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!
 A: 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
$$a_n^2+a_{n+1}^2=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+4}+(\sqrt2-1)^{2n+4}\right)$$
$$=\frac{1}{8}\left((\sqrt2+1)^{2n+2}+(\sqrt2-1)^{2n+2}+(\sqrt2+1)^{2n+2}(3+2\sqrt2)+(\sqrt2-1)^{2n+2}(3-2\sqrt2)\right)$$
$$=\frac{1}{2\sqrt2}\left((\sqrt2+1)^{2n+3}+(\sqrt2-1)^{2n+3}\right)=a_{2n+2}$$
A: Define $f(x)$ by
$$f(x)= \frac{1}{1-2x-x^2} = \sum_{n=0}^{\infty} a_n x^n$$
so
$$f(x) -2x f(x) -x^2 f(x) = 1$$
From this equation we deduce $a_0=1$, $a_1=2$, and $a_{n+2} -2a_{n+1} -a_n = 0$ for $n \ge 0$.  It follows by induction on $n$ that if we let
$$A = \begin{pmatrix}
0 &1 \\
1 &2 \end{pmatrix}$$
then
$$A^{n+2} =\begin{pmatrix}
a_n &a_{n+1} \\
a_{n+1} &a_{n+2} \end{pmatrix}$$ for $n \ge 0$.
Now we can compute
$$A^{n+2} \cdot A^{n+2} = \begin{pmatrix}
a_n^2+a_{n+1}^2  & a_n a_{n+1} + a_{n+1} a_{n+2} \\
a_n a_{n+1} + a_{n+1} a_{n+2} & a_{n+1}^2 a_{n+2}^2
\end{pmatrix}$$
On the other hand,
$$A^{n+2} \cdot A^{n+2} = A^{2n+2+2} = \begin{pmatrix}
a_{2n+2} & a_{2n+3} \\
a_{2n+3} & a_{2n+4} \end{pmatrix}$$
Comparing the upper-left elements in these two matrices, we see
$$a_n^2+a_{n+1}^2 = a_{2n+2}$$
