I am searching the parities (or sign) of the power permutations $x^k$ in $GF(2^n)$ for some $n$ and we need $gcd(k,2^n-1) = 1$ to be permutation.
I will use them in S-boxes in cryptographic algorithms, so I need small degrees permutations. The degrees of this permutations are nothing but Hamming weight of $k$. i.e. the number of 1's of the number, in binary notation of $k$. As an example $x^3$ is quadratic, $x^7$ is cubic etc.
OK, then I compile some codes and realize all quadratic permutations have parity even in $GF(2^{4m})$ for some int $m$.
In the rest of finite fields, so in $GF(2^{4m+1}),GF(2^{4m+2}),GF(2^{4m+3})$, I find quadratic and odd permutations.
Is there a mathematics behind this, i am asking since i could not see it, although it seems so. And if you show me a way, it will be great.