I have a question about primitive root. The Problem is :
Prove that $3$ is a primitive root of $7^k$ ($k\ge 1$)
I try using mathematical induction on k. But, I don't know how to solve it. If you give me some hints, then I'll appreciate it :) Thank you.
Using Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$.,
Let me denote ord$_ma$ to be Multiplicative order of $a\pmod m$
show that ord$_73=6=7-1$ i.e., $3$ is a primitive root $\pmod7$
Again, from this, we can prove ord$_{(7^2)}3=6$ or $7\cdot6$
But $\displaystyle3^6=729\not\equiv1\pmod{7^2}\implies$
ord$\displaystyle_{(7^2)}3=6\cdot7=\phi(7^2)$ i.e., $3$ is a primitive root $\pmod{7^2}$
