If $g$ is a primitive root modulo $37$, which of the numbers $g^2, g^3,.., g^8$ is a primitive root modulo 37?

Question

If $g$ is a primitive root modulo $37$, which of the numbers $g^2, g^3,.., g^8$ is a primitive root modulo $37$?

This problem is a problem bothering me. Any help would be much appreciated.

Which of the exponents is coprime with $;\varphi(37);$ ? – Timbuc Dec 03 '14 at 04:51 — Timbuc, Dec 03 '14 at 04:51

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

OR

ord$_pa=d\implies$ ord $_p(a^k)=\dfrac d{(k,d)}$

Now $g$ is primitive root $\pmod{37}\iff $ord$_{37}(g)=\phi(37)=36$

$\implies$ord $_{37}(g^k)=\dfrac{36}{(k,36)}$

So, we need $(k,36)=1$ to keep ord$_{37}(g^k)=36$

edited Apr 13 '17 at 12:21

Community

answered Dec 03 '14 at 04:55

lab bhattacharjee

when you say $(k, 36 )$ do you mean $gcd(k, 36)$ ? – Pasie15 Dec 03 '14 at 17:20
@Pasie15, Definitely. This is a common symbol for GCD – lab bhattacharjee Dec 03 '14 at 17:54

1 Answers1