An algebra problem by Ilham Saiful Fauzi

Note that putting $a=0$ , $\displaystyle f(0)(f(b)-1)=b \implies f(b)=\frac{b}{c}+1$ where $f(0)=c$

Since $f(x)=\frac{x}{c}+1$ , substituting this into the equation we have $(\frac{a}{c}+1)(\frac{b}{c}+1) - (\frac{ab}{c}+1)=a+b$ . Solving upon we get $c^2=c$ but $c\ne 0$ otherwise $f(x)$ won't be defined and therefore $c=1$ which makes $f(x)=x+1$ . $f(2016)=\boxed{2017}$