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I solved this problem:

Find a complete set of mutually incongruent primitive roots of $13$.

I know that there are $\phi(\phi(13))=4$ primitive roots of 13, which are $2,6,7,$ and $11$. However, I just found these by individually computing powers of every number less than $13$ modulo $13$. Is there a better way to do this problem?

Also, can I use this information to find a primitive root of $13^{901}$? I know that if I can show that $r$ is a primitive root of $13$ and also of $13^2=169$, then it's a primitive root of $13^t$ for any $t\in\mathbb{N}$, but there is a lot of computation involved in this.

MathQuestion
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    Once you have one primitive root $r$ mod 13, the others are just $r^m$ where $m$ runs through the values prime to 12. Also, if $r$ is a primitive root mod 13, then either $r$ or $r+13$ is a primitive root mod $13^2$. Some computation is unavoidable. – Gerry Myerson Dec 08 '15 at 06:27
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    Let's see whether $2$ works. Note that $2^{12}\not\equiv 1\pmod{169}$, easy calculation. So $2$ is a primitive root of $13^2$. – André Nicolas Dec 08 '15 at 06:32
  • Oh, I see. If $2^{12}\not\equiv 1\pmod{169}$ and $2^{13}\not\equiv 1\pmod{169}$, then $2$ must be a primitive root of $13^2$ since $\phi(169)=156=12\cdot 13$. Thank you! – MathQuestion Dec 08 '15 at 07:01
  • Related : http://math.stackexchange.com/questions/31679/if-g-is-a-primitive-root-of-p2-where-p-is-an-odd-prime-why-is-g-a-prim – lab bhattacharjee Dec 17 '15 at 16:32

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