# Primitive roots

• Sep 11th 2008, 05:26 PM
Kiwi_Dave
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:

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

My attempt:

$m*ind_gx \equiv ind_gg$ (mod $\phi(p)$)
$\therefore m*ind_gx \equiv p-1 \equiv 0$ (mod $p-1$)
$\therefore ind_gx \equiv 0$ (mod $p-1$)
$\therefore x \equiv 1$ (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, $\phi(p))$ = m
• Sep 11th 2008, 05:53 PM
ThePerfectHacker
Quote:

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:

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

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