I was trying to prove the following identity of Fibonacci using induction.
$$F_{n+1}^2-F_{n}F_{n+2}=(-1)^n, \forall n \in\mathbb{N}$$
My proof goes as follows-
We will proceed by mathematical induction.If $n=1$ then we have $F_{2}^2-F_{1}F_{3}=1-2=-1$ and $(-1)^1=-1$.Therefore the identity holds for $n=1$.
Suppose that $F_{k+1}^2-F_{k}F_{k+2}=(-1)^k$ for some $k \in\mathbb{N}$ (inductive hypothesise). Notice that: $$ F_{(k+1)+1}^2-F_{k+1}F_{(k+1)+2} = F_{k+2}^2-F_{k+1}F_{k+3} \\ =F_{k+2}^2-F_{k+1}(F_{k+2}+F_{k+1})$$ [since $F_{k+3}=F_{k+2}+F_{k+1}$ motivation was to generate $F_{k+1}^2$]. $$=F_{k+2}^2-F_{k+1}^2-F_{k+1}F_{k+2} \\=F_{k+2}^2-F_{k+1}^2-(F_{k+2}-F_{k})F_{k+2}$$ [since $F_{k+2}=F_{k+1}+F_k$ motivation was to generate $F_{k}F_{k+2}$] $$=F_{k+2}^2-F_{k+1}^2-F_{k+2}^2+F_kF_{k+2} \\=-(F_{k+1}^2-F_kF_{k+2}) \\ = -(-1)^k$$ [by induction hypothesis]
Notice that $(-1)^k$ gives $1$ at even values of $k$ and $-1$ at odd values of $k$. Multiplying by $-1$ reverses this order. Also If $k$ is even then $(-1)^{k+1}$ gives $-1$ and if $k$ is odd then $(-1)^{k+1}$ gives $1$. Therefore $-(-1)^k$ is equivalent to $(-1)^{k+1}$. Therefore, we can replace $-(-1)^k$ by $(-1)^{k+1}$. As desired.
is this proof correct? if not can you share a hint?
