# Solution

Conjecture and prove a formula for the product $F_{i-1} \cdot F_{i+1}$, where $F_i$ represents the $i^{th}$ Fibonacci number.

We conjecture that

$$F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$$

We can prove this via strong induction...

First, we deal with the basis step. When $n=1$, we have $F_0 \cdot F_2 = 1 \cdot 2 = 2$, and $F_1^2 + (-1)^2 = 1 + 1 = 2$. These values being equal, the basis step is established.

Now, we turn our attention to the inductive step.

We need to show that if $F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$ when $n \le k$, then $F_k \cdot F_{k+2} = F_{k+1}^2 + (-1)^{k+2}$.

To this end, consider the following...

$$\begin{array}{rcl} F_k \cdot F_{k+2} &=& (F_{k-2} + F_{k-1})(F_k + F_{k+1})\\\\ &=& F_{k-1} F_{k+1} + F_{k-1} F_k + F_{k-2} F_k + F_{k-2} F_{k+1}\\\\ &=& \left[ F_k^2 + (-1)^{k+1} \right] + F_{k-1} F_k + \left[ F_{k-1}^2 + (-1)^k \right] + F_{k-2} F_{k+1}\\\\ &=& F_k^2 + F_{k-1} F_k + F_{k-1}^2 + F_{k-2} F_{k+1}\\\\ &=& (F_k^2 + 2 F_{k-1} F_k + F_{k-1}^2) -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& (F_k + F_{k-1})^2 -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 -F_{k-1} F_k + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 - F_{k-1} (F_{k-2} + F_{k-1}) + F_{k-2} F_{k+1}\\\\ &=& F_{k+1}^2 + F_{k-2} (F_{k+1} - F_{k-1}) - F_{k-1}^2\\\\ &=& F_{k+1}^2 + F_{k-2} F_k - F_{k-1}^2\\\\ &=& F_{k+1}^2 + F_{k-1}^2 + (-1)^k - F_{k-1}^2\\\\ &=& F_{k+1}^2 + (-1)^k\\\\ &=& F_{k+1}^2 + (-1)^{k+2} \end{array}$$

Hence, by the principle of mathematical induction, the following holds for all integers $n \ge 1$: $$F_{n-1} \cdot F_{n+1} = F_n^2 + (-1)^{n+1}$$ QED.