Primitive Root Theorem

**mndi1105** · November 28th 2008, 08:42 AM

Let m be an integer with m >2. If a primitive root modulo m exists, prove that the only incongruent solutions of the congruence x^2=1 mod m are x=1 mod m and x= -1 mod m.

**ThePerfectHacker** · November 28th 2008, 09:27 AM

Originally Posted by mndi1105

Let m be an integer with m >2. If a primitive root modulo m exists, prove that the only incongruent solutions of the congruence x^2=1 mod m are x=1 mod m and x= -1 mod m.

If $x^2\equiv 1(\bmod m)$ then write $x\equiv r^y$ where $r$ is a primitive root.
Therefore, $r^{2y}\equiv r^0 (\bmod m)\implies 2y\equiv 0(\bmod \phi(m) )$ .
Thus, what can you conclude?

Thread: Primitive Root Theorem

LinkBack

Thread Tools

Display

Primitive Root Theorem

Similar Math Help Forum Discussions

Prove if a is a primitive root mod p^2, then it is a primitive root mod p

abstract algebra [primitive root mod n, euler theorem, fermat theorem]

Primitive Root/Wilson's Theorem

Primitive root

primitive root

Search Tags