What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$? What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
I know that $289 = 17 \times 17$
so would it be $2^8\equiv 256\bmod17 =1$
and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

Define the set $H \subset {\displaystyle (\mathbb {Z} /289\mathbb {Z} )^{\times }}$ by
$\tag 1 H = \bigr\{[a + 17m] \,\large \mid \, \normalsize  a \in \{-1,+1\} \text{ and } 0 \le m \lt 17\bigr\}$
It is easy to show that $H$ contains exactly $34$ elements.
Proposition 1: The set $H$ is closed under multiplication.
Proof
Consider,
$\quad (a + 17m)(b+17n) = ab + 17(an +bm) + mn\cdot 17^2$
while dividing $an +bm$ by $17$ to get the non-negative residue. $\quad \blacksquare$
So we can state (see bullet $1$ of this elementary group theory)
Proposition 2: The set $H$ forms a group of order $34$.
Continuing,
Proposition 3: The element $[16]$ generates $H$.
Proof
The order of $[16]$ must divide $34$.
The order of $[16]$ is not equal to $2$. Moreover, by applying the binomial theorem we can write
$\quad 16^{17} = \bigr((-1) + 17\bigr)^{17} = (-1)^{17} + \binom{17}{16}(-1)^{16}\cdot 17^{1} + K\cdot 17^2 \equiv -1 \pmod{289}$
and so the order of $[16]$ must be $34$. $\quad \blacksquare$
There are two methods we can use here to finding the order of $[2]$.
Method 1:
Since $[2]^4 = [16]$ and $[2] \notin H$ the order of $[2]$ is strictly greater than $34$. Also, with this fact and
$\quad [2]^{136} = [16]^{34} = [1]$
we must conclude that the order of $[2]$ is either $68$ or $136$.
Now
$\quad [2]^{68} = [16]^{17} \ne [1]$
and we therefore conclude that the order of $[2]$ is $136$.
Method 2
Since $[2]^1, [2]^2, [2]^3 \notin H$ and $[2]^4 = [16] \in H$ we can employ the group theory found here and conclude that the order of $[2]$ is $4 \times 34 = 136$.
