Prove that $\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$ 
Let $\alpha,\beta,\gamma$ be the roots of $x^3-3x^2+1 = 0$. Prove that $$\alpha^n+\beta^n+\gamma^n \equiv 4^n+5^n+(-6)^n \pmod{17}$$ where $n$ is an integer.

It sort of makes sense why they are congruent since we can say $x^3-3x^2+1 \equiv (x-4)(x-5)(x+6) \pmod{17}$, but since $\alpha,\beta,\gamma$ aren't integers, how do we prove this?

Hint $\ $ Both sides satisfy the associated recurrence $\,f_{n+3} = \cdots$ with same initial conditions, thus they are equal for all $\,n\,$ by a trivial induction (i.e.  uniqueness theorem for recurrences).
To obtain the associated recurrence  sum the following for $\,a\,$ over all $\,3\,$ roots.
$$ a^{n+3}-3a^{n+2} +a^n =\, a^n(a^3-3a^2+1) = 0,\ \ {\rm  for\ any\ root}\ \ a$$
yielding $\ \  \alpha^{n+3} +\beta^{n+3} + \gamma^{n+3} - 3 (\alpha^{n+2} +\beta^{n+2} + \gamma^{n+2}) + (\alpha^{n} +\beta^{n} + \gamma^{n}) \ =\  0$
therefore  $\,\ f_{n+3} - 3f_{n+2} + f_n = 0,\ $ for $\ f_n = \alpha^{n} +\beta^{n} + \gamma^{n}$
TIP $ $ note that $\,4,5, -6\,$ are the roots of the polynomial mod $17.$
