Since \(\displaystyle \mathbb{Z}_{17}\)-{0} is a finite group of order 16 , so for every nonzero \(\displaystyle x\in \mathbb{Z}_{17}\) we have \(\displaystyle x^{16} = 1\) , so the solution is \(\displaystyle \mathbb{Z}_{17}\)-{0} .

I got that because \(\displaystyle 16=-1\) in the field, we have the equation now being \(\displaystyle x^{-1}=1\), and seeing as all elements in the field have multiplicative inverses all elements within \(\displaystyle Z_{17}\) are solutions.

I got that because \(\displaystyle 16=-1\) in the field, we have the equation now being \(\displaystyle x^{-1}=1\), and seeing as all elements in the field have multiplicative inverses all elements within \(\displaystyle Z_{17}\) are solutions.

you can't say \(\displaystyle x^{16}=x^{-1}\) , as I said \(\displaystyle x^{16}=1\) and from \(\displaystyle x^{-1}=1\) you can just conclude that \(\displaystyle x=1\) and nothing else !

you can't say \(\displaystyle x^{16}=x^{-1}\) , as I said \(\displaystyle x^{16}=1\) and from \(\displaystyle x^{-1}=1\) you can just conclude that \(\displaystyle x=1\) and nothing else !

Thats not what I said at all.
The solutions would be 1,2,...16. NOT just 1.
This is because \(\displaystyle x^{-1}=1\) for \(\displaystyle x=1,2...,16\)

Thats not what I said at all.
The solutions would be 1,2,...16. NOT just 1.
This is because \(\displaystyle x^{-1}=1\) for \(\displaystyle x=1,2...,16\)

from \(\displaystyle x^{-1}=1\) we have \(\displaystyle x=1\) because the only element whose inverse is 1 is itself 1 . Moreover since the order of each element divides the order of the group and the order of this group is 16 then \(\displaystyle x^{16}=1\) and \(\displaystyle x^{16}=x^{17}x^{-1}\) but \(\displaystyle x^{17}=x\) and so \(\displaystyle x^{16}=x.x^{-1}=1\) and as you see it won't be \(\displaystyle x^{-1}\).