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Given the facts:

(a) 2 is primitive root modulo 101

(b) $2^{50}+1$ isn't divisible by $101^{2}$

I have been asked to show that 2 is a primitive root modulo $101^{101}$.

How do I do that?

I started by defining $m:=\operatorname{ord}_{101^{101}}(2)$.

Then we get $2^{m}=1 \mod 101$. And from (a) we get $100\mid m$.

So in fact $m=100k$, and we would like to show that $k=101^{100}$, therefore $m=101^{100}\cdot100=\phi(101^{101})$.

That's what I have.

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    $\begingroup$ What have you tried? See How to ask a good question. $\endgroup$ – Saad Jun 22 at 13:06
  • $\begingroup$ If you just search for "primitive roots $\pmod {p^n}$" you'll find many useful results. $\endgroup$ – lulu Jun 22 at 13:14
  • $\begingroup$ @Saad question updated $\endgroup$ – Daniel Jun 22 at 13:18
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    $\begingroup$ You've got that $2$ is a primitive root modulo $101$. Then, in number theory, either $a$ or $a+p$ is a primitive root modulo $p^2$. So check to see whether $103$ is a primitive root modulo $101^2$. If not, then $2$ is a primitive root modulo $101^2$ and furthermore, is a primitive root modulo $101^{n}$ which again, is another result that can be proved. $\endgroup$ – thesmallprint Jun 22 at 13:48
  • $\begingroup$ math.stackexchange.com/questions/31679/… $\endgroup$ – lab bhattacharjee Jun 24 at 18:04

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