Here are some interesting conjectures I would like to prove. For all positive integers $a=bc,m,n$ the following expressions are integers:
$$c\sum_{k=1}^{am}k\left\{\frac{kbn}{am}\right\}$$
$$\frac{1}{c}\sum_{k=1}^{bm}k\left\{\frac{kan}{bm}\right\}$$
where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part operator.
Can anyone prove these?
The statement is not true as long as the following code is valid for small $b,c,m,n$ (arithmetic errors, etc.):
pastebin c++ code
It outputs lots of counter-examples for both formulas.
Though your original formula (multiples of $6$) from your deleted question, seems to be true.
First counter example of formula one (according to code) is $(b,c,m,n) = (1,1,2,1)$. On paper we have: $$ c\sum_{k=1}^{am}k\left\{\frac{kbn}{am}\right\} = \\ 1\cdot \sum_{k=1}^2 (k \cdot \left\{\frac{k}{2}\right\}) = \\ 1\cdot \{\frac{1}{2}\} + 2\cdot\{\frac{2}{2}\} = \\ \frac{1}{2} $$ There's a counterexample, QED.