Floor Function Power!

For each $k$ denote $m_{k}$ and $n_{k}$ the remainders of $k\pmod{m}$ and $k\pmod{n}$ respectively. Then we have $\biggl\lfloor\frac{k}{m}\biggr\rfloor + \left\lfloor\frac{k}{n}\right\rfloor \equiv m \left\lfloor\frac{k}{m}\right\rfloor + n \cdot \left\lfloor\frac{k}{n}\right\rfloor = k-m_{k}+ k-n_{k} \equiv m_{k}+n_{k} \pmod{2}$

Now since $(m,n)=1$ by the Chinese Remainder theorem the map $k \mapsto (m_{k},n_{k})$ is a bijection so we have $\sum_{k=0}^{mn-1} (-1)^{\lfloor\frac{k}{m}\rfloor + \lfloor\frac{k}{n}\rfloor} =\sum_{k=0}^{mn-1} (-1)^{m_{k}+n_{k}} = \sum_{i=0}^{m-1}\sum_{j=0}^{n-1} (-1)^{i+j} =\left(\sum_{i=0}^{m-1} (-1)^{i}\right) \cdot \left(\sum_{j=0}^{n-1} (-1)^{j}\right) = 1$