Strong Induction for Fibonacci number related identity $f_{n+m} = f_{m}f_{n+1} + f_{m-1} f_n$

Question

Let $f_n$ be the $n^{th}$ Fibonacci number. Let $m$ be a fixed strictly positive integer.

Prove by strong induction that for all $n\ge 0$, $$f_{n+m} = f_{m}f_{n+1} + f_{m-1} f_n$$

edit: $f_{n+m} = f_{m}f_{n+1} + f_{m-1} f_n$ instead of $f_{n-m} = f_{m}f_{n+1} + f_{m-1} f_n$
(so sorry lol)

Answer 1

Hint: \begin{align*} f_{n+1-m} & =f_{n-m}+f_{(n-1)-m}\\ & =f_{m}f_{n+1}+f_{m-1}f_{n}+f_{m}f_{n}+f_{m-1}f_{n-1}\\ & =f_{m}(f_{n+1}+f_{n})+f_{m-1}(f_{n}+f_{n-1})\\ & =\cdots \end{align*}