Question

Prove that for positive integers a and b with no common divisor, \[\sum_{i=1}^{b-1} \left \lfloor {ai \over b} \right \rfloor = \sum_{j=1}^{a-1} \left \lfloor {bj \over a} \right \rfloor\].
I have seriously no idea where to start. Any hints?

Answer 1

does all ai's have no common division factor with b?
and same for bi's??

Answer 2

I dunno. All that's given is that a and b have no common divisor.

Answer 3

\[\sum_{i=1}^{b-1} \left \lfloor {ai \over b} \right \rfloor =\left \lfloor {a \over b} \right \rfloor \sum_{i=1}^{b-1}i = \left \lfloor {a \over b} \right \rfloor {{b(b-1)}\over2} \]