Prove that $\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$ is always odd for any natural $n$. 
Prove that $$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
is always odd for any natural $n$.

I attempted to write the binomial expansion and sum it so the root numbers cancel out, and wanted to factorise it but didn't know how. I also attempted to use induction but was not sure how to proceed.

HINT:
Say $\left(\frac{3+\sqrt{17}}{2}\right)=a$ and $\left(\frac{3-\sqrt{17}}{2}\right)=b$.
Now observe that:
$$\left(\frac{3+\sqrt{17}}{2}\right)^n + \left(\frac{3-\sqrt{17}}{2}\right)^n$$
$$=a^n+b^n$$
$$=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+b^{n-2})$$
$$=\color{red}{3\cdot\left[\left(\frac{3+\sqrt{17}}{2}\right)^{n-1}+ \left(\frac{3-\sqrt{17}}{2}\right)^{n-1}\right]+ 2\cdot \left[\left(\frac{3+\sqrt{17}}{2}\right)^{n-2} + \left(\frac{3-\sqrt{17}}{2}\right)^{n-2}\right]}$$
Now use strong induction and see what you can do.
P.S. $3 \times \mathrm{odd} + 2\times \mathrm{odd} = \mathrm{odd + even} = \mathrm{odd}$
Hope this helps you.
