Past exam of the short options course of the University of Oxford / 2023 / 1 / a / i / Solution

The multiplicative group ${\mathbb{F}}_p^{\times }$ is a cyclic group of order $p-1$, so choose a primitive root $g$. Every nonzero residue is uniquely $g^j$, with $j$ taken modulo $p-1$. The condition becomes $(p-1)\mid kj$. Writing $d=\gcd (k,p-1)$, its solutions are $j=0,(p-1)/d,\dots ,(d-1)(p-1)/d$. Zero is not a solution. Therefore the number isThis includes $p=2$. The cyclic group argument gives the exact count, whereas the bound of $k$ roots for a nonzero degree-$k$ polynomial over a field would only give an upper bound.