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can you offer a proof for this proposition?

$$\frac{1}{d}\sum_{o(\chi )|d} {\chi (u) = \left\{ {\begin{array}{*{20}{c}} {1 \quad \text{if u is a dth power resedue modulo p}}\\ {0 \quad \text{otherwise}} \end{array}} \right.}$$

$o(\chi)=d$ means order of $\chi$ modulo $p$ is $d$.

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  • $\begingroup$ Why is this proposition not famous, too? $\endgroup$ – Dietrich Burde Sep 6 '16 at 19:58
  • $\begingroup$ see what I did there using a discrete logarithm modulo $p$ for expliciting the $\chi$ modulo a prime $p$ (if $p$ is not prime, the concept of discrete logarithm becomes a little more complicated) $\endgroup$ – reuns Sep 6 '16 at 20:01
  • $\begingroup$ Oh, I don't want to reject it. But why analytic proof? What about an algebraic proof? $\endgroup$ – Dietrich Burde Sep 6 '16 at 20:02
  • $\begingroup$ not convinced you can avoid the discrete logarithm ; it is a bijection / group isomorphism $h : (\mathbb{F}_p^*,\times) \to (\mathbb{Z}_{p-1},+)$. So $u = x^d$ becomes $h(u) = d \ h(x)$. And the $\chi$ become the discrete Fourier basis on $\mathbb{Z}_{p-1}$ $\endgroup$ – reuns Sep 6 '16 at 20:07
  • $\begingroup$ @user356319 what do you mean ? and if you don't use the discrete logarithm then you have to use it implicitely, namely that $(\mathbb{F}_p^*,\times)$ is a cyclic group isomorphic to $(\mathbb{Z}_{p-1},+)$ $\endgroup$ – reuns Sep 6 '16 at 20:19

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