Proving this by induction for all n > 0... First, we show that this is true for n = 1 and n = 2. n=1: F2 FO=2*1=2, while (F1)^2 -(-1)^1=1- (-1)=2 n = 2: F3 * F1 = 3*1=3, while (F2)^2 -(-1)^2=4- 1 = 2. Next, we assume that the claim is true for n = k and use this to prove that it is also true for n = k+1: F(k+2) * F(k) = (F(k+1) + F(k)) * (F(k+1) - F(k-1)), using the Fibonacci recurrence rule = F(k+1)^2 - F(k+1) F(k-1) + F(k+1) F(k) - F(k) F(k-1) = F(k+1)^2 - [F(k)^2-(-1)^k] + [F(k+1) F(k) - F(k) F(k-1)], by the inductive hypothesis = F(k+1)^2 - [F(k)^2-(-1)^k] + F(k) [F(k+1) - F(k- 1)] = F(k+1)^2 [F(k)^2 -(-1)^k] + F(k) [F(k)], by the Fib. recurrence = F(k+1)^2 (-1)^(k+1). This completes the inductive step, and hence the proof