Honsberger's Identity

Theorem
Let $F_k$ be the $k$th Fibonacci number.

Then:
 * $\displaystyle \forall m,n : F_{m+n} = F_{m-1} F_n + F_m F_{n+1}$

Proof
From the initial definition of Fibonacci numbers, we have:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall m,n : F_{m+n} = F_{m-1} F_n + F_m F_{n+1}$

Basis for the Induction
$P \left({1}\right)$ is the case $F_{m+1} = F_{m-1} F_1 + F_m F_2$, which holds.

$P \left({2}\right)$ is the case $F_{m+2} = F_{m-1} F_2 + F_m F_3$, which also holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ and $P \left({k-1}\right)$ are true, where $k > 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle F_{m+k} = F_{m-1} F_k + F_m F_{k+1}$ and $\displaystyle F_{m+k-1} = F_{m-1} F_{k-1} + F_m F_k$

Then we need to show:
 * $\displaystyle F_{m+k+1} = F_{m-1} F_{k+1} + F_m F_{k+2}$

Induction Step
This is our induction step:

So $P \left({k}\right) \land P \left({k-1}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m,n : F_{m+n} = F_{m-1} F_n + F_m F_{n+1}$