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I am trying to show that for any integer $a$, $$e(a/q) = \sum_{d|q, d|a} \dfrac{1}{ϕ(q/d)} \sum_{χ \ (mod \ q/d)} χ(a/d) τ(χ).$$ First I considered the case $(a,q)=1$ and the mentioned equality holds. But for $(a,q)>1$, I tried anything that I knew from the analytic number theory course but I failed. Any useful hint would be appreciated.

The Gauss sum $τ(χ)$ of $χ$ is defined to be $τ(χ) = \sum_{a=1}^q χ(a) e(a/q)$.

Because I have worked on this exercise a lot and checked many theorems I knew, even a little but useful hint might be helpful, thanks!

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  • $\begingroup$ Perhaps you need to use the result in Section 9.2.1 Exercise 1.(a) to prove the result in Section 9.2.1 Exercise 1.(b)? $\endgroup$
    – Steven Clark
    Commented Apr 12 at 20:23
  • $\begingroup$ @StevenClark I proved (a) and that is when (a,q)=1 which is special case of (b). One needs to show (b) is true for (a,q)>1 and in that case other terms appear since there are more divisors d than d=1. $\endgroup$
    – Ali
    Commented Apr 13 at 2:41

2 Answers 2

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By part (a) of the same exercise, there is

$$ \sum_{d|q,d|a}{1\over\phi(q/d)}\sum_{\chi\pmod{q/d}}\chi(a/d)\tau(\chi) =\sum_{\substack{d|q,d|a\\(q/d,a/d)=1}}e\left(a/d\over q/d\right). $$

Since $(q/d,a/d)=1$ iff $d=(a,q)$, we see that the right hand side simplifies to $e(a/q)$.

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Write $q'=q/d$ $a'=a/d$. $$e(a/q)=\sum _{n=1\atop {n\equiv a(q)}}^qe(n/q)=\sum _{d|q}\sum _{n=1\atop {(n,q)=d\atop {n\equiv a(q)}}}^qe(n/q)=\sum _{d|q,a}\sum _{n=1\atop {(n,q')=1\atop {n\equiv a'(q')}}}^{q'}e(n/q').$$ Now proceed as you did for $(q,a)=1$.

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