Express $F_{m+n}$ in terms of $F_{m-1}, F_{m}, F_{n}$, and $F_{n+1}$.
We would like to express $F_{m+n}$ as a sum of products of the form $F_{k} F_{l}$ with $k$ close to $m$ and $l$ close to $n$. We realize that it is the terms to the power of $k-l$ that need to be eliminated, and it is advantageous to choose $k$ and $l$ in each product so that $k-l$ is constant. Hence the idea of calculating $F_{m-1} F_{n}$ and $F_{m} F_{n+1}$. We find:

$$
\begin{gathered}
F_{m-1} F_{n}=\frac{1}{5}\left(\phi^{m+n-1}+\left(-\phi^{-1}\right)^{m+n-1}-(-1)^{n}\left(\phi^{m-n-1}+\left(-\phi^{-1}\right)^{m-n-1}\right)\right) \\
F_{m} F_{n+1}=\frac{1}{5}\left(\phi^{m+n+1}+\left(-\phi^{-1}\right)^{m+n+1}-(-1)^{n+1}\left(\phi^{m-n-1}+\left(-\phi^{-1}\right)^{m-n-1}\right)\right)
\end{gathered}
$$

To eliminate the parasitic terms, we add the preceding equalities. At the same time, we transform all terms with powers "close" to $m+n$ into terms with power $m+n$, even if it means introducing multiplicative factors. We finally obtain:

$$
\left.F_{m-1} F_{n}+F_{m} F_{n+1}=\frac{1}{5}\left(\left(\phi+\phi^{-1}\right) \phi^{m+n}-\left(\phi+\phi^{-1}\right)\left(-\phi^{-1}\right)^{m+n}\right)\right)
$$

Then, using the fact that $\phi+\phi^{-1}=\sqrt{5}$ and simplifying, we finally get $F_{m-1} F_{n} + F_{m} F_{n+1} = F_{m+n}$.

The Fibonacci sequence has many interesting arithmetic properties.