Peter wrote: The Fibonacci sequence $\{F_{n}\}$ is defined by \[F_{1}=1, \; F_{2}=1, \; F_{n+2}=F_{n+1}+F_{n}.\] Show that $F_{mn}-F_{n+1}^{m}+F_{n-1}^{m}$ is divisible by $F_{n}^{3}$ for all $m \ge 1$ and $n>1$. Once again let's use induction for $m$. 1. For $m=1$ everything is obvious. 2. Let's suppose that we have proved the statement for $m=k$. 3. And finally let's prove it for $m=k+1$. \[F_{nk+n}=F_{nk}F_{n+1}+F_{nk-1}F{n}\equiv F_{n+1}(F_{n+1}^{k}-F_{n-1}^{k})+F_{nk-1}F_{n}\] => \[F_{n(k+1)}-F_{n+1}^{k+1}+F_{n-1}^{k+1}\equiv F_{nk-1}F_{n}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}\] As we have proved here:http://www.problem-solving.be/pen/viewtopic.php?t=415 \[F_{nk-1}\equiv F_{n-1}^{k}(mode F_{n}^{2})\] => \[F_{n}F_{nk-1}\equiv F_{n}F_{n-1}^{k}(mode F_{n}^{3})\] => \[F_{nk-1}F_{n}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}\equiv F_{n}F_{n-1}^{k}-F_{n+1}F_{n-1}^{k}+F_{n-1}^{k+1}=0\] we are done!