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 $F_{mn-1}-F_{n-1}^{m}$ is divisible by $F_{n}^{2}$ for all $m \ge 1$ and $n>1$. Once again let's use the property of Fibonacci sequence that we proved here: http://www.problem-solving.be/pen/viewtopic.php?t=413 So we have that $F_{a+b}=F_{a}F_{b+1}+F_{a-1}F_{b}$ Let's lead the solution by induction for $m$ 1. For $m=1$ we have that $F_{n}^{2}|F_{n-1}-F_{n-1}=0$ 2. Let's suppose that we have proved the statement of the problem for $m=k$ 3. Now let's procced to prove that $F_{n}^{2}|F_{n(k+1)-1}-F_{n-1}^{k+1}$. We have that $F_{n(k+1)-1}=F_{nk-1}F_{n+1}+F_{nk-2}F_{n}\equiv F_{n-1}^{k}F_{n+1}+(F_{nk}-F_{nk-1})F_{n}(modF_{n}^{2})$ From the property proven here: http://www.problem-solving.be/pen/viewtopic.php?t=414 we have that $\gcd(F_{nk};F_{n})=F_{n}$ then $F_{n}|F_{nk}$. thus \[F_{n-1}^{k}F_{n+1}+(F_{nk}-F_{nk-1})F_{n}\equiv F_{n-1}^{k}F_{n+1}-F_{n-1}^{k}F_{n}=F_{n-1}^{k+1}(modeF_{n}^{2})\] so \[F_{n}^{2}|F_{n(k+1)-1}-F_{n-1}^{k+1}.\] We are done!