# number thoery

• Nov 22nd 2008, 07:53 AM
bigb
number thoery
Let m be within integers and with m>2, If a primitive root modulo m exists, prove that the only incongruent solutions of the congruence x^2 ≡1mod(m)are x ≡1mod(m) and x ≡-1mod(m)
• Nov 22nd 2008, 02:59 PM
ThePerfectHacker
Quote:

Originally Posted by bigb
Let m be within integers and with m>2, If a primitive root modulo m exists, prove that the only incongruent solutions of the congruence x^2 ≡1mod(m)are x ≡1mod(m) and x ≡-1mod(m)

If $x$ is solution then $x\equiv r^y$ where $r$ is primitive root. Then it means $r^{2y}\equiv 1 (\bmod m)$. Therefore, $2y \equiv 0 (\bmod \phi (m))$.
This congruence just has two solutions $y=0, \tfrac{\phi(m)}{2}$.
These two correspond to $x=\pm 1$.