>>460
Can be done in a couple of ways; summing $\lfloor \frac{ak}{m} \rfloor \lfloor \frac{bk}{m} \rfloor$ or computing $\int_0^1 \{ax\}\{bx\}$ works.

For(1),We will show that $\lambda = \int_{0}^{1}\left \{ ax \right \} \left \{ bx \right \}dx$ satisfied.  In fact,we would show that $\left [ \frac{ab\left ( k-1 \right )}{m} ,\frac{abk}{m}\right ]\cap\mathbb{Z}= \emptyset$ 
 then we are done. This $\Longleftrightarrow\left \{ \frac{ak}{m} \right \} \left \{ \frac{bk}{m} \right \} -m\int_{\frac{k-1}{m} }^{\frac{k}{m} }\left \{ ax \right \}\left \{ bx \right \} dx=\mathrm {O}\left ( \frac{1}{m} \right )$ and $\Longleftrightarrow \sup_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} -\inf_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} =\mathrm {O} \left ( \frac{1}{m} \right )$  It's obviously right. 

$\blacksquare$