0
$\begingroup$

Let $d$ be a positive integer, and suppose $c$ is an integer such that $\gcd(c,d) = 1$. Then the following identity holds:

$$\displaystyle \left \lvert \{b \pmod{d} : b^2 \equiv c \pmod{d} \}\right \rvert = \sum_{\substack{\chi \pmod{d} \\ \chi^2 = \chi_0}} \chi(c).$$

What is the correct analogue for the right hand side when we replace the left hand side with

$$\displaystyle \left \lvert \{b \pmod{d} : b^k \equiv c \pmod{d}\}\right \rvert$$

with $k \geq 3$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

If $\gcd(c,d) = 1$ we have

$$\displaystyle \left \lvert \{b \pmod{d} : b^k \equiv c \pmod{d} \}\right \rvert = \sum_{\substack{\chi \pmod{d} \\ \chi^k = \chi_0}} \chi(c).$$

This is not really anything to do with numbers - it works in any finite abelian group, and here we are applying it to the multiplicative group of invertible residue classes mod $c$.

It's a consequence of orthogonality of characters for the group $G/ G^k$.

Improve this answer
$\endgroup$
2
  • 5
    $\begingroup$ Just to elaborate on Will's solution: let $G_d:=(\mathbb{Z}/d\mathbb{Z})^{\times}$ and $f_d(c)=\#\{ b \bmod d: b^k \equiv c \bmod d\}$ be a function on $G_d$. By orthogonality, $f_d(c) = \sum_{\chi \bmod q} \overline{\chi(c)} \widehat{f_d}(c),$ where $$\widehat{f_d}(c)=|G_d|^{-1}\sum_{x \in G_d} f_d(x)\chi(x).$$ Now we can compute $\sum_{x \in G_d} f_d(x)\chi(x) = \sum_{y \in G_d} \chi(y^k) = \sum_{y \in G_d} \chi^k(y)$. If $\chi^k=\chi_0$, this is $|G_d|$. Otherwise, this vanishes by orthogonality. $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 8, 2021 at 19:04
  • $\begingroup$ ($\widehat{f_d}(c)$ should be $\widehat{f_d}(\chi)$.) $\endgroup$
    – Ofir Gorodetsky
    Commented Dec 8, 2021 at 19:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .