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If $p$ is a prime, show that the product of the $\phi(p-1)$ primitive roots of $p$ is congruent modulo $p$ to $(-1)^{\phi(p-1)}$.

I know that if $a^k$ is a primitive root of $p$ if gcd$(k,p-1)=1$.And sum of all those $k's$ is $\frac{1}{2}p\phi(p-1)$,but then I don't know how use these $2$ facts to show the desired result.
Please help.

  • 3
    Hint: If $a$ is a primitive root, so is $a^{-1}.$2012-07-12
  • 0
    Use $a^p=a \mod p$.2012-07-12
  • 2
    What Geoff says. And remember that $\phi(n)$ is even more often than not.2012-07-12

3 Answers 3