Does $i^n=j^n$ for $i, j \in GF(2^q)$ and $i \neq j$ for some $n<2^q-1$ Let $i, j \in GF(2^q)$ and $i \neq j$ and $i,j\neq0$.

Is that possible that $i^n=j^n$ for some $n$ such that $0 &lt; n &lt; 2^q-1$?

I am looking for a proof if the answer is no, or for a method to find $n$ if the answer is yes.
Actually, the answer is "such an $n$ exists iff neither $i, j$ are 0".

We have $x^{2^q-1} = 1$ for $x \ne 0$ (because multiplication in $GF(2^q)$ over the nonzero elements forms a group of order $2^q-1$), hence if $i, j \ne 0$, $n=2^q-1$ is one answer (if not necessarily the minimal one).

If we add the simple observation that if one of $i, j$ is 0, then $i^n = j^n$ cannot happen; this gives us the full answer for all possible values $i, j$.

If you want to extend the question to find the minimal $n$, then you can do this if you know the factorization of $2^q-1$.  That's because if $i^n = j^n$, then $n$ must be a divisor of $2^q-1$; hence you can just search through the divisors of $2^q-1$.

If $2^q−1$ is a Mersenne prime, then no such $0 &lt; n &lt; 2^q-1$ will exist, no matter what $i,j$ are. However, if it is composite, then such $i,j$ will exist (and, in fact, you can find such an $i,j$ pair for every nontrivial divisor $n$ of $2^q−1$). If $g$ is a generator, then two such values are $g^{a\cdot(2^q−1)/n}$, $g^{b\cdot(2^q−1)/n}$ for $0\le a,b&lt;n$