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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What can $\displaystyle m$ be? - do not forget that what you have there is equivalent to $\displaystyle \left. m \right|\left[ {\left( {x - 1} \right) \cdot \left( {x + 1} \right)} \right] $
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