Summation of Dedekind sums to zero I'm trying to show that:
$$\sum_{a=0}^b\text{GCD}(a,b)s(a,b)=0$$
More generally, can we also show:
$$\sum_{a=0}^b\text{GCD}(a,b)^ls(a,b)=0$$
where $s$ is the Dedekind sum. Any ideas?
 A: 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 


*

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

*$s(-a,b)=-s(a,b)$, and 

*$s(ka,kb)=s(a,b)$ 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. 
A: 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:
$$\sum_{k=1}^b\sum_{a\in S_d}\left\{\frac{ka}{b}\right\}=\frac{1}{2}\sum_{k=1}^b\sum_{a\in S_d}\chi\left(\frac{ka}{b}\right)$$
where
$$\chi(x)=\begin{cases}0 & x\in\mathbb{Z}\\1 & x\notin\mathbb{Z}\end{cases}$$
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:
$$|S_d|\frac{b(b_0-1)}{2b_0}=|S_d|\frac{b-d}{2}$$
A: 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. 


*

*$((z + x)) = ((x))$, where $z$ is integer

*$ 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: $∑_{a=0}^b GCD(a, b) s(a, b) = 0$
