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if $n$ is primitive root for odd prime $p$, then $-n$ is also a primitive root for $p$ $\iff p = 1 \text{ mod } 4$

I am having trouble solving this question, can someone show to me how to solve it?

Thank you.

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  • $\begingroup$ If $n$ is a primitive root, then $-1\equiv n^{(p-1)/2}\pmod p$. $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 20:16
  • $\begingroup$ I know that $1 = n^{(p-1)} \text{ mod p}$, but why $-1 = n^{(p-1)/2} \text{ mod p}$ ? $\endgroup$ – Enllwx Nov 1 '18 at 20:32

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