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Math Help - the sum of a^m

  1. #1
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    the sum of a^m

    if we have any positive integer m ,
    the sum of a^m = -1 if (q-1)|m or 0 otherwise.
    Note, the sum is over a belongs to Fq , where Fq is a field.
    I appreciate your help,
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  2. #2
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    Quote Originally Posted by Mike12 View Post
    if we have any positive integer m ,
    the sum of a^m = -1 if (q-1)|m or 0 otherwise.
    Note, the sum is over a belongs to Fq , where Fq is a field.
    I appreciate your help,


    Hint: 1) \displaystyle{(q-1)\mid m\Longrightarrow m=t(q-1)\Longrightarrow \sum\limits_{a\in\mathbb{F}_q}a^m=\sum\limits_{a\i  n\mathbb{F}_q}(a^{q-1})^t}

    2) \displaystyle{(q-1)\nmid m\Longrightarrow m=t(q-1)+r\,,\,\,0<r<q-1\Longrightarrow \sum\limits_{a\in\mathbb{F}_q}a^m=\sum\limits_{a\i  n\mathbb{F}_q}(a^{q-1})^t\cdot a^s} .

    But there exists b\in\mathbb{F}_q,\,\,s.t.\,\,b^s\neq 1 (why?), so

    \displaystyle{b^s\sum\limits_{a\in\mathbb{F}_q}a^s  =\sum\limits_{a\in\mathbb{F}_q}(ba)^m , and since we have a group here the last sum runs over

    all the elements of \mathbb{F}_q^* and zero, so \displaystyle{b^s\sum\limits_{a\in\mathbb{F}_q}a^m  =\sum\limits_{a\in\mathbb{F}_q}a^m} , thus...

    Tonio
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  3. #3
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    Thanks Tonio,
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