First, define the Fibonacci numbers:
Let $F_n$ be the sequence of Fibonacci numbers, given by
$F_0 = 0, F_1 = 1, \text{ and}\quad F_n = F_{n-1} + F_{n-2}\; \text{ for}\; n \geq 2.$
Hints:
$(1)$ Use the definition of $F_n$.
- E.g., $F_{n+1} = F_{n} + F_{n-1}$
$(2)$ You can use the following good-to-know identities:
i) $F_{n-1}^2 + F_n^2 = F_{2n}$.
ii) $F_{n-1}F_n + F_n F_{n+1} = F_{2n+1}$.
Note that the above identities follow from the more general identity:
$(I)$: For $n, m \in \mathbb{N}$: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$
Proof of $(I)$:
Fix $n \in \mathbb{N}$. We shall use induction on $m$. For $m=1$, the right-hand side of the equation becomes $F_{n-1}F_{1} + F_{n}F_{2} = F_{n-1} + F_{n},$ which is equal to $F_{n+1}$. When $m=2$, the equation is also true.( I hope you can prove this!).
Now assume, that the result is true for $k=3,4, \cdots , m$. We want to show that the result is true for $k=m+1$. $ \text{For} \ k=m-1 \ \text{we have} \quad F_{n+m-1} = F_{n-1}F_{m-1} + F_{n}F_{m},\,\text{ and}$ $ \text{For} \ k = m \ \text{we have} \quad F_{m+n}=F_{n-1}F_{m} + F_{n}F_{m+1}$ Adding both the sides you will get $F_{m+n-1} + F_{m+n} = F_{m+n+1} = F_{n-1}F_{m+1} + F_{n}F_{m+2},$ $\text{so,}\;\; F_{m+n}=F_{n-1}F_m + F_{n}F_{m+1}$
Identities (i) and (ii) follow from $(I)$ by putting $m = n$ and manipulating the expressions.