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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Originally Posted by paupsers 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? $\displaystyle g^{nu} \equiv g^i(mod p) \iff nu \equiv i (mod (p-1))$ I suppose $\displaystyle (g,p)=1\Longrightarrow g^{p-1}=1\!\!\pmod p\mbox{ , by Fermat's Little Theorem }\Longrightarrow $ $\displaystyle \Longrightarrow g^{nu}=g^{i}\!\!\pmod p\iff g^{nu-i}=1\!\!\pmod p\iff nu-i=0\!\!\pmod{p-1}$ Tonio
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.
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