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Let $w$ be a primitive root. That is $w$ has order $p-1 \mod p$. Evaluate $$\left(\frac{w}{p}\right)$$ This is the task I am faced with so far I have only $w^{p-1}\equiv 1 \mod p$

I am having some difficulty with this problem could anyone point me in the right direction thanks. I normally like to note more of my thoughts but I have been thinking about this for a while and nothing of substance has appeared?

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Well if $w$ has order $p - 1,$ then since $(\frac{w}{p}) \equiv w^{\frac{p-1}{2}} \pmod {p}, (\frac{w}{p})$ won't be 1. I think you can deduce what $w$ on $p$ should be.

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  • $\begingroup$ So it would be $-1$ then right? $\endgroup$ – Ryan S Dec 14 '16 at 17:40
  • $\begingroup$ Yes. Because it can't be 0. $\endgroup$ – green frog Dec 14 '16 at 17:44
  • $\begingroup$ But rather than eliminating possible choices, $p - 1$ has to show up in the power sequence of $w,$ and it can only show up at $w^{\frac{p-1}{2}}$ otherwise $w$ cannot be a primitive root. $\endgroup$ – green frog Dec 14 '16 at 17:46
  • $\begingroup$ Okay many thanks! $\endgroup$ – Ryan S Dec 14 '16 at 17:46

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