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Let $\chi: (\mathbb Z/N\mathbb Z)^{\times} \rightarrow \mathbb C^{\times}$ be a character, consider

$a=\frac{1}{N}\sum_{i=1}^N \chi(i)i$

where $\chi(n)=0$ if $n$ is not coprime to $N$.

If $\chi$ is the Legendre symbol assigned to an imaginary quadratic field, then $a$ is an algebraic integer up to power of $2$ and $3$, because $a$ is essentially the class number by the class number formula.

What about general case?

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Presumably the question is whether there exists natural numbers $\ell,m$ such that $2^\ell\cdot3^ma$ is an algebraic integer.

A bit of testing gave the following example. Let's use $N=25$. In that case $2$ is a generator of $\Bbb{Z}_N^*$. Let's try a quartic character defined by $\chi(2^t)=i^t$. I will denote by $s(t)$ the smallest positive remainder of $2^t$ modulo $25$ The sum becomes $$ \begin{aligned} \sum_{t=0}^{19}i^ts(t) &=1+2 i-4-8 i+16+7 i-14-3 i+6+12 i\\ &\quad-24-23 i+21+17 i-9-18 i+11+22 i-19-13 i\\ &=-15-5i. \end{aligned} $$ Implying that $a=-(3+i)/5$ which is not of the required form.


If the numerator were $\phi(N)$ instead of $N$ then it might be related to an inner product of group characters, but I'm not sure about that either. I picked $N=25$ for my first test because then $\phi(N)$ has prime factors $>3$. Why do you think these sums would have that form?

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  • $\begingroup$ Thank you! Because the class number formula claim it's such integer in special cases. $\endgroup$ – sawdada Dec 24 '18 at 20:54
  • $\begingroup$ If $N$ is a prime number, maybe that's true. $\endgroup$ – sawdada Dec 24 '18 at 20:55

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