TITLE: Every sixth polynomial shares a factor of $(a^2-6)$
QUESTION [12 upvotes]: I currently looking at the polynomials you get from the series expansion of 
$$
\frac{1-x^2}{1-ax+2x^2}=1+a x
 +(a^2-3) x^2
 +(a^3-5a) x^3
 +\underbrace{(a^4-7a^2+6)}_{(a-1)(a+1)(a^2-6)} x^4+\dots
$$
W|A helped here...
What I found is that from $x^{4}$ onwards, every sixth polynomial shares a factor of $(a^2-6)$. How to prove that?

REPLY [9 votes]: I will prove what you want kind of indirectly using Chebyshev polynomials of second kind. Actually once you formulate the question in terms of these polynomials this mysterious $a^2-6$ becomes quite meaningful. Let me start by  The generating function of Chebyshev polynomial of second kind, $U_n$ (which you can take as the definition of these polynomials):
$$
\frac{1}{1-2ax+x^2}=\sum_{n=0}^\infty U_n(a)x^n
$$
So your function which I will call $I(x,a)$ is nothing but
$$
I(x,a):=\frac{1-x^2}{1-ax +2x^2}\xrightarrow{t=\sqrt{2}x}\left[1-\frac{t^2}{2}\right]\sum_{n=0}^\infty U_n\left(\frac{a}{\sqrt{8}}\right)t^n$$
We are only interested in the value of this function when $a=\pm \sqrt{6}$. But then $\frac{a}{2\sqrt{2}}=\cos \pi/6$ or $=\cos(\pi-\pi/6)$ (You can already see where every sixth term pattern is coming from). Then your function becomes (I defined $X:=a/\sqrt{8}$)
$$
I(X,t):=U_0(X) + tU_1(X)+
t^2\sum_{n=0}^\infty \left[U_{n+2}(X)-\frac{1}{2}U_n(X)\right]t^n
$$
We are only interested in $n=6k+2$ and want to prove that
$$
2U_{4+6k}\left(\cos \frac{\pi}{6}\right)=
U_{2+6k}\left(\cos \frac{\pi}{6}\right)
$$
Now using yet another property of Chebyshev polynomial of second kind, in general if $z=\cos \psi$, then
$$
U_m(z) = \frac{\sin((m+1)\psi)}{\sin \psi}
$$
In our case then one can easily show with identity above that $
U_{4+6k}\left(\cos \frac{\pi}{6}\right)=
(-1)^k
$ and $
U_{2+6k}\left(\cos \frac{\pi}{6}\right)=2(-1)^k
$. This show that $a=+\sqrt{6}$ is a root for every sixth polynomial after $x^4$ in the expansion of $I(x,a)$. Similarly you can check the $a=-\sqrt{6}$ is also a root.
These identities about Chebyshev polynomials are surprisingly easy to prove actually. For a fast reference about Chebyshev polynomials, look here, or wikipedia.