# Primitive roots as roots of equations.

Take $g$ to be a primitive root $\pmod p$, and $n \in \{0, 1,\ldots,p-2\}$ write down a necessary sufficient condition for $x=g^n$ to be a root of $x^5\equiv 1\pmod p$ . This should depend on $n$ and $p$ only, not $g$.

How many such roots $x$ of this equation are there? This answer may only depend on $p$.

At a guess for the first part I'd say as $g^{5n} \equiv g^{p-1}$ it implies for $x$ to be a root $5n \equiv p-1 \pmod p$. No idea if this is right and not sure what to do for second part. Thanks for any help.

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Hint. In any abelian group, if $a$ has order $n$, then $a^r$ has order $n/\gcd(n,r)$.

(Your idea is fine, except that you got the wrong congruence: it should be $5n\equiv p-1\pmod{p-1}$, not modulo $p$; do you see why?)

For the second part, you'll need to see what you get from the first part. That will help you figure it out.

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 Thanks for help so far! Yeah I see it should be p-1 but still not sure on how many such roots. From your hint I think the answer is gcd(5, p-1) this shows how many n

$g^{m} \equiv 1 \mod p$ if and only if $\mathrm{ord}_p(g)$ divides $m$.

Since $g$ is primitive root, we get that $p-1=\mathrm{ord}_p(g)$ has to divide $5n$.

Can you finish the problem now?

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