TITLE: Fibonacci addition law $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
QUESTION [24 upvotes]: Question: Let $F_n$ the sequence of Fibonacci numbers, given by $F_0 = 0, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. Show for $n, m \in \mathbb{N}$: $$F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$$
My (very limited) attempt so far: after creating a small list of the values $F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55$ i can see that yes it does seem to work for instance $F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9$. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this?

REPLY [5 votes]: There are several good answers already, but I thought I would add the following derivation because it is one of the few uses I know for the sum property of permanents; namely,
If $A$, $B$, and $C$ are matrices with identical entries except that one row (column) of $C$, say the $k^{th}$, is the sum of the $k^{th}$ rows (columns) of $A$ and $B$, then $\text{ per } A + \text{ per } B = \text{per } C$.
Start with the matrices $\begin{bmatrix} F_n & F_{n-1} \\ F_0 & F_1 \end{bmatrix}$ and $\begin{bmatrix} F_n & F_{n-1} \\ F_1 & F_2 \end{bmatrix}$.  Since $F_0 = 0$ and $F_1 = F_2 = 1$, they have permanents $F_n$ and $F_n + F_{n-1} = F_{n+1}$, respectively.  Applying the Fibonacci recurrence and the permanent sum property,  we have $\text{ per } \begin{bmatrix} F_n & F_{n-1} \\ F_2 & F_3 \end{bmatrix} = F_{n+2}$.  By continuing to construct new matrices whose second rows are the sums of the second rows of the previous two matrices, this process continues until we have $F_n F_{m+1} + F_{n-1}F_m = \text{ per} \begin{bmatrix} F_n & F_{n-1} \\ F_m & F_{m+1} \end{bmatrix} = F_{n+m}.$
For more on this approach (but with determinants), see this paper I wrote a few years ago: "Fibonacci Identities via the Determinant Sum Property," The College Mathematics Journal, 37 (4): 286-289, 2006.