Given $P_n(x)=\frac1{2^n}\left[(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n\right]$, prove $P_n(x)-xP_{n-1}(x)+\frac{1}{4}P_{n-2}(x)=0$ 
Given the expression
$$P_n(x)=\dfrac{1}{2^n}\left[\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n \right]$$ Prove that $P_n(x)$ satisfies the identity
$$P_n(x)-xP_{n-1}(x)+\frac{1}{4}P_{n-2}(x)=0$$ and that $P_n(x)$ is polynomial in $x$ of degree $n$.

My solution:
I put the $P_n(x)$ in given identity and I obtained $0=0$ $\implies$ $P_n(x)$ satisfy this.
Is there any other way to solve this?
I also tried using binomial expansion but couldn't proceed further.
 A: Hint: prove that $\,P_n(\cos \theta)=\frac{1}{2^{n-1}}\cos n\theta\,$ then use the trig identities for $\,\cos (n+1)\theta\,$ to derive the recurrence. (Related: Chebyshev polynomials of the first kind.)
A: First of all, I would write
$$P_n(x) = \left(\frac{x+\sqrt{x^2-1}}{2}\right)^n + \left(\frac{x-\sqrt{x^2-1}}{2}\right)^n = \alpha^n + \beta^n$$
and realise that $\alpha$ and $\beta$ are the roots of the polynomial $Q(T) = T^2-xT +\frac{1}{4} = (T-\alpha)(T-\beta)$.
Then, if you define the linear operator $S$ which associates to each $a_n$ in a sequence the following term, that is $S(a_n) = a_{n+1}$, then your identity can be written as:
$$P_{n+2} - xP_{n+1} + \frac{1}{4}P_n = S^2(P_n) - xS(P_n) + \frac{1}{4}P_n = (S^2 - xS + \frac{1}{4})P_n = Q(S)P_n = 0$$
And, as we saw before, $Q(S) = (S-\alpha)(S-\beta)$. But it is clear that
$$(S-\alpha)(\alpha^n) = S(\alpha^n) - \alpha^{n+1} = \alpha^{n+1}-\alpha^{n+1} = 0$$
And
$$(S-\beta)(\beta^n) = S(\beta^n) - \beta^{n+1} = \beta^{n+1}-\beta^{n+1} = 0$$
And so $Q(S)(\alpha^n) = (S-\beta)(S-\alpha)(\alpha^n) = (S-\beta)0 = 0$ and similarly you show that $Q(S)(\beta^n) = 0$, and therefore $Q(S)(\alpha^n + \beta^n) = Q(S)(\alpha^n) + Q(S)(\beta^n) = 0$.
The identity is proved (By the way, if I recall correctly, every linear recurrence equations is solved this way).
The function $\alpha^n + \beta^n$ is a polynomial in $x$, because it is a symmetric polynomial of the roots of $Q(x)$, and therefore, by the fundamental theorem on symmetric polynomials, it can be expressed as a polynomial in the elementary symmetric functions on $\alpha$ and $\beta$, which are the coefficients of $Q(S)$, that is, 1, -x and $\frac{1}{4}$.
That the degree of this polynomial is $n$ can be seen without the binomial expansion. Simply compute the limit
$$\lim_{x\rightarrow \infty} \frac{P_n(x)}{x^n}$$
And check it is not $\infty$ or $0$. I think it's $1$, that is, all $P_n$ are monic polynomials.
A: Your problem is of a type of generalized Fibonacci sequence described here. To wit, $f_n=af_{n-1}+bf_{n-2}$ has a solution of the form
$$
f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n)
$$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.
Thus, given the specifics of your problem, we can see that $\left(f_1-\frac{af_0}{2}\right)=0$ and $f_0=2$. Furthermore, it's apparent that $a=x$ and $b=-1/4$. And voila! There is your solution.
