Find (if there exist) a primitive root $\bmod 8.$

Question

Recall.

• Definition $1$. $Ord_n(a)$ is the smallest number $m$ st. $\quad a^{m} \equiv 1(\bmod n)$
• Definition $2$. We say $r$ is a primitive root modulo $n$ if $\operatorname{Ord}_{n} (r)=\phi(n)$
• Note: $\left\{1,r, r^2,..., r^{\phi(n)-1}\right\}=U_n$

Find (if there exist) a primitive root $\bmod 8.$

I couldn't find a primitive root, how can I show there isn't any primitive root mod 8?

Answer 1

Hint If you assume by contradiction that $r$ is a primitive roots, then $r$ is odd and hence $$r^2-1=(r-1)(r+1)$$ is the product of two consecutive even integers. One of those is divisible by $4$ and the other one is even, thus $$r^2-1 \equiv 0 \pmod{8}$$