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