Let $n$ be a positive integer such that there exists a polynomial $f(x) \in \mathbb{F}_2[x]$ of degree n such that $f(x)\cdot x^n f(1/x) = 1+x+\cdots + x^{2n}\in \mathbb{F}_2[x]$. Prove that the order of 2 modulo 2n+1 is odd.
Note that we just need to show that there exists some odd positive integer d so that $2^d \equiv 1\mod (2n+1)$, since then the order of 2 will divide d. There might be some useful property about roots of an irreducible polynomial in $\mathbb{F}_p[x]$ for a prime p. I know that $\mathbb{F}_p[x]$ is a UFD (since $K[x]$ is a Euclidean domain whenever K is a field). Also, $x^{2n+1} - 1 = (x-1) f(x)x^n f(1/x)$. So if $\alpha$ is a root of $x^{2n+1}-1$ that's not 1, then $f(\alpha) \alpha^n f(1/\alpha) = 0$. So either $\alpha$ or $1/\alpha$ is a root of $f$.