how to find p-th primitive root of unity in GF(2^m)

Question

The definition of quadratic residue codes involves finding a primitive p-th root of unity in some finite extension field of $GF(2)$. 2 is a quadratic residue of prime $p$. By brute force search I found solutions for $p=7, \zeta=Z(2^3)$ and $p=23,\zeta=Z(2^{11})^{89}$; $Z(2^m)$ is a primitive element in $GF(2^m)$. My question is how is this solved in general? In particular, what is the solution for $p=47$?

Answer 1

This question was answered to some satisfaction in the comment by Jyrki Lahtonen, as written by the poster in another comment.

From the comments:

1. "In every group, if an element $a$ has order $n$, then $a^k$ has order $n / \text{gcd}(n,k)$. If $\zeta$ is a primitive element of the field $GF(2^{23})$ it has order $2^{23}-1$. As $$\frac{2^{23}-1}{47} = 178481,$$ we conclude that $\zeta^{178481}$ is a root of unity of oder 47."

2. "See for example this old thread and others linked to it."

3. "Compare: The exponent $89$ is similarly gotten as $(2^{11}-1)/23$."