Proving the Fibonacci identity $(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$ Prove that for two natural numbers $m$ and $k$, where $m>k$ the following identity holds:

$$(−1)^{m−k}(F_{m+k+1}F_{m−k−1}−F_{m+k}F_{m−k}) =F_{k}^2+F_{k+1}^2$$

Here the exercise comes with a hint:
The constant is $F^2 _m$, try to substitute $k=0$ in.
Okay fair enough, let us try this, this will give us:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}F_{m}) =F_{0}^2+F_{1}^2$$
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =F_{0}^2+F_{1}^2=0^2+1^2=1$$
We can now use a identity by Cassini, namely that for $m>0=k$ we have:
$$ (−1)^{m}(F_{m+1}F_{m−1}−F_{m}^2) =(-1)^m(-1)^m=(-1)^{2m}=1 \checkmark$$
Okay great, so our identity works whenever $k=0$, but I do not see how this helps with a generalisation.

Edit/idea: Did I just write down the base case for an approach via induction on $k$ perhaps?

I found another approach that is also very satisfying.
We will use the fact that we can write the Fibonacci $Q$-matrix:
$$Q^{m+k} Q^{k-m}=Q^k Q^k $$
$$\begin{pmatrix} F_{m+k+1} & F_{m+k} \\
F_{m+k} & F_{m+k-1} \end{pmatrix} \begin{pmatrix} F_{k-m+1} & F_{k-m} \\
F_{k-m} & F_{k-m-1} \end{pmatrix} =
\begin{pmatrix} F_{k+1} & F_{k} \\
F_{k} & F_{k-1} \end{pmatrix} \begin{pmatrix} F_{k+1} & F_{k} \\
F_{k} & F_{k-1} \end{pmatrix} $$
We now compute the matrix product but are only interested in the first entry:
$$\begin{pmatrix} F_{m+k+1}F_{k-m+1}+F_{m+k}F_{k-m} & \dots \\
 \dots &  \dots \end{pmatrix}  =
\begin{pmatrix} F^2_k+ F^2_{k+1} & \dots \\
 \dots & \dots \end{pmatrix}  $$
    This gives us the following equality:
    $$ F_{m+k+1}F_{-(m-k-1)}+F_{m+k}F_{-(m-k)} =  F^2_k+ F^2_{k+1} $$ Which is almost what we desire. Notice that as $m>k$, we actually know that $F_{m-k}$ and $F_{k-m-1}$ have negative indices, we will use the extension of the Fibonacci sequence to negative integers and the corresponding identity: $F_{-n}= (-1)^{n+1} F_n $, After applying this to our equality, we get:
        $$ F_{m+k+1}\cdot (-1)^{m-k-1+1} F_{m-k-1}+F_{m+k}\cdot (-1)^{m-k+1}F_{m-k} =  F^2_k+ F^2_{k+1} $$
        Now we simplify this to:
            $$  (-1)^{m-k} (F_{m+k+1} F_{m-k-1})- (-1)^{m-k}(F_{m+k} F_{m-k}) =  F^2_k+ F^2_{k+1} $$
$$  (-1)^{m-k} (F_{m+k+1} F_{m-k-1}-F_{m+k} F_{m-k}) =  F^2_k+ F^2_{k+1} $$
    As desired $\square$.
