Statement on the Fibonacci sequence 
Question: Let $n,m,\in\mathbb{N^*}$ with $n>1$ and let $u_n$ denote the $n$-th  term of the Fibonacci sequence, then $$u_{n+m}=u_{n-1}u_m+u_nu_{m+1}$$

I know these theorems:

*

*Two consecutive terms of the Fibonacci sequence are relatively prime.

*The Fibonacci sequence, we have $u_n\mid u_m\Longleftrightarrow n\mid m$

*In the Fibonacci sequence, we have $(u_m,u_n)=u_{(m,n)}$$$$$I tried:

I) $n=2$ then $$u_{m+2}=u_1u_m+u_2u_{m+1}\Longrightarrow u_{m+2}=u_m+u_{m+1}\;\;\;OK$$
  
  II) Hypothesis: $$u_{n+m}=u_{n-1}u_m+u_{n}u_{m+1}$$
III) Thesis: $$u_{n+m+1}=u_{n}u_m+u_{n+1}u_{m+1}$$
Show:$$u_{n+m+1}=\underbrace{u_{n+m}}_{\text{hypot.}}+u_{n+m-1}=\\=u_{n-1}u_m+u_{n}u_{m+1}+u_{n+m-1}$$ and now?

 A: One slick way -- which actually was supposed to be a motivating example for the linear algebra course I am currently teaching, but I already spent the first two lectures on other motivating examples without getting to it -- is to make use of matrix multiplication.  Namely, if 
$M = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]$
then for all positive integers $n$,
$M^n = \left[ \begin{array}{cc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right]$.
This is very straightforward to prove by induction.  And it is amazing how many of the basic identities about Fibonacci numbers drop right out of it.  Yours certainly does, simply using $M^{m+n} = M^m M^n$.  
A: Let's prove this using induction:
First we'll prove that it holds for one case and the we'll just add 1 to m and 1 to n:
Step 1:Basis
$n = 2, m = 1$
$$U_3 = U_1*U_1 + U_2*U_2$$
$$U_3 = 1*1 + 1*1$$
$$U_3 = 2\text{, which is true}$$
$n = 2, m = 2$
$$U_4 = U_1*U_2 + U_2*U_3$$
$$U_4 = 1*1 + 1*2$$
$$U_4 = 3\text{, which is true}$$
$n = 3, m = 1$
$$U_4 = U_2*U_1 + U_3*U_2$$
$$U_4 = 1*1 + 2*1$$
$$U_4 = 3\text{, which is true}$$
Step 2:Inductive hypothesis
In the basis we've shown that it works for at least two consecutive numbers, so the statement holds for at least one pair of $U_{n+m}$ and $U_{n+m+1}$
Step 3: Inductive step
Now we'll show that it holds for every 3 consecutive number:
We have:
$$U_{n+m+2} = U_{n+m+1} + U_{n+m}$$
$$U_{n+m+2} = U_{n-1}*U_{m+1} + U_n*U_{m+2} + U_{n-1}*U_m + U_n*U_{m+1}$$
$$U_{n+m+2} = U_{n-1}(U_m + U_{m+1}) + U_n(U_{m+1} + U_{m+2})$$
$$U_{n+m+2} = U_{n-1}*U_{m+2} + U_n*U_{m+3}$$
Q.E.D.
We've proven that if it holds for $U_{k}$ and $U_{k+1}$, then it holds also for $U_{k+2}$. Then if it holds for $U_{k+1}$ and $U_{k+2}$ it also holds for $U_{k+3}$ and so on.

Edit:
I prove it, but I don't think that this answer would satisfy you.
First let me explain the method to you. First we prove that it work for the smallest values of $n$ and $m$, the will prove that it's true for $n$ and $m+1$. Knowing that this is true, we'll show that this is work also for $n$ and $m+2$. So the idea is:
Let $n$ be constant term, you can choose any number $\ge 2$ in the basis we prove that the statement holds for $m = 1$ and $m = 2$. We'll use this fact to prove that it also holds for $m = 3$. Then we'll use the fact that for $m = 2$ and $m = 3$ the statement is true, so we'll simularly prove it holds for $m = 4$ and so on to infinity.
Basis
n = 2,m = 1
$$U_3 = U_1*U_1 + U_2*U_2$$
$$U_3 = 1*1 + 1*1$$
$$U_3 = 2\text{, which is true}$$
$n = 2, m = 2$
$$U_4 = U_1*U_2 + U_2*U_3$$
$$U_4 = 1*1 + 1*2$$
$$U_4 = 3\text{, which is true}$$
Inductive hypothesis
Statement holds for at least one par of consecutive Fibonacci numbers, let those be: $U_{n+m}$ and $U_{n+(m+1)}$.
Inductive step
We need to prove:
$$U_{n+(m+2)} = U_{n-1}*U_{m+2} + U_n*U_{m+3}$$
$$U_{n+(m+2)} = U_{n-1}(U_m + U_{m+1}) + U_n*(U_{m+1} + U_{m+2})$$
$$U_{n+(m+2)} = U_{n-1}*U_m + U_{n-1}*U_{m+1} + U_n*U_{m+1} + U_n*U_{m+2}$$
If we rearrange the terms we'll and make a substitition we'll get:
$$U_{n+(m+2)} = U_{n+(m+1)} + U_{n+m}$$
Which is obviously true.
Hope this make it clearer.
A: I tried:

I) $n=2$ then $$u_{m+2}=u_1u_m+u_2u_{m+1}\Longrightarrow u_{m+2}=u_m+u_{m+1}\;\;\;OK$$
II) Hypothesis: $$u_{n+m}=u_{n-1}u_m+u_{n}u_{m+1}$$
III) Thesis: $$u_{n+m+1}=u_{n}u_m+u_{n+1}u_{m+1}$$
Show:$$u_{n+m+1}=\underbrace{u_{n+m}}_{\text{hypot.}}+u_{n+m-1}=\\=u_{n-1}u_m+u_{n}u_{m+1}+u_{n+m-1}$$ and now?

