For all positive integers $i$, let $F_{i}$ denote the $i^{{th}}$ Fibonacci number, with $F_{1}=F_{2}=1$. We will show by induction that the identity

holds for all positive integers $n\geq 2$. When $n=2$, we can substitute in the values for $F_{1}$, $F_{2}$ and $F_{{3}}$ yielding the statement $2\times 1-1^{2}=(-1)^{2}$, which is true. Now suppose that the theorem is true when $n=m$, for some integer $m\geq 2$. Recalling the recurrence relation for the Fibonacci numbers, $F_{{i+1}}=F_{i}+F_{{i-1}}$, we have

by the induction hypothesis. So we get $F_{{m+2}}F_{m}-F_{{m+1}}^{2}=(-1)^{{m+1}}$, and the result is thus true for $n=m+1$. The theorem now follows by induction.