Cleverly finding Primitive Roots

Question

From Number Theory we have the theorem that:

There are primitive roots mod $n$ if and only if $n = 1,2,4,p^k, 2p^k$, where $p$ is an odd prime.

The question is, for such general $n$ given as above (with $n \neq p$), how does one find primitive roots without having to brute-force? Perhaps the Chinese Remainder Theorem is relevant? I'm excluding the case $n=p$ because it already has a well-written answer Finding a primitive root of a prime number .

Well I doubt that there might be a way to figure out a primitive root of an integer, but I remember sometimes back in my olympiad training my teacher told us this is some kind of brute force training in early college, some of his cases were $625$ and $2401$. Regarding Chinese Remainder Theorem, i don't think it is that relevant, because you only have one modulo to consider, but you may comment your opinion down here. — Nikola Tolzsek, Dec 30 '20 at 11:12
If $g$ is a primitive root modulo $p$, then $g$ or $g+p$ is a primitive root modulo every power of $p$ — J. W. Tanner, Dec 30 '20 at 11:14
@J.W.Tanner Great answer! Is there an easy way I can take care of the case $2p^k$? — Albert, Dec 30 '20 at 11:17
If $g$ is a primitive root modulo $p^k$, then either $g$ or $g + p^k$ is a primitive root modulo $2p^k$ — J. W. Tanner, Dec 30 '20 at 11:24
Wow, that's a strong statement. Can I find a proof/reference somewhere? — Albert, Dec 30 '20 at 11:26

Albert · Accepted Answer · 2020-12-30T11:55:23.763

Thanks to everyone who commented! From the comments, the answer can be found:

For $n = p^k$, firstly find a primitive root modulo $p$ as in Finding a primitive root of a prime number. Then, use that if $g$ is a primitive root mod $p$, then $g$ or $g+p$ is a primitive root mod $p^k$ for any $k \in \mathbb{Z}_{\ge 1}$. Then for $n = 2p^k$, denoting by $g'$ the primitive root for $p^k$, either $g'$ or $g'+p^k$ is a primitive root mod $2p^k$.

But if anyone wants to post their own methods or proofs of the propositions used (I actually don't know the proofs), please do so!

Cleverly finding Primitive Roots

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