Roots of a special polynomial I have a question that seems tricky to me. So, let $P_n$ be the polynomial defined by $P_n(x+1/x)=x^n+x^{-n}$. And let $Q$ be some polynomial with $\sup_{x\in [-2,2]}|Q(x)|<2$. Then $P_n-Q$ has at least $n$ different roots. I have no idea yet. It's clear that $P_n(x)\notin (-2,2)$ für $x \notin (-2,2)$ but I don't see any use of it. I would be grateful for any idea.
 A: Hint:
Let $t=x+1/x$. Resolving the equation with respect to $x$ one obtains: 
$$
x_\pm=\frac{t\pm\sqrt{t^2-4}}{2}.
$$
Notice: $x_+=x^{-1}_-$.
Thus:
$$
P_n(x)=\left(\frac{x+\sqrt{x^2-4}}{2}\right)^n+\left(\frac{x-\sqrt{x^2-4}}{2}\right)^n.\tag{1}
$$
Observe that upon binomial expansion of both summands the odd powers of $\sqrt{x^2-4}$ will cancel, so that the function (1) is indeed a polynomial of order $n$.
Next observe that there can be no real roots of the polynomial outside the interval $(-2,2)$ as both summands are real and have the same sign. Inside the interval one can use the substitution $x=2\cos t$ to obtain:
$$
P_n(2\cos t)=e^{itn}+e^{-itn}=2\cos nt,\text{ with } 0\le t\le\pi.  
$$
The function obviously has $n$ distinct real roots $t_k=\frac{2k+1}{2n}\pi$ with $k=0..(n-1)$. Recalling that the power of $P_n(x)$ is $n$ they represent the complete list of roots of the polynomial.
Besides the following inequality holds for all $-2\le x \le 2$:
$$
-2\le P_n(x)\le 2,
$$
with limits being attained at $x=2\cos\frac{2k+1}{n}\pi$ and $x=2\cos\frac{2k}{n}\pi$, respectively.
