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#1

$Old$ July 7th, 2009, 10:35 PM

silentbob $silentbob is offline$

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$Default$ 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

$Old$ July 11th, 2009, 06:59 AM

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What can $m$ be?

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

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