Considering the following series expansion
$$ \frac{1}{{1 - 2x - x^2 }} = \sum\limits_{n = 0}^\infty {a_n } x^n $$
prove that $ \forall n,\,\exists m $ such that $ a_n ^2 + a_{n + 1} ^2 = a_m $
I tried proving this using the coefficients computed by Wolfram but it didn't really help. Any ideas?
First find the zeros of $1-2x-x^2$:
$$x=\frac{-2\pm\sqrt{4+4}}2=1\pm\sqrt2\;.$$
Let $\alpha=1+\sqrt2$ and $\beta=1-\sqrt2$; then
$$(1-\alpha x)(1-\beta x)=1-(\alpha+\beta)x+\alpha\beta x^2=1-2x-x^2\;,$$
so you can decompose $\frac1{1-2x-x^2}$ into partial fractions with denominators $1-\alpha x$ and $1-\beta x$ as
$$\frac1{1-2x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\;.$$
I’ll leave it to you to calculate $A$ and $B$. Then observe that
$$\begin{align*} \frac{A}{1-\alpha x}+\frac{B}{1-\beta x}&=A\sum_{n\ge 0}\alpha^nx^n+B\sum_{n\ge 0}\beta^nx^n\\\\ &=\sum_{n\ge 0}\left(A\alpha^n+B\beta^n\right)x^n\;, \end{align*}$$
so $a_n=A\alpha^n+B\beta^n$, and
$$a_n^2+a_{n+1}^2=\left(A\alpha^n+B\beta^n\right)^2+\left(A\alpha^{n+1}+B\beta^{n+1}\right)^2\;.\tag{1}$$
Expand $(1)$ and collect terms, and with a little work you can find $m$ (in terms of $n$) such that it equals
$$A\alpha^m+B\beta^m=a_m\;.$$