I was wondering how to prove that
$$f(n+m+2) = f(n+1)f(m+1) + f(n)f(m)$$
where $f$ is the fibonacci sequence and n, m are positive integers.
Can be this done with induction?
I'm lost with this method of proof, because there are two variables.
Any idea or suggestion is welcome.
Can be this done with induction?
It can. More specifically, it can be done with strong induction on two variables. First I suggest looking at https://math.stackexchange.com/a/7665/146030 and thinking of why, in both cases, the first three statements implies the fourth.
We will prove the claim that
$$f(n+m+2)=f(n+1)f(m+1)+f(n)f(m).$$
To begin we define the fibonacci sequence as
\begin{align} f(0)&=0 \\ f(1)&=1 \\ f(n)&=f(n-1)+f(n-2), \text{for } n\ge2. \end{align}
When $n=0$ and $m=0$ then
\begin{align} f(n+m+2) &= f(2) \\ &= 1 \\ &= 1 \cdot1 + 0\cdot0 \\ &= f(1)f(1)+f(0)f(0) \\ &= f(n+1)f(m+1)+f(n)f(m) \end{align}
and so the statement is true when $n=m=0$.
To prove the statement true for all nonnegative $n,m$, we first induct on $n=k$ for a fixed $m$. Assume the statement true for all $0\leq k\leq n$. We now prove the statement for $k+1$.
\begin{align} f((k+1)+m+2) &= f(k+m+3) \\ &= f(k+m+2) + f(k+m+1) \\ &= f(k+m+2) + f((k-1)+m+2) \\ &= \big[f(k+1)f(m+1)+f(k)f(m)\big] + \big[f(k)f(m+1)+f(k-1)f(m)\big] \\ &= \big[f(k+1)f(m+1)+f(k)f(m+1)\big] + \big[f(k)f(m)+f(k-1)f(m)\big] \\ &= \big[f(k+1)+f(k)\big]f(m+1) + \big[f(k)+f(k-1)\big]f(m) \\ &= f(k+2)f(m+1) + f(k+1)f(m) \\ &= f((k+1)+1)f(m+1) + f(k+1)f(m) \end{align}
And so by mathematical induction the statement is true for all $n$ and that fixed $m$. We can see that a similar inductive proof works for a fixed $n$ and $m=k$. Thus we can conclude the statement is true.
I'm afraid this statement is impossible to prove as it is wrong. Indeed this can be seen with $n=1, m=0$: $$f(1+0+2)=f(3)=f(2)+f(1)=2$$ whereas for your proposition $$f(1+0+2)=f(1+1)f(0+1)+f(0)f(1)=f(2)f(1)+f(0)f(1)=1 \cdot 1+0 \cdot 1=1$$
As for Trevor Fancher's induction, it is completely right except that his basic step is incomplete.
(I shall refer to the statement as $P(n,m)$). Indeed what he does is prove that for a fixed p, $P(n,p)$ and $P(n+1,p)$ true $\Rightarrow P(n+2,p)$ true $\forall n \in \mathbb{N}$. By symetry he also proves that for a fixed p, $P(p,m)$ and $P(p,m+1) \Rightarrow P(p,m+2)$ $\forall m \in \mathbb{N}$. Then to conclude the proof what we do is:
But in this proof we only see that $P(0,0)$ is true, not $P(0,1)$,$P(1,0)$ nor $P(1,1)$ hence this proof is incorrect as they are necessary to conclude the proof (and thus for the induction to be valid).
I believe the true similar statement you were looking to prove is $$f(n+m+1)=f(n+1)f(m+1)+f(n)f(m)$$ which is proven is a quasi-identical manner.
Your proposition is not true (as it can be seen for $P(0,1) : f_{0+1+2} ≠ f_{0+1}f_{1+1} + f_{0}f_{1}$). Some elements are missing to build such a proof in a rigourous way : You use both ranks $−1$ and $$. Then your inductive step is based on two consecutive ranks, means we want to use this for our induction : $P(k,n) \land P(k+1,n)\implies P(k+2,n)$. It comes that you have to prove $P(0,n)\implies P(1,n)$: which is false. One should have proved that the proposition holds for $P(0,0), P(0,1), P(1, 0), P(1,1)$ to be able to use 2 consecutive ranks in the induction. The correct proposition is rather $f_{m+n+1}$ = $f_{m+1}f_{n+1} + f_mf_n$ $(Analysis II w/ Anna) > All$
