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Math Help - one more primitive root question

  1. #1
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    one more primitive root question

    if m has a primitive root, then the only solutions to x^2 congruent to 1 (mod m) are x is congruen to +-1
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  2. #2
    Super Member PaulRS's Avatar
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    What can m be? - do not forget that what you have there is equivalent to \left. m \right|\left[ {\left( {x - 1} \right) \cdot \left( {x + 1} \right)} \right]<br />
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