Well... the ring of polynomials over a field an Euclidean domain, so in particular $\mathbb{F}_2[X]$ is a UFD. Further, in $\mathbb{F}_2[X]$ you don't have to worry about the uniqueness of factorisation up to a unit, because the only unit is $1$. – A.P. Mar 19 '15 at 14:42

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