= 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 $j=0,(p-1)/d,\ldots,(d-1)(p-1)/d$. Zero is not a solution. Therefore the number is
$$
\boxed{\gcd(k,p-1)}.
$$
This 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.
Back to article page
