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.
On
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!
Let $a,b \in \Bbb{R}$. Then each can be writ $a = a_i + a_f$, sim. for $b$, where $a_i = $ integer part, $a_f$ = fractional part. Then $ab = a_i b_i + a_i b_f + a_f b_i + a_f b_f$. Notice that if $(c \geq 0) \in \Bbb{Z}$ then $\{c + a\} = \{a\}$. Thus $\{ab\} = \{a_ib_f + a_f b_i + a_f b_f\}$. Notice that the fractional parts are $b_f = d/e$ with $d \lt e, d, e \in \Bbb{Z}$, sim. for $a$. Let's do some rewriting:
$\{a_ib_f + \dots + a_f b_f\} = \{a_i d/e + b_i f/g + (df)/(eg)\}$.
Okay, cancel that analysis but keep it just in casder e, since the numbers we are dealing with are $k \in \Bbb{Z}$ and a fraction $a/b$. So First assume $a(b-1) \lt b \equiv a = 1$. Then we have $\{a/b\} = a/b$ and we have $\{k a/b\} = \frac{\text{remainder of }ka/b}{b}$. But when you sum all the remainders over $k = 1 \dots b$, you get $$ \{a/b\} + \{2a/b\} + \dots + \{(b-1)a/b\}, \ \ \text{ since } \{ba/b\} = 0 $$
Now when you multiply a fraction by a constant $k \in \Bbb{Z}$ you get for instance $2 3/5 = 6/5 = 1 + 1/5$ or $k a/b = $ I give up! :)