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?

Here is a proof based upon Binets formula
\begin{align*}
F_k=\frac{\varphi^k-\psi^k}{\varphi-\psi}\qquad k\geq 0\tag{1}
\end{align*}
where $\varphi=\frac{1}{2}\left(1+\sqrt{5}\right), \psi=\frac{1}{2}\left(1-\sqrt{5}\right)=-1/\varphi$.

We start with the right-hand side of OPs formula and obtain
  \begin{align*}
F_k^2=\left(\frac{\varphi^k-\psi^k}{\varphi-\psi}\right)^2&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2\left(\varphi\psi\right)^k+\psi^{2k}\right)\\
&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\tag{2}\\
\end{align*}
  Putting $k\to k+1$ in (2) we get
  \begin{align*}
\color{blue}{F_k^2+F_{k+1}^2}&=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}-2(-1)^k+\psi^{2k}\right)\\
&\qquad+\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k+2}-2(-1)^k+\psi^{2k+2}\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{3}
\end{align*}

And now the left-hand side.

We obtain
  \begin{align*}
F_{m+k}F_{m-k}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{m+k}-\psi^{m+k}\right)\left(\varphi^{m-k}-\psi^{m-k}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\varphi^{m+k}\psi^{m+k}-\varphi^{m-k}\psi^{m+k}+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}-\left(\varphi\psi\right)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)+\psi^{2m}\right)\\
&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\tag{4}
\end{align*}
  Replacing $k$ with $k+1$ in (4) we get
  \begin{align*}
F_{m+k+1}F_{m-k-1}&=\frac{1}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\tag{5}
\end{align*}
The left-hand side of OPs formula can now be rewritten using (4) and (5) as
  \begin{align*}
\color{blue}{(-1)^{m-k}}&\color{blue}{\left(F_{m+k+1}F_{m-k-1}-F_{m+k}F_{m-k}\right)}\\
&=\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}+(-1)^{m-k}\left(\varphi^{2k+2}+\psi^{2k+2}\right)\right)\\
&\qquad-\frac{(-1)^{m-k}}{\left(\varphi-\psi\right)^2}\left(\varphi^{2m}+\psi^{2m}-(-1)^{m-k}\left(\varphi^{2k}+\psi^{2k}\right)\right)\\
&\,\,\color{blue}{=\frac{1}{(\varphi-\psi)^2}\left(\varphi^{2k}\left(1+\varphi^2\right)+\psi^{2k}\left(1+\psi^2\right)\right)}\tag{6}
\end{align*}
  A comparison of (3) and (6) shows OPs identity is valid.

