Proof Cassinis identity with induction and Fibonacci sequence

Let $p_1,p_2,p_3,...$ be the numbers from the Fibonacci sequence $(1,1,2,3,5,...)$

Proof Cassini's identity:

$p^2_{n+1}-p_n*p_{n+2}=(-1)^n$, where n is a natural number

I have tried to prove it by induction.

First I let $n=1$.

$1^2-1*2=(-1)^1\Rightarrow 1-2=-1\Rightarrow-1=-1$

As it's true for $n=1$ it must be true for all $n$ if it's true for $n+1$. So what I have to proof is that

$p^2_{n+2}-p_{n+1}p_{n+3}=(-1)^{n+1}$

I am uncertain how to proceed from here and if I'm correct so far.

Solution 1:

$$p_{n+2}^2-p_{n+1}p_{n+3}=p_{n+2}(p_{n}+p_{n+1})-p_{n+1}(p_{n+1}+p_{n+2})=\cdots.$$

Solution 2:

we denote the sequence by $F_{k}$ and we assume that the identity for $n=k$ $$F_{k+1}F_{k-1}-F_k^2=(-1)^{k+1}$$ plugging $$F_{k-1}=F_{k+1}-F_k$$ in the given equation then we have $$F_{k+1}(F_{k+1}-F_k)-F_k^2=(-1)^{k+1}$$ $$F_{k+1}^2-F_{k+1}F_k-F_k^2=(-1)^{k+1}$$ $$F_{k+1}^2-F_k(F_{k+1}+F_k)=(-1)^{k+1}$$ $$F_{k+1}^2-F_kF_{k+2}=(-1)^{k+1}$$ and this is the identitiy for $n=k+1$