Another Fibonacci identity: $F_{2n-1} = F_{n}^2 + F_{n-1}^2$ Here's a problem that is leading me in circles.  
Consider the Fibonacci number $F_n$ defined by $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n \geq 2$.

Prove that $F_{2n-1} = F_{n}^2 + F_{n-1}^2$.

Tried an induction proof that lead me nowhere fast.  So far we have only been given limited identities, and they are all basically summation formulas for the odd terms, or even terms, or all terms up to n.  I also though that maybe if it was a difference, the difference of squares could factor to something, but I'm just not getting it.  
 A: The identity may be derived from the interesting fact that
$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right ).$$
From this, we may observe that
$$\begin{align} \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^m \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^n &= \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^{m+n} \\ &= \left ( \begin{array} \\ F_{m+1} & F_m\\F_m & F_{m-1} \\ \end{array} \right ) \left ( \begin{array} \\ F_{n+1} & F_n\\F_n & F_{n-1} \\ \end{array} \right )\\ &= \left ( \begin{array} \\ F_{m+n+1} & F_{m+n}\\F_{m+n} & F_{m+n-1} \\ \end{array} \right ) \end{align} $$
Therefore we can say that, for example,
$$F_{m+n-1} = F_m F_n + F_{m-1} F_{n-1}$$
Plug in $m=n$ and the desired identity follows.
EDIT
The first identity quoted above follows from putting the Fibonacci recurrence into matrix form:
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) \left ( \begin{array} \\F_{k+1} \\ F_k \\ \end{array} \right )$$
which may be immediately verified.  We may repeat this matrix multiplication $k$ times to get
$$\left ( \begin{array} \\F_{k+2} \\ F_{k+1} \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\F_{2} \\ F_1 \\ \end{array} \right ) = \left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right )^k \left ( \begin{array} \\1 \\ 1 \\ \end{array} \right )$$
Noting that $F_{k+2}= F_{k+1}+ F_k$ and $ F_{k+1} = F_k +  F_{k-1}$, the stated identity follows.
A: Using Binet's Fibonacci Number Formula,
$F_r=\frac{a^r-b^r}{a-b}$ where $a,b$ are the roots of $x^2-x-1=0\implies ab=-1,a+b=1$
So, $$F_n^2+F_{n-1}^2=\frac{(a^n-b^n)^2+(a^{n-1}-b^{n-1})^2}{(a-b)^2}$$
$$=\frac{a^{2n}+b^{2n}+a^{2n-2}+b^{2n-2}-2(ab)^{n-1}(ab+1)}{(a-b)^2}$$
$$=\frac{a^{2n-1}\left(a+\frac1a\right)+b^{2n-1}\left(b+\frac1a\right)}{(a-b)^2}\text{ as } ab=-1$$
$$=\frac{a^{2n-1}(a-b)+b^{2n-1}(b-a)}{(a-b)^2} \text{ as } \frac1a=-b,\frac1b=-a$$
$$=\frac{a^{2n-1}-b^{2n-1}}{(a-b)}=F_{2n-1}$$
