I'm trying to show that:


More generally, can we also show:


where $s$ is the Dedekind sum. Any ideas?

  • 2
    $\begingroup$ Isn't it easiest to just show that $s(-a,b)=-s(a,b)$? $\endgroup$ Commented May 7, 2014 at 7:30

3 Answers 3


NOTE: OP has changed the question drastically, several times. The answer below relates to one of the versions of the question, and may not have any relevance to the current version.

A couple of standard (and easily verified) properties of the Dedekind sum are

  1. if $i\equiv j\pmod b$ then $s(i,b)=s(j,b)$,

  2. $s(-a,b)=-s(a,b)$, and

  3. $s(ka,kb)=s(a,b)$ for nonzero integer $k$.

It would appear that the conjecture follows instantly from these properties.

A reference for these properties is the Rademacher-Grosswald book I have mentioned in other questions about Dedekind sums.

  • $\begingroup$ You must mean this book? I couldn't find the reference in your other posts. $\endgroup$
    – Alexander Gruber
    Commented May 11, 2014 at 17:31
  • $\begingroup$ @Alexander, yes, that's the book. I think ruadath has been busy deleting Dedekind sum questions, which might explain what happened to my other posts. $\endgroup$ Commented May 11, 2014 at 22:57
  • 1
    $\begingroup$ That's between you and your advisor. Once people have posted mathematical content to this website, I don't think you have the moral right to delete it. $\endgroup$ Commented May 12, 2014 at 23:41

Alright, I have made a key observation in determining how we go about proving the above. We need to prove the following statement. Let $S_d=\{a<b\mid \gcd(a,b)=d\}$. Then we have:

$$\sum_{a\in S_d}s(a,b)=0$$

Sounds like a much more reasonable line of attack. Indeed, the question then reduces to showing that:

$$\sum_{k=1}^b\sum_{a\in S_d}\left\{\frac{ka}{b}\right\}=\frac{1}{2}\sum_{k=1}^b\sum_{a\in S_d}\chi\left(\frac{ka}{b}\right)$$


$$\chi(x)=\begin{cases}0 & x\in\mathbb{Z}\\1 & x\notin\mathbb{Z}\end{cases}$$

We can rewrite the above

$$\sum_{k=1}^b\sum_{a\in S_d}\left\{\frac{ka_0}{b_0}\right\}=\frac{1}{2}\sum_{k=1}^b\sum_{a\in S_d}\chi\left(\frac{ka_0}{b_0}\right)$$

where $a_0,b_0$ are coprime. After commuting the above sums, it is clear that both sides have value:



Let us change the order of summation from increasing to decreasing: $ ∑_{a=b}^0 GCD(a, b) s(a, b) = ∑_{a=0}^b GCD(b - a, b) s(b - a, b) = {∑_{a=0}^b GCD(b - a, b) ∑_{i=0} ^b (({i \over b})) (( {(b-a) i \over b}))} = ∑_{a=0}^b GCD(b - a, b) ∑_{i=0} ^b (({i \over b})) (( i - {a i \over b})).$

Let us note some properties of $((x)) $ function.

  1. $((z + x)) = ((x))$, where $z$ is integer
  2. $ 0 < x < 1$, $((x)) + ((-x))) = x - 0 + {1 \over 2} + (-x) - 1 + {1 \over 2} = 0.$, it implies $((x)) = -((-x))$

So: $∑_{a=0}^b GCD(b - a, b) ∑_{i=0} ^b (({i \over b})) (( i - {a i \over b})) = ∑_{a=0}^b GCD(a, b) ∑_{i=0} ^b (({i \over b})) (-(({ai \over b}))) = {-∑_{a=0}^b GCD(a, b) s(a, b)}$

So: $∑_{a=0}^b GCD(a, b) s(a, b) = 0$

  • $\begingroup$ Moreover, using that idea it could be proven that $∑_{a=b}^0 s(a,b) = 0$ $\endgroup$
    – Rodion
    Commented May 14, 2014 at 15:18

You must log in to answer this question.

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