Polynomial discrete mathematics I ran into this question:
Let $p$ be a prime number. We will work on $\mathbb{Z}_{p}$.
Let $d$ be a divisor of $p-1$, $(p-1)/d=r$.
Show that the equation $x^{d}=1$ has exactly $d$ solutions on $\mathbb{Z}_{p}$
and they are: $1^r$, $2^r$, ..., $(p-1)^r$.
Thanks in advance,
Yaron

hint: $\mathbb{Z}_{p}$ is field 
and i know  $\mathbb{Z}_{p}-\{0\}$ is group with multiplication.
hence $$\forall  x \in \mathbb{Z}_{p}-\{0\}=\{\overline 1,...,\overline {p-1}\} :x^{\phi( p)}=1$$($o(a)||\mathbb{Z}_{p}-\{0\}|=\phi( p)$)
then $x^{p-1}=1$ and $x^{dr}=1$  and clearly answer is :$\{ 1^r,..., ({p-1})^r\}$
