finding equivalent hyperovals

Question

If $H=D(x^k)$ is a hyperoval, then $D(x^t)$ is a hyperoval equvalent to $H$ for $t=1/k$, $1-k$, $1/(1-k)$, $k/(1-k)$ and $(k-1)/k$. If I consider the Segre Hyperoval $D(x^6)$ with $q = 32 = 2^5$, how can I define $x^{1/6}$, $x^{-5}$, $x^{-1/5}$, $x^{-6/5}$, and $x^{5/6}$. How can I transform these exponents to integers?

Answer 1

You want to use the fact that $x^{31} = 1$ for all $x \neq 0$, so you compute the exponents $\bmod{31}$. So, for example, $5^{-1} = -6 = 25$, and $6^{-1} = -5 = 26$.