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Math Help - Finding Root of a Polynomial

  1. #1
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    Finding Root of a Polynomial

    Let p be a prime number. Find all roots of x^(p-1) in Z_p



    I have this definition.
    Let f(x) be in F[x]. An element c in F is said to be a root of multiplicity m>=1 of f(x) if (x-c)^m|f(x), but (x-c)^(m+1) does not divide f(x).

    I'm not sure if I use this idea somehow or not.
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  2. #2
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    Quote Originally Posted by kathrynmath View Post
    Let p be a prime number. Find all roots of x^(p-1) in Z_p



    I have this definition.
    Let f(x) be in F[x]. An element c in F is said to be a root of multiplicity m>=1 of f(x) if (x-c)^m|f(x), but (x-c)^(m+1) does not divide f(x).

    I'm not sure if I use this idea somehow or not.

    If you meant \mathbb{Z}_p:=\mathbb{Z}/p\mathbb{Z} , then we have here a field, and in it x^r=0\Longleftrightarrow x=0\,,\,\,\forall r\in\mathbb{N} ,

    so now you can solve your problem.

    Tonio
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    If you meant \mathbb{Z}_p:=\mathbb{Z}/p\mathbb{Z} , then we have here a field, and in it x^r=0\Longleftrightarrow x=0\,,\,\,\forall r\in\mathbb{N} ,

    so now you can solve your problem.

    Tonio
    I think, perhaps, the OP meant the polynomail x^{p-1}-1...otherwise the question is boring and we are left wondering why the exponent was p-1; why is it not just arbitrary?

    If this is the case, then a simple application of Fermat's Little Theorem does the trick.
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  4. #4
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    No the question was just x^(p-1)
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  5. #5
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    I don't understand how you ot x^r=0
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by kathrynmath View Post
    I don't understand how you ot x^r=0
    The reason is that \mathbb{Z}_p is a field. As a field has no zero divisors, then you are immediately done. If you have no idea what a field is, then continue reading...

    If x \in \{1, \ldots, p-1\} then gcd(x, p)=1, as p is prime. Therefore, there exists a, b \in \mathbb{Z} such that ax+bp=1 (using the Euclidean algorithm, etc.) That is, there exists a^{\prime} \in \mathbb{Z}_p such that a^{\prime}x \equiv 1 \text{ mod } p ( a^{\prime} is just a \text{ mod } p).

    So, if x^r=0 then 1=a^{-r}x^r=0, a contradiction.
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