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Math Help - Quadratic Gauss Sum

  1. #1
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    Quadratic Gauss Sum

    Let the quadratic Gauss sum be defined by

    G(m;n)=\underset{r=1}{\overset{n}{\sum}}\omega^{mr  ^{2}}, where \omega=e^{2\pi i/n}.

    I have managed to show that \omega^{mr^{2}}=\omega^{ms^{2}}, whenever mod\: n)" alt="r\equiv s\mod\: n)" />.

    My problem is that I don't know how to deduce that G(m;n)=\{\sum}\omega^{mr^{2}}, where the summation extends over ANY complete set of residues.

    Could anybody help me with this deduction please?
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  2. #2
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    Quote Originally Posted by Cairo View Post
    Let the quadratic Gauss sum be defined by

    G(m;n)=\underset{r=1}{\overset{n}{\sum}}\omega^{mr  ^{2}}, where \omega=e^{2\pi i/n}.

    I have managed to show that \omega^{mr^{2}}=\omega^{ms^{2}}, whenever mod\: n)" alt="r\equiv s\mod\: n)" />.

    My problem is that I don't know how to deduce that G(m;n)=\{\sum}\omega^{mr^{2}}, where the summation extends over ANY complete set of residues.

    Could anybody help me with this deduction please?

    If S:=\{s_1,...s_n\} is a complete set of residues modulo n, then \displaystyle{\forall i=0,1,...,n\,\,\exists !s_{j(i)}\in S} s.t.

    \displaystyle{s_{j(i)}=i\!\!\pmod n\Longrightarrow \omega^{mi^2}=\omega^{ms_{j(i)}^2}} and thus the corresponding sums are equal...

    Tonio
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