Let $\omega(x)=\dfrac{ax+b}{cx+d}$ , where $a,b,c,d$ are real numbers.

Given that: $\omega(\omega(\omega(1)))=1$ and $\omega(\omega(\omega(2)))=2015$ .

Find the sum of possible values of $\omega(1)$ .

The answer is 1.000.

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Note that w(x) is strictly monotonous. Suppose that w(x) is strictly crescent (without losing of generality). If w(1)>1 , hence we have w(w(1))>w(1) (because it is strictly crescent) and that implies that w(w(w(1)))>w(1)>1, which is an absurd. Following the same path, if w(1)<1 we also have an absurd. Therefore the only suitable solution is that one where w(1)=1.