Question: Show that $3$ is primitive root $\bmod17 $ and find the indexes of all integers $a$ such that $\gcd (a,17)=1$ relative to $3$.
Answer: First of all, $17$ is an odd prime number. So, there are exactly $\phi(\phi(17))=\phi(16)=8$ primitive roots $\bmod17 $.
We can easily find (I don't want to waste your time with this) that $\text{ord}_{17}(3)=16=\phi(17)$.
Now, the set $\{3^i:i=0,1,2,3,...,15\}$ is a reduced residue system $(\bmod 17)$.
We want to find all the indexes of $1,2,...,16$ relative to $3$.
With simple calculations
- $1\equiv 3^{0} \bmod 17 \iff \text{ind}_3 1 =0 $,
- $2\equiv 3^{14} \bmod 17 \iff \text{ind}_3 2 = 14 $
- $3\equiv 3^{1} \bmod 17 \iff \text{ind}_3 3 = 1 $
- etc...
and we find the indexes of the numbers prime to $17$ between and $16$.
My question is, shall we say something for the indexes of numbers over $16$, and, if yes, what?
Thank you and I apologize for my English.