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Let $\chi_{k}(n)$ denote the Kronecker symbol $(k|n)$ and $\mathbb{Z}_{k}^{\times}$ the group of units modulo $|k|$. Under which assumptions on $k \in \mathbb{Z} \setminus \{0 \}$ can one conclude the following?

  1. $\sum_{d \in \mathbb{Z}_{k}^{\times}} \chi_{k}(d) = 0$;

  2. $\left| \sum_{d \in \mathbb{Z}_{k}^{\times}} d \chi_{k}(d) \right| \geq |k|$; and,

  3. $k$ divides $\left| \sum_{d \in \mathbb{Z}_{k}^{\times}} d \chi_{k}(d) \right|$.

Clearly, $3.$ implies $2.$ if the sum is non-zero.

NB: This is not a homework problem.

Update: If $k > 0$ is congruent to $0$ or $3$ modulo $4$, then $\sum_{d \in \mathbb{Z}_{k}^{\times}} \chi_{-k}(d) = 0$. However, this particular sum also vanishes if $k$ is one of the following: ${18, 25, 126, 150, 162, 234, 242, \dots}$ Is this sequence known?

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2 Answers 2

For negative $k$, your 2nd and 3rd sums are very closely related to the class number of the field ${\bf Q}(\sqrt k)$, see this link. And the class number is always a positive integer. And I think you want $|k|$ on the right in your second formula.

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Have you done any kind of representation theory? I have done very little number theory, but after googling Kronecker symbol it just seems like we can discuss the more general question about why, for example, given a non-trivial character $\chi$ on a group $G$ do we have that $\displaystyle \sum_{g\in G}\chi(g)=0$. Am I missing something? For example, the first follows by taking $\langle \chi_{\text{triv}},\chi_k\rangle$ and using the orthogonality relations.

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No, this doesn't work. Take for example, $\sum_{d \in \{1,3 \}} \chi_{4}(d) = \chi_{4}(1) + \chi_{4}(3) = 1 + 1 = 2$. –  user02138 Nov 4 '11 at 3:47
    
I apologize, I should have just not said anything. I hope your problem works out (P.S. have you tried looking in Ireland and Rosen?) –  Alex Youcis Nov 4 '11 at 3:51

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