This is just one line from a larger proof about nth power solutions, but I don't understand how/why it's true. Any help?

$\displaystyle g^{nu} \equiv g^i(mod p) \iff nu \equiv i (mod (p-1))$

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- Apr 3rd 2011, 09:58 AMpaupsersWhy does this congruence hold?
This is just one line from a larger proof about nth power solutions, but I don't understand how/why it's true. Any help?

$\displaystyle g^{nu} \equiv g^i(mod p) \iff nu \equiv i (mod (p-1))$ - Apr 3rd 2011, 02:21 PMtonio
- Apr 3rd 2011, 03:16 PMBruno J.
This isn't true. For example $\displaystyle (-1)^2 \equiv 1 \mod p$, but $\displaystyle 2 \neq 0 \mod p-1$ for $\displaystyle p>2$.

It's true if $\displaystyle g$ is a primitive root.