Compare and Predict

$f(x+2)(1-f(x))=1+f(x)$ , by algebraic manipulation we get:

$(1).$ $f(x+2)=\frac{1+f(x)}{1-f(x)};$

$(2).$ $f(x)=\frac{f(x+2)-1}{f(x+2)+1}.$

• From $(1)$ :

$\begin{aligned} f(x+4) & = \frac{1+f(x+2)}{1-f(x+2)} \\ & = \frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}} \\ & = \frac{\left ( \frac{2}{1-f(x)} \right )}{\left ( \frac{-2f(x)}{1-f(x)} \right )} \\ & =-\frac{1}{f(x)}. \end{aligned}$

• From $(2)$ :

$\begin{aligned} f(x-4) & =\frac{f(x-2)-1}{f(x-2)+1} \\ & =\frac{\frac{f(x)-1}{f(x)+1}-1}{\frac{f(x)-1}{f(x)+1}+1} \\ & =\frac{\left ( \frac{-2}{f(x)+1} \right )}{\left( \frac{2f(x)}{f(x)+1} \right)} \\ & =-\frac{1}{f(x)}. \end{aligned}$

Therefore $f(x+4)=f(x-4),$ $\Rightarrow f(x+8)=f(x).$

Thus, $f(2017)=f(252 \times 8 +1)=f(1)= \boxed{2016}.$

$\begin{aligned} f(x+2)(1-f(x)) & = 1+f(x) \\ f(x+2) & = \frac {1+f(x)}{1-f(x)} \\ f(x+4) & = \frac {1+f(x+2)}{1-f(x+2)} = \frac {1+\frac {1+f(x)}{1-f(x)}}{1-\frac {1+f(x)}{1-f(x)}} = - \frac 1{f(x)} \\ f(x+8) & = - \frac 1{f(x+4)} = - \frac 1{-\frac 1{f(x)}} = f(x) \end{aligned}$

$\implies f(x+8) = f(x)$ , that is a $f(x)$ is function with a period 8, therefore, $f(2017) = f(2017 \text{ mod }8) = f(1) = \boxed{2016}$ .