TITLE: A sum of fractional parts.
QUESTION [20 upvotes]: I am looking to evaluate the sum $$\sum_{1\leq k\leq mn}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\} .$$
Using matlab, and experimenting around, it seems to be $\frac{(m-1)(n-1)}{4}$ when $m,n$ are relatively prime.  How can we prove this, and what about the case where they are not relatively prime?
Conjecture:  Numerically, it seems that for any $m,n$ we have $$\sum_{1\leq k\leq mn}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\} =\frac{(m-1)(n-1)}{4}+C(\gcd(m,n))$$
where $C(\gcd(m,n))$ is some constant depending only on the $\gcd(m,n)$.
Additionally:  Can we sum this even when it is not a complete interval? Suppose that $0<a<b<mn,$ do we have an exact form for $$\sum_{a\leq k\leq b}\left\{ \frac{k}{m}\right\} \left\{ \frac{k}{n}\right\}.$$ 
Remark: In the one variable case we have $$\sum_{1\leq k\leq n}\left\{ \frac{k}{n}\right\} =\frac{n-1}{2}$$ the sum over an interval $a,b$ has an explicit form.

REPLY [3 votes]: Analysis of a small example $m=4, n=6$ with $d=\mathrm{gcd}(4,6)=2$ indicates how to derive the general case.
\begin{align*}
\sum_{1\leq k\leq 24}&\left\{\frac{k}{4}\right\}\left\{\frac{k}{6}\right\}\\
&=\color{blue}{\frac{1}{4}\cdot\frac{1}{6}+\frac{2}{4}\cdot\frac{2}{6}+\frac{3}{4}\cdot\frac{3}{6}
+\frac{0}{4}\cdot\frac{4}{6}+\frac{1}{4}\cdot\frac{5}{6}+\frac{2}{4}\cdot\frac{0}{6}}\\
&\,\,\qquad\color{blue}{+\frac{3}{4}\cdot\frac{1}{6}+\frac{0}{4}\cdot\frac{2}{6}+\frac{1}{4}\cdot\frac{3}{6}
+\frac{2}{4}\cdot\frac{4}{6}+\frac{3}{4}\cdot\frac{5}{6}+\frac{0}{4}\cdot\frac{0}{6}}\\
&\qquad+\frac{1}{4}\cdot\frac{1}{6}+\frac{2}{4}\cdot\frac{2}{6}+\frac{3}{4}\cdot\frac{3}{6}
+\frac{0}{4}\cdot\frac{4}{6}+\frac{1}{4}\cdot\frac{5}{6}+\frac{2}{4}\cdot\frac{0}{6}\\
&\qquad+\frac{3}{4}\cdot\frac{1}{6}+\frac{0}{4}\cdot\frac{2}{6}+\frac{1}{4}\cdot\frac{3}{6}
+\frac{2}{4}\cdot\frac{4}{6}+\frac{3}{4}\cdot\frac{5}{6}+\frac{0}{4}\cdot\frac{0}{6}\\
&=2\left(\frac{1}{4}\cdot\frac{1}{6}\color{blue}{+\frac{2}{4}\cdot\frac{2}{6}}+\frac{3}{4}\cdot\frac{3}{6}
\color{blue}{+\frac{0}{4}\cdot\frac{4}{6}}+\frac{1}{4}\cdot\frac{5}{6}\color{blue}{+\frac{2}{4}\cdot\frac{0}{6}}\right.\\
&\qquad\left.+\frac{3}{4}\cdot\frac{1}{6}\color{blue}{+\frac{0}{4}\cdot\frac{2}{6}}+\frac{1}{4}\cdot\frac{3}{6}
\color{blue}{+\frac{2}{4}\cdot\frac{4}{6}}+\frac{3}{4}\cdot\frac{5}{6}\color{blue}{+\frac{0}{4}\cdot\frac{0}{6}}\right)\tag{1}\\
&=2\left(\frac{0}{4}+\frac{2}{4}\right)\left(\frac{0}{6}+\frac{2}{6}+\frac{4}{6}\right)
+2\left(\frac{1}{4}+\frac{3}{4}\right)\left(\frac{1}{6}+\frac{3}{6}+\frac{5}{6}\right)\tag{2}
\end{align*}

Generalising these steps we obtain for non-negative integers $m,n$ and $d=\mathrm{gcd}(m,n)$
  \begin{align*}
&\color{blue}{\sum_{1\leq k\leq mn}}\color{blue}{\left\{\frac{k}{m}\right\}\left\{\frac{k}{n}\right\}}\\
&\quad=d\sum_{k=1}^{\frac{mn}{d}}\left\{\frac{k}{m}\right\}\left\{\frac{k}{n}\right\}\tag{3}\\
&\quad=d\sum_{j=0}^{d-1}\left(\sum_{p=0}^{\frac{m}{d}-1}\frac{j+pd}{m}\right)
\left(\sum_{q=0}^{\frac{n}{d}-1}\frac{j+qd}{n}\right)\tag{4}\\
&\quad=d\sum_{j=0}^{d-1}\left(\frac{m}{d}\cdot\frac{j}{m}+\frac{d}{m}\cdot\frac{1}{2}\left(\frac{m}{d}-1\right)\frac{m}{d}\right)
\left(\frac{n}{d}\cdot\frac{j}{n}+\frac{d}{n}\cdot\frac{1}{2}\left(\frac{n}{d}-1\right)\frac{n}{d}\right)\tag{5}\\
&\quad=d\sum_{j=0}^{d-1}\left(\frac{j}{d}+\frac{1}{2}\left(\frac{m}{d}-1\right)\right)
\left(\frac{j}{d}+\frac{1}{2}\left(\frac{n}{d}-1\right)\right)\\
&\quad=d\left(\frac{1}{d^2}\cdot\frac{1}{6}(d-1)d(2d-1)
+\frac{1}{2}\cdot\left(\frac{m}{d}-1\right)\frac{1}{d}\cdot\frac{1}{2}(d-1)d\right.\\
&\qquad\left.+\frac{1}{2}\cdot\left(\frac{n}{d}-1\right)\frac{1}{d}\cdot\frac{1}{2}(d-1)d
+\frac{1}{4}\left(\frac{m}{d}-1\right)\left(\frac{n}{d}-1\right)\right)\tag{6}\\
&\quad=\frac{1}{6}(d-1)(2d-1)+\frac{1}{4}(m-d)(d-1)\\
&\qquad +\frac{1}{4}(n-d)(d-1)+\frac{1}{4}(m-d)(n-d)\\
&\quad\,\,\color{blue}{=\frac{1}{4}(m-1)(n-1)+\frac{1}{12}\left(d^2-1\right)}
\end{align*}

Comment:

In (3) we use as in (1) that the $mn$ summands are $d=gcd(m,n)$ copies of the first $\frac{mn}{d}$ summands.
In (4) we see correspondingly to (2) in groups of $\frac{mn}{d}$ elements $d$ copies of size $\frac{m}{d}\cdot\frac{n}{d}$.
In (5) we calculate the inner sums using $\sum_{j=0}^n j=\frac{1}{2}n(n+1)$ and $\sum_{j=0}^n j^2=\frac{1}{6}n(n+1)(2n+1)$.
In (6) we calculate the outer sum using the summation formulas as in (5).