# Thread: Solution to equation in field

1. ## Solution to equation in field

Find all solutions to the equation $x^{16} = 1$ in the field $Z_{17}$.

A solution or a starting point would be greatly appreciated.

2. Originally Posted by featherbox
Find all solutions to the equation $x^{16} = 1$ in the field $Z_{17}$.

A solution or a starting point would be greatly appreciated.
Since $\mathbb{Z}_{17}$-{0} is a finite group of order 16 , so for every nonzero $x\in \mathbb{Z}_{17}$ we have $x^{16} = 1$ , so the solution is $\mathbb{Z}_{17}$-{0} .

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

4. Originally Posted by featherbox
I got that because $16=-1$ in the field, we have the equation now being $x^{-1}=1$, and seeing as all elements in the field have multiplicative inverses all elements within $Z_{17}$ are solutions.
you can't say $x^{16}=x^{-1}$ , as I said $x^{16}=1$ and from $x^{-1}=1$ you can just conclude that $x=1$ and nothing else !

5. Originally Posted by xixi
you can't say $x^{16}=x^{-1}$ , as I said $x^{16}=1$ and from $x^{-1}=1$ you can just conclude that $x=1$ and nothing else !
Thats not what I said at all.
The solutions would be 1,2,...16. NOT just 1.
This is because $x^{-1}=1$ for $x=1,2...,16$

Why can't $x^{16}=x^{-1}$?

6. Originally Posted by featherbox
Thats not what I said at all.
The solutions would be 1,2,...16. NOT just 1.
This is because $x^{-1}=1$ for $x=1,2...,16$

Why can't $x^{16}=x^{-1}$?
from $x^{-1}=1$ we have $x=1$ 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 $x^{16}=1$ and $x^{16}=x^{17}x^{-1}$ but $x^{17}=x$ and so $x^{16}=x.x^{-1}=1$ and as you see it won't be $x^{-1}$.