<s>[INST] Proving Generating Function holds a specific recurrence. Consider the generating function $$\dfrac{1}{1 − 2x − x^2} = \sum_{n=0}^{\infty}a_nx^n$$
Prove that for each integer $n \ge 0$,
$$a_n^2+a_{n+1}^2 = a_{2n+2}$$
Hint: Find a $2 \times 2$ matrix $A$ such that
$$A^{n+2} =\begin{bmatrix}a_n &a_{n+1}\\
a_{n+1} &a_{n+2}\end{bmatrix}$$
and consider the top left entry of the matrix product $A^{n+2}A^{n+2}$.
Looking at the hint, I think about how we used matrices and eigenvalues to find the closed form expression of recurrence relations, but I can only really do that for stuff of the form $a_n=a_{n-1}+a_{n-2}$, and I'm not sure how the matrix product fits in there.
Instead, I tried to find the generating function for the recurrence listed above as:
$$\begin{align*}
a_n^2+a_{n+1}^2 &= a_{2n+2}\\
\left(x^n\right)^2a_n^2+\left(x^n\right)^2a_{n+1}^2 &= \left(x^n\right)^2a_{2n+2}\\
\sum_{n=0}^{\infty}\left(x^n\right)^2a_n^2+\sum_{n=0}^{\infty}\left(x^n\right)^2a_{n+1}^2 &= \sum_{n=0}^{\infty}\left(x^n\right)^2a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \sum_{n=0}^{\infty}x^{2n}a_{2n+2}\\
A(x)^2+\dfrac{A(x)^2}{x^2} &= \dfrac{A(x)}{x^2}\\
x^2A(x)^2+A(x)^2&=A(x)\\
A(x)=\dfrac{1}{x^2+1}
\end{align*}$$
But that's clearly not what we wanted. What mistakes did I make, and how does the hint fit into all of this? Thanks!
 [/INST] Too long for a comment
$$\frac{1}{1 − 2x − x^2}=\frac{1}{(1+\sqrt2+x)(\sqrt2-1-x)}=\frac{1}{2\sqrt2}\left(\frac{1}{1+\sqrt2+x}+\frac{1}{\sqrt2-1-x}\right)$$
$$=\frac{1}{2\sqrt2}\sum_{n=0}^\infty\left(\frac{x^n}{(\sqrt2-1)^{n+1}}+(-1)^n\frac{x^n}{(\sqrt2+1)^{n+1}}\right)$$
Then, given that $\frac{1}{(\sqrt2+1)^{n+1}}=\frac{(\sqrt2-1)^{n+1}}{(\sqrt2+1)^{n+1}(\sqrt2-1)^{n+1}}=(\sqrt2-1)^{n+1}$
$$a_n=\frac{1}{2\sqrt2}\left(\frac{1}{(\sqrt2-1)^{n+1}}+(-1)^n\frac{1}{(\sqrt2+1)^{n+1}}\right)=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{n+1}+(-1)^n(\sqrt2-1)^{n+1}\right)$$
and
$$a_{n+1}=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{n+2}+(-1)^{n+1}(\sqrt2-1)^{n+2}\right)$$
Then
$$a_n^2+a_{n+1}^2=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+4}+(\sqrt2-1)^{2n+4}\right)$$
$$=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+2}(3+2\sqrt2)+(\sqrt2-1)^{2n+2}(3-2\sqrt2)\right)$$
$$=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{2n+3}+(\sqrt2-1)^{2n+3}\right)=a_{2n+2}$$
</s>