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
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.