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

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*}
