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Math Help - number thoery

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    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)
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    Quote Originally Posted by bigb View Post
    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.
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