Prove that $x_1^n+x_2^n$ is an integer and is not divisible by $5$ If $x_1$ and $x_2$ are the roots of the polynomial $x^2-6x+1$ then  , for every non-negative integer, prove that $x_1^n+x_2^n$  is an integer and is not divisible by $5$ .
My trying:
$ x_1 = 3+2\sqrt{2}$ and $ x_1 = 3-2\sqrt{2}$
So $ x_1^n +x_2^n = (3+2\sqrt{2})^n + (3-2\sqrt{2})^n$ = $ \dfrac{(3+2\sqrt{2})^{2n}+1}{(3+2\sqrt{2})^n}$
Then what should I do to solve this problem ?  

We have $x_1 + x_2 = \text{ Sum of roots }=6$. From the equation, we have
$$x_1^2 -6x_1 + 1 = 0 \text{ and }x_2^2 -6x_2 + 1 = 0$$
Adding both we get
$$x_1^2 + x_2^2 = 6\underbrace{(x_1+x_2)}_{\text{Integer}} - 2 = \text{Integer}$$
Now use strong induction and make use of the fact that
$$x_1^{n+2} -6x_1^{n+1} + x_1^n = 0 \text{ and }x_2^{n+2} -6x_2^{n+1} + x_2^n = 0$$
i.e.,
$$x_1^{n+2} + x_2^{n+2} = 6(x_1^{n+1}+x_2^{n+1}) - (x_1^n + x_2^n)$$
Use the same idea to show that
$$x_1^n + x_2^n \equiv \begin{cases} 1 \pmod5 & n\equiv 1 \pmod{4}\\ 
4 \pmod5 & n\equiv 0, 2 \pmod{4}\\ 3 \pmod5 & n\equiv 3 \pmod{4}\\\end{cases}$$
