How to show that 2 is primitive root modulo $101^{101}$?

Asked Jun 22 '20 at 13:03

Active Jun 22 '20 at 14:09

Viewed 469 times

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.

edited Jun 22 '20 at 14:09

Bernard

asked Jun 22 '20 at 13:03

Daniel

1

What have you tried? See How to ask a good question. – Ѕᴀᴀᴅ Jun 22 '20 at 13:06
If you just search for "primitive roots $\pmod {p^n}$" you'll find many useful results. – lulu Jun 22 '20 at 13:14
@Saad question updated – Daniel Jun 22 '20 at 13:18
3

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. – thesmallprint Jun 22 '20 at 13:48
https://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 Jun 24 '20 at 18:04

0 Answers0