1. ## Primitive roots

Suppose g is a primitive root modulo p (a prime) and suppose m|p-1 (1<m<p-1). How many solutions are there to the congruence:

$\displaystyle x^m-g \equiv 0$ (mod p)

My attempt:

$\displaystyle m*ind_gx \equiv ind_gg$ (mod $\displaystyle \phi(p)$)
$\displaystyle \therefore m*ind_gx \equiv p-1 \equiv 0$ (mod $\displaystyle p-1$)
$\displaystyle \therefore ind_gx \equiv 0$ (mod $\displaystyle p-1$)
$\displaystyle \therefore x \equiv 1$ (mod p)

Hence g=1 which is a contradiction so there are no solutions.

I don't believe my own answer! I have not used the statement that
m|p-1 which tells me that g.c.d(m,$\displaystyle \phi(p))$ = m

2. Originally Posted by Kiwi_Dave
Suppose g is a primitive root modulo p (a prime) and suppose m|p-1 (1<m<p-1). How many solutions are there to the congruence:

$\displaystyle x^m-g \equiv 0$ (mod p)
If there is a solution $\displaystyle x$ then $\displaystyle x\equiv g^y~ (p)$.
Thus, $\displaystyle g^{my} \equiv g^1 ~ (p)$.
Therefore, $\displaystyle my\equiv 1 ~ (p-1)$.
Now $\displaystyle \gcd(m,p-1) =m$ and $\displaystyle m\not | 1$.
Thus, there are no solutions to this congruence.