Dedekind Sum Integrality Result Can we prove the following is always an integer?
$$6b\sum_{k=1}^bk\left\{\frac{ka}{b}\right\}$$
where $\{x\}=x-\lfloor x\rfloor$ denotes the fractional part operator.
UPDATE:
Through the calculation of several identities and going through heavy casework, I have proven the above formula; the proof, however, is too lengthy and messy to post here.

For $a = 1$ we have $\sum_{k =1}^b k\{ka/b\} = 1\{1/b\} + 2 \{2/b\} + \dots + (b-1)\{(b-1)/b\} = \sum_{k=1}^b k^2/b = 1/b \sum$(integers).  And $\gcd(b,6)/\gcd(1,6) = \gcd(b,6)$ an integer. And the sum of squares formula from:
sum of sequence of squares article
has a $6$ on the bottom and a $b$ factor on the top, so I'm sure it has something to do with it!
