= Solution
Fix $n\geq1$ and use <mathematical induction> on $m$ with two initial cases, as appropriate for the second-order <Fibonacci recurrence>. At $m=0$, the right side is $F_{n-1}F_0+F_nF_1=F_n$. At $m=1$, it is $F_{n-1}+F_n=F_{n+1}$.
If the identity holds at $m$ and $m+1$, then
$$
\begin{aligned}
F_{n+m+2}
&=F_{n+m+1}+F_{n+m}\\
&=F_{n-1}(F_{m+1}+F_m)+F_n(F_{m+2}+F_{m+1})\\
&=F_{n-1}F_{m+2}+F_nF_{m+3}.
\end{aligned}
$$
Hence the <induction> establishes
$$
\boxed{F_{n+m}=F_{n-1}F_m+F_nF_{m+1}\quad(n\geq1,\ m\geq0)}.
$$
Back to article page
