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Thread: Solution to equation in field

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by featherbox View Post
    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} .
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by featherbox View Post
    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 !
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  5. #5
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    Quote Originally Posted by xixi View Post
    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 $\displaystyle x^{16}=x^{-1}$?
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  6. #6
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    Quote Originally Posted by featherbox View Post
    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 $\displaystyle x^{-1}$.
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