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

$256 \equiv 1 \pmod {17}$ but $256\not \equiv 1 \pmod {289}$ which we need.
But not $289 = 17\times 17$ so $\phi (289) = 17\cdot16$ so $2^{17\cdot 16}\equiv 1\pmod {289}$ by Eulers theorem.
But the order might be something smaller that divides $17\cdot 16$.
We can figure that $2^8 = 17*15 + 1 \equiv 17*(-2) + 1\pmod{17^2}$ so
$2^{16} \equiv 17^2 *4 + 2*(-2)*17 + 1 \equiv -67 \pmod {289}$.
So the order of $2$ is not $16$ and thus not anything that divides $16$.  So the order of $2$ will be a multiple of $17$.  be a multiple of $17$ that divides $16*17$.
And $2^{17} \equiv -8*17+2$
$2^{2*17} \equiv (-8*17+2)^2 \equiv -32*17+ 4\equiv 2*17+4 \equiv 38\pmod{289}$.
$2^{4*17} \equiv 4^2*17^2 + 16*17 + 4^2 \equiv 16*17 +16\equiv 18*16\equiv 1*(-1)\equiv -1 \pmod {289}$.
And so $2^{8*17}\equiv (-1)^2 \equiv 1 \pmod {289}$.
So the order of $2$ is $8*17= 136$.
