Question: For any real $t$ and $n \in\mathbb{N}$, I want to show that $p\_n(x)=x^{n+1}\cos [(n-1)t] - x^n\cos (nt) - x\cos (t) +1$ is divisible by $q(x) = x^2 - 2x\cos (t) +1$. As a hint I have: *complex numbers*.

I don't know how to attack the problem using complex numbers. In the case $n=1$, $p\_1(x)=q(x)$. Also I compute and check by hand the case $n=2$. Then I start thinking about induction, but the idea failed.

Any idea how this can be solved via complex numbers? Thank you in advance.

Answer: 
Here is a rolling start. Use the quadratic formula on the quadratic to get the roots

$$x = {\cos(t)\pm\sqrt{4\cos^2(t) - 4}\over 2} 
=\cos(t)\pm i\sin(t) = e^{\pm it}. $$

Are these roots of the $n$th degree polynomial?
This is equivalent to $z^{\pm1}$ being roots, with $z:=e^{it}$. We only need check $\pm=+$, since $p\_n$ has real coefficients, so its nonreal roots are in conjugate pairs. So just simplify$$p\_n(z)=z^{n+1}(z^{n-1}+z^{1-n})/2-z^n(z^n+z^{-n})/2-z(z+z^{-1})/2+1.$$