If i define $f(m,n)=$ $$\sum_{1\leq k\leq mn}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\} .$$

Then prove $$f(m+n,n) - f(m,n) =\frac{n^2-n}{4}$$ for all $m$ and $n$.

This question came from part of answer from this question: A sum of fractional parts.


Since $k$ goes from $1$ to $mn$, then the pairs $((k \bmod m),(k \bmod n))$ will meet all cases once.

thus the product is

$$\sum_{1\le k\le m}\left\{\frac{k}{m}\right\}\sum_{1\le t\le n}\left\{\frac{t}{n}\right\}=\frac{m-1}{2}\frac{n-1}{2}\;.$$


  • $\begingroup$ $\{a\}=a-[a]$, where [a] is the largest integer less than or equal to a. $\endgroup$ – Yimin May 4 '12 at 3:44

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