I am trying assignment problems about properties of Dedekind sums and I am unable to solve this question.
I tried by putting qh, qk instead of h, k in definition of S(h, k) but I am only getting upto this S(qh, qk) = $\sum_{r modqk } [r/(qk) -0.5]((hr/k)) $ . But I need to prove it equal to S(h, k) .
Can someone please help! !
We have \begin{align} s(qh, qk) &= \sum_{r=1}^{qk-1} \left(\left(\frac{r}{qk}\right)\right)\left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{m=0}^{q-1} \sum_{r = mk + 1}^{(m+1)k-1} \left(\left(\frac{r}{qk}\right)\right)\left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{m=0}^{q-1} \sum_{r = 1}^{k-1} \left(\left(\frac{mk + r}{qk}\right)\right)\left(\left(\frac{h(mk+r)}{k}\right)\right)\\ &= \sum_{m=0}^{q-1} \sum_{r = 1}^{k-1} \left(\left(\frac{mk + r}{qk}\right)\right)\left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{r = 1}^{k-1} \sum_{m=0}^{q-1} \left(\left(\frac{mk + r}{qk}\right)\right) \left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{r = 1}^{k-1} \sum_{m=0}^{q-1} \left(\frac{mk + r}{qk} - \frac{1}{2}\right) \left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{r = 1}^{k-1} \left(\frac{r}{k} - \frac{1}{2}\right) \left(\left(\frac{hr}{k}\right)\right)\\ &= \sum_{r=1}^{k-1} \left(\left(\frac{r}{k}\right)\right)\left(\left(\frac{hr}{k}\right)\right). \end{align} We are done.