Proving $a_{n+1}^2=a_n·a_{n+2}+(-1)^n$ for Fibonacci sequence $\{a_n\}$ Let $a_1,a_2,a_3,...,a_n$ be the sequence of Fibonacci numbers.
Then we are required to prove by induction that $$a_{n+1}^2=a_n \cdot a_{n+2} + (-1)^n$$
My Attempt:
Initially putting a few values like $n=1,2 \: \text{or}\ 3$ makes it certain that we can move ahead.
Following the conventional way we can assume that the proposition holds true for some $k$ s.t.$$a_{k+1}^2=a_k \cdot a_{k+2} + (-1)^k  \tag1$$
To prove that the proposition holds for all values of $n$, we need to prove for the same for $n=k+1$ too.
In short we need to prove that:
$$a_{k+2}^2=a_{k+1} \cdot a_{k+3} + (-1)^{k+1}  \tag2$$
We know that 
$$ a_{k+2}^2=(a_{k+3}-a_{k+1})^2$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_{k+1}^2-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}^2+a_k \cdot a_{k+2} + (-1)^k-2\cdot a_{k+3}\cdot a_{k+1}$$
$$\Rightarrow a_{k+2}^2=a_{k+3}[a_{k+3}- 2\cdot a_{k+1}]+a_k \cdot a_{k+2} + (-1)^k \tag{*}$$
It is easy to see that $$a_{k+3}-a_{k+1}=a_{k+2} \tag3$$ 
And $$a_{k+2}-a_{k+1}=a_{k} \tag4$$
Adding equation $3$ and $4$, we get$$a_{k+3}-2a_{k+1}=a_{k} \tag5$$
Putting eq$(5)$ in eq$(*)$ 
$$a_{k+2}^2=a_{k+3}[a_k]+a_k \cdot a_{k+2} + (-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k[a_{k+3}+a_{k+2}]+(-1)^k $$
$$\Rightarrow a_{k+2}^2=a_k(a_{k+4})+(-1)^k$$
This I indeed completely different from something that I wanted to prove. Is there any way to get ahead of this or is there something wrong with what I have done?
 A: Induction step $n\to n+1$
$$\begin{align}a_{n+1}^2-a_n \cdot a_{n+2} &= a_{n+1}^2-a_n \cdot (a_{n+1}+a_n) \\
& =a_{n+1}(a_{n+1}-a_n)-a_n^2 
\\&= a_{n+1}a_{n-1}-a_n^2 
\\&= (a_n^2-a_{n+1}a_{n-1})
\\&= -(-1)^{n-1}\\
&= (-1)^n\end{align}$$
A: A sneaky linear algebra way. Let $$A_{n}=\begin{pmatrix}a_{n+1}&a_{n}\\a_{n+2}&a_{n+1}\end{pmatrix}$$
Then show that $$A_{n+1}=\begin{pmatrix}0&1\\ 1&1\end{pmatrix}A_n$$
using the recurrence relation.
Then show $$\det A_{n+1} = -\det A_n\tag{1}$$
Finally, the only induction you need is to show, using (1), that $$\det A_{n}=(-1)^{n-1}\det A_1$$
which makes the induction simple.
A: Your original statement is equivalent to
$$a_{n+1}^2=a_n(a_{n+1}+a_n)+(-1)^n,$$
that is
$$a_{n+1}^2-a_na_{n+1}-a_n^2=(-1)^n.\tag{1}$$
To prove $(1)$ all you need to do is to prove it for $n=1$ and verify
$$a_{n+2}^2-a_{n+1}a_{n+2}-a_{n+1}^2=-
(a_{n+1}^2-a_na_{n+1}-a_n^2)^.\tag{2}$$
This is an easy induction.
One can prove $(2)$ just by substituting $a_{n+1}+a_n$
for $a_{n+2}$ on the LHS.
