I know how to find primitive roots of relatively small primes. But how can one possibly find a primitive root modulo $5^{10}$? I can't test every single number less than $5^{10}$. Is there a fast way to locate just one primitive root of $5^{10}$?
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It’s $2$ or $3$ $\bmod 5$. – BAI Oct 19 '17 at 11:59
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http://mathworld.wolfram.com/PrimitiveRoot.html – Darío A. Gutiérrez Oct 19 '17 at 12:18
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Find a primitive root mod $5$. I notice $3$ is a primitive root mod $5$.
Either $3$ or $3 + 5 = 8$ will be a primitive root mod $25$. A quick check shows that $3$ is still a primitive root mod $25$.
A primitive root mod $25$ will be a primitive root mod $5^k$ for all $k \geq 1$.
This works because of a few fundamental ideas concerning primitive roots.
Lemma: If $g$ is a primitive root mod an odd prime $p$, then either $g$ or $g + p$ is a primitive root mod $p^2$.
Proof: This is the theorem in this answer, although the phrasing is a little bit different.
And also,
Lemma: If $g$ is a primitive root mod $p^2$ where $p$ is an odd prime, then $g$ is a primitive root mod $p^k$ for all $k \geq 1$.
Proof: See this question and its answers.
davidlowryduda
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