How to prove this identity related to dedekind sums I am studying analytic number theory and this question was asked in my Mid-semester question paper.
Edit 1: Definition of dedekind sum
https://mathworld.wolfram.com/DedekindSum.html
Question:
10. If $\,p\,$ is prime prove that
$$ (p+1)\,s(h,k) = s(p\,h,k)+\sum_{m=0}^{p-1} s(h+m\,k,p\,k). $$
I used the definition of s(h, k) in both s(ph, k) and sum involving s(h+mk,pk) in hope of getting LHS.but couldn't.
Another thing to note is that there are p terms in sum and 1 in RHS and s(h, k) is p+1 times on LHS.
But I was not able to use that also.
It is my request to you to help in proving this identity.
Original image: https://i.stack.imgur.com/bVqDv.jpg
 A: Definition:
$$((x)) = \left\{\begin{array}{ll}
                   0 & x\ \mathrm{is\ an\ integer}, \\
                   x - \lfloor x\rfloor - \frac{1}{2} & \mathrm{otherwise}
                 \end{array}
\right.$$
and
$$s(h,k) = \sum_{r=1}^{k-1} \left(\left(\frac{r}{k}\right)\right)\left(\left(\frac{hr}{k}\right)\right).$$
Fact 1: $s(ph, pk) = s(h, k)$.
(See: An exercise related to properties of dedekind sums)
Fact 2: If $p$ is a prime, $q$ is an integer, $p \nmid q$, and $x$ is rational, then
$$\sum_{t=0}^{p-1} \left(\left( \frac{x + qt}{p}\right)\right) = ((x)). $$
(The proof is given at the end.)
Now, we have
\begin{align}
&\sum_{m=0}^{p-1} s(h + mk, pk)\\
=\ &
\sum_{m=0}^{p-1} \sum_{r=1}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)\Big(\Big(\frac{(h + mk) r}{pk}\Big)\Big)\\
=\ & \sum_{r=1}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)\\
=\ & \sum_{r = 1, \, p\, \nmid\, r}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)
+ \sum_{r = 1, \, p\, \mid \, r}^{pk - 1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\sum_{m=0}^{p-1}\Big(\Big(\frac{\frac{hr}{k} + mr}{p}\Big)\Big)\\
=\ & \sum_{r = 1, \, p\, \nmid\, r}^{pk-1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\Big(\Big(\frac{hr}{k}\Big)\Big)
+ \sum_{v=1}^{k-1} \Big(\Big(\frac{v}{k}\Big)\Big)
p\Big(\Big(\frac{hv}{k}\Big)\Big) \tag{1}\\
=\ & \sum_{r=1}^{pk - 1} \Big(\Big(\frac{r}{pk}\Big)\Big)
\Big(\Big(\frac{hr}{k}\Big)\Big) - \sum_{v=1}^{k-1} \Big(\Big(\frac{vp}{pk}\Big)\Big)
\Big(\Big(\frac{hvp}{k}\Big)\Big)
+ \sum_{v=1}^{k-1} \Big(\Big(\frac{v}{k}\Big)\Big)
p\Big(\Big(\frac{hv}{k}\Big)\Big)\\
=\ & s(ph, pk) - s(ph, k) + ps(h,k)\\
=\ & s(h,k) - s(ph,k) + ps(h,k) \tag{2}\\
=\ & (p+1)s(h,k) - s(ph,k)
\end{align}
where we have used Fact 2 in (1), Fact 1 in (2).
We are done.
$\phantom{2}$
Proof of Fact 2: Since $\{1, 2, \cdots, p-1\} = \{q, 2q, \cdots, (p-1)q\}\ (\mathrm{mod}\ p)$,
we have $\sum_{t=0}^{p-1} \left(\left( \frac{x + qt}{p}\right)\right) = \sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right)$.
Let $f(x) = \sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right)$.
Clearly, $f(x+1) = f(x)$. Since $((0))=0$ and $((y)) = y - \frac{1}{2}$ for $y\in (0, 1)$, we have
$f(0) = \sum_{t=0}^{p-1} \left(\left( \frac{t}{p}\right)\right)
= \sum_{t=1}^{p-1} \left( \frac{t}{p} - \frac{1}{2}\right) = 0$.
For $0< x < 1$, we have
$\sum_{t=0}^{p-1} \left(\left( \frac{x + t}{p}\right)\right) =
\sum_{t=0}^{p-1} \left(\frac{x + t}{p} - \frac{1}{2}\right) = x - \frac{1}{2} = ((x))$. We are done.
