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Let $p$ be an odd prime. How can Ihow that $a$ is a primitive root modulo $p$ iff $a^{(p-1)/q}\ncong 1 \pmod{p}$ for all prime divisors $q$ of $p-1$. Thanks

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$\LaTeX$ tip: to get the (mod p), use \pmod{p}. It will supply the spacing, the parentheses, and the roman typeface mod. –  Arturo Magidin Feb 1 '12 at 5:48
    
@ArturoMagidin, thanks for the latex tip! –  Edison Feb 1 '12 at 14:50

1 Answer 1

up vote 4 down vote accepted

I hope you can see one direction: if $a^{(p-1)/q}\equiv1\pmod p$ then $a$ is not a primitive root.

Now suppose $a$ is not a primitive root. Then $a^d\equiv1\pmod p$ for some proper divisor $d$ of $p-1$. So all you have to show is that every proper divisor of $p-1$ is a divisor of $(p-1)/q$ for some prime divisor $q$ of $p-1$.

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ah, yes, thank you for that. –  Edison Feb 1 '12 at 5:33

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