somebody please help me with this one:
If r is a primitive root of p^2, p being an odd prime. show that the solutions of the congruence X^(p-1)=1 (mod p^2) are precisely the integer r^p, r^(2p),......r^((p-1)p)
thanks
Check that all of these are solutions by substituion and that $\displaystyle r^i \not \equiv r^j ~ (\bmod p^2)$ for $\displaystyle i\not = j$.
Now if $\displaystyle x$ is solution then $\displaystyle x\equiv r^y$ and so $\displaystyle r^{(p-1)y} \equiv 1(\bmod p^2)$. Therefore, $\displaystyle (p-1)y \equiv 0 (\bmod p(p-1))$.
Now $\displaystyle \gcd (p-1,p(p-1)) = p-1$ so there are $\displaystyle p-1$ solutions to this congruence. Which means we found all of them.