PROBLEM: Let p be a prime an let k be a positive divisor of p-1. Show that the congruence
x^k≡1 (mod p)
has exactly k solutions (mod p).
Thanks!
Let $\displaystyle \alpha$ be a primitive root, and let $\displaystyle \beta = \alpha^{\frac{p-1}{k}}$. Then $\displaystyle 1^k = \beta^k \equiv (\beta^2)^k = \dots = (\beta^{k-1})^k$. (It's easy to see that all of $\displaystyle 1, \beta, \beta^2,\dots, \beta^{k-1}$ are distinct, so we have $\displaystyle k$ solutions.) Moreover if $\displaystyle x$ is a solution, we can write $\displaystyle x=\alpha^y$ for some least nonnegative integer $\displaystyle y$; and then since $\displaystyle \alpha^{yk}=1$ we must have $\displaystyle p-1 \mid yk \Rightarrow yk=m(p-1)$ and then we have that $\displaystyle x=\alpha^{\frac{p-1}{k}m}$ is one of our solutions above.