# 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 $0 < n < 2^q-1$?

I am looking for a proof if the answer is no, or for a method to find $n$ if the answer is yes.

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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 $0 < n < 2^q-1$ 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 $0\le a,b<n$

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Thanks poncho! Actually, I am looking for the case $i \neq 0$ and $j \neq 0$. I also made an error on the bounds of $n$. It should have been $0 < n < 2^q-1$. You are right that $n=2^q-1$ is one answer. But I am not sure if a smaller $n$ with the desired property exists. Could you provide a reason or a counter example, at least. – Helium Apr 27 '14 at 7:05
@Mohsen: look at my last paragraph; if $2^q-1$ is a Mersenne prime, then no such $n$ 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 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 $0 < a,b < n$ – poncho Apr 27 '14 at 11:37
I think I got it. Your comment made it clear to me. I'll leave another comment if I had any other question later. Thanks a lot. – Helium Apr 27 '14 at 12:32
@poncho: Your comment adds a lot to the answer, you might want to consider editing it in – figlesquidge May 30 '14 at 9:21
@poncho: You should be able to answer this crypto.stackexchange.com/questions/16481/… – Helium May 30 '14 at 22:27