I was wondering how to show that $a^5$ is a primitive root mod $p$ implies that $a$ is a primitive root mod $p$. I've showed that $a^{(p-1)/2} \equiv -1 \pmod {p}$ using Legendre symbols, but I don't think this implies that $a$ is a primitive root.
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If $a^5$ is a primitive root, then by definition all elements mod $p$ are powers of $a^5$, that is, are of the form $(a^5)^k=a^{5k}$. But then all elements are powers of $a$, and $a$ is a primitive root.
(This can only happen when $(p-1,5)=1$, but is not the point here.)
lhf
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We have $\langle a^5\rangle\subseteq \langle a\rangle\subseteq(\mathbb Z/p\mathbb Z)^\times$. Hence $\langle a^5\rangle=(\mathbb Z/p\mathbb Z)^\times$ implies $\langle a^5\rangle= \langle a\rangle=(\mathbb Z/p\mathbb Z)^\times$.
Hagen von Eitzen
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If $a$ were not a primitive root, then there would exist some $n<p-1$ such that $a^n=1$. But then $a^{5n}=1$ and $a^5$ would not be a primitive root either.
ajotatxe
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The contradiction is that $(a^5)^n=1$, but the order of $a^5$ should be greater than $n$. – ajotatxe Dec 01 '14 at 18:11
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Like Order of elements modulo p
OR
If $\operatorname{ord}_ma=10$, find $\operatorname{ord}_ma^6$ ,
more generally, ord$_pa=d\implies$ ord $_p(a^k)=\dfrac d{(k,d)}$
lab bhattacharjee
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