Got a bijection?

i=1nABi+j=1mBAj=f(n)f(m)\large\sum _{ i=1 }^{ n }{ \left\lfloor \frac { A }{ B } i \right\rfloor } +\sum _{ j=1 }^{ m }{ \left\lfloor \frac { B }{ A } j \right\rfloor } = f(n)f(m)

Let A,BA,B be odd natural numbers that are relatively prime, and let n=B12n= \frac { B-1 }{ 2 } , m=A12m=\frac { A-1 }{ 2 } such that the equation above is fulfilled. If ff is a linear function with f(1)>0 f(1) > 0 , what is the sum of the coefficients of f(x)f(x)?

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