An algebra problem by Jansen Wu

For $f(n) = f(n-1) - f(n-2)$ , the characteristic equation is $r^2 - r + 1 = 0$ ; $\implies r = \dfrac {1 \pm i\sqrt 3}2 = e^{\pm \frac \pi 3i}$ .

$\begin{aligned} \implies f(n) & = {\color{#3D99F6}c_1} e^{\frac \pi 3i} + {\color{#3D99F6}c_2} e^{-\frac \pi 3i} & \small \color{#3D99F6} \text{where }c_1, \ c_2 \text{ are constants.} \\ f(0) & = 0 \\ c_1 + c_2 & = 0 & \small \color{#3D99F6} \implies c_2 = - c_1 \\ f(1) & = 1 \\ c_1 \left(e^{\frac \pi 3i} - e^{-\frac \pi 3i} \right) & = 1 \\ -2c_1 \sin \frac \pi 3 & = 1 \\ \implies c_1 & = - \frac 1{\sqrt 3} \\ \implies f(n) & = \frac 2{\sqrt 3} \sin \frac {n\pi}3 \\ f(2017) & = \frac 2{\sqrt 3} \sin \frac {2017\pi}3 \\ & = \frac 2{\sqrt 3} \sin 672 \frac 13\pi \\ & = \frac 2{\sqrt 3} \sin \frac \pi 3 \\ & = \frac 2{\sqrt 3} \cdot \frac {\sqrt 3}2 \\ & = \boxed{1} \end{aligned}$

First few terms are 0, 1, 1, 0, -1, -1, 0, 1, . . . As every term is dependent on the previous two terms, the series has a period 6. As $2017 \equiv 1 (mod 6)$ , the value of $f(2017)=1$