P4132算不出的等式

\[\begin{align*} &\sum_{k=1}^{\frac{p-1}{2}} \biggl\lfloor\frac{kq}{p}\biggr\rfloor+\sum_{k=1}^{\frac{q-1}{2}} \biggl\lfloor\frac{kp}{q}\biggr\rfloor\\ =&\sum_{k=1}^{\frac{p-1}{2}} \biggl(\frac{kq-kq\ mod\ p}{p}\biggr)+\sum_{k=1}^{\frac{q-1}{2}} \biggl\lfloor\frac{kp}{q}\biggr\rfloor\\ =&\sum_{k=1}^{\frac{p-1}{2}}\frac{kq}{p}-\sum_{k=1}^{\frac{p-1}{2}}\frac{kq\ mod\ p}{p}+\sum_{k=1}^{\frac{q-1}{2}} \biggl\lfloor\frac{kp}{q}\biggr\rfloor\\ \because&\sum_{k=1}^{\frac{p-1}{2}}\frac{kq\ mod\ p}{p}\\ =&\frac{\sum_{k=1}^{\frac{p-1}{2}}kq-\sum_{k=1}^\frac{q-1}{2}(p\cdot(\frac{p-1}{2}-\lfloor\frac{kp}{q}\rfloor))}{p}\\ =&\sum_{k=1}^{\frac{p-1}{2}}\frac{kq}{p}-\sum_{k=1}^\frac{q-1}{2}(\frac{p-1}{2}-\lfloor\frac{kp}{q}\rfloor)\\ \therefore &\text{ 原式}=\frac{p-1}{2}\cdot\frac{q-1}{2} \end{align*} \]