This is worse than it looks

Note that since both $f(x) = f(x-1) \cdot f(x+1)$ and $f(x+1) = f(x) \cdot f(x+2)$ , we have $f(x) = f(x-1) \cdot f(x) \cdot f(x+2) \rightarrow$ either $f(x) = 0$ or $f(x-1) \cdot f(x+2) = 1 \rightarrow f(x) \cdot f(x+3) = 1$ for all $x$ . Since $f(x)=0$ seems not to be the maximum number of possible solutions, as it only yields $1$ , we ignore it for now. So we have $f(x) \cdot f(x+3) = 1 \rightarrow f(x) = \dfrac{1}{f(x+3)}$ for all $x$ . By this same logic, we have $f(x+3) = \dfrac{1}{f(x+6)}$ and $f(x) = f(x+6)$ . Thus, we also have $f(x-1) = f(x+5)$ and all other variants in this form. Now, we can achieve a maximum of $6$ possible values. Also, the sextuplet ${f(1), f(2), f(3), f(4), f(5), f(6)}$ will be in the form ${a, b, c, \dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}}$ . Thus, for any ${a, b, c}$ s.t. $a \ne b$ , $b \ne c$ , $a \ne c$ , and none of ${a,b,c}$ are equal to $1$ or $0$ , we will achieve the maximum $6$ different values. Thus, the answer is $\boxed{6}$ .