Let $p$ be an odd prime. Prove that $a^2$ is never a primitive root $\pmod p$

I came across this question while studying primitive roots. I know it has something to do with the fact that if the order of $a$ is $m$ then for every $k \in \mathbb{Z}$, the order of $a^k$ is $m/(m,k)$. The question is as follows:

Let $p$ be an odd prime. Prove that $a^2$ is never a primitive root $\pmod{p}$.

I would appreciate any help. Thank you.

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Okay. So, what would $m/\gcd(m,k)$ have to be for $a^2$ to be a primitive root? – Hurkyl Mar 19 '12 at 1:13

If $(a,p)=1$, then $1\equiv a^{p-1}=(a^2)^\frac{p-1}{2}$ implies that $\text{ord}_p(a^2)\le\frac{p-1}{2}$.

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$g$ is a primitive root modulo $p$ if and only if the order of $g$ modulo $p$ is $p-1$.

If the order of $a$ is not $p-1$, then $a^2$ has order less than or equal to the order of $a$, hence is not a primitive root.

If the order of $a$ is $p-1$ then what is the order of $a^2$?

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Your hunch is correct: $\rm\:(m,k)>1\:\Rightarrow\:ord(a^k) = m/(m,k) < m = ord(a).\:$ Since $\rm\:a^k\:$ has smaller order than $\rm\:a,\:$ it doesn't have maximal order in $\rm\left<a\right>.\:$ Your problem is the special case $\rm\:2 = k\ |\ m$

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The order of $a^2$ is no more than $\frac{p-1}{2}$ since $a^{p-1} \equiv 1\pmod{p}$ by Fermat's Little Theorem.

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