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

From this, for $d∣p−1,$ the polynomial $x^d−1\equiv0\pmod p$ has exactly d roots mod p.
As $\frac{p-1}r=d, (x^r)^d=x^{r\cdot d}=x^{p-1}$ which is $\equiv1\pmod p$ using Fermat's Little Theorem if $(x,p)=1$
Now, $x$ can assume $p-1$ in-cogruent values.
So, for integer $y, 1\le y\le p-1$ $y^r, (y,p)=1\implies y$ is a solution of $x^d\equiv1\pmod p$
