# Thread: Solution to equation in field

1. ## Solution to equation in field

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

A solution or a starting point would be greatly appreciated.

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

A solution or a starting point would be greatly appreciated.
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} .

3. 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.

4. Originally Posted by featherbox
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 !

5. Originally Posted by xixi
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 $\displaystyle x=1,2...,16$

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

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 $\displaystyle x^{-1}=1$ for $\displaystyle x=1,2...,16$

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