Proving this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ by induction Where $n \in \mathbb{N}$ 
and
$$
F_n = \begin{cases}
0                 & \text{ if } n = 0 \\
1                 & \text{ if } n = 1 \\
F_{n-1} + F_{n-2} & \text{ if } n > 1
\end{cases}
$$
This is basically describing the famous Fibonacci sequence.
If we try the base case with 1, it works (for $0$ I am not sure...)
When $n = 1$
$2 \cdot0 - 1^2 = (-1)^1 = -1$
For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.
Now, for the inductive step, we try to prove for $n+1$, so for  $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.
Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even). 
Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).
Or simply:
$$
(-1)^n = -(-1)^{n+1}
$$
$$
1 \cdot (-1)^n = -(-1)\cdot(-1)^n
$$
$$
1 = -(-1)
$$
Which is true.
I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...
 A: The identity may be derived from the interesting fact that
$$\left ( \begin{array} \\ 1 & 1\\1 & 0 \\ \end{array} \right ) ^k = \left ( \begin{array} \\ F_{k+1} & F_k\\F_k & F_{k-1} \\ \end{array} \right )$$
This may be proved via induction.
The result follows from taking the determinant of both sides.
A: Your proof is completely incorrect. You need to prove why $$F_{n+2} F_{n} - F_{n+1}^2 = -(F_{n+1} F_{n-1} - F_{n}^2)$$ to prove that $F_{n+2} F_{n} - F_{n+1}^2 = (-1)^{n+1}$.
You can check the base case. For the inductive step, we have
\begin{align}
F_{n+2} F_{n} - F_{n+1}^2 & = (F_{n+1}+F_{n})F_{n} - F_{n+1}^2 & \left(\because F_{n+2} = F_{n+1} + F_{n} \right)\\
& = F_{n+1} F_{n} + F_{n}^2 - F_{n+1}^2\\
& = F_{n+1} F_{n} - F_{n+1}^2 + F_{n}^2\\
& = F_{n+1}(F_{n}-F_{n+1}) + F_{n}^2\\
& = -F_{n+1} F_{n-1} + F_{n}^2 & \left(\because F_{n+1} = F_{n} + F_{n-1}\right)\\
& = -\left(F_{n+1}F_{n-1} - F_{n}^2\right)
\end{align}
Now you are done.
A: I think you have the right idea; you wrote down the important equality
$$-(F_{n+1}\cdot F_{n-1}-F_n^2)=F_{n+2}\cdot F_n-F_{n+1}^2$$
but you got there going the wrong way - that is to say you assumed the thing you were trying to prove to get there. The rest of your proof from there is a non-sequitir which proves that $(-1)^{n+1}=-(-1)^n$, but that's not what you wanted.
What you ought to do is take the expression
$$F_{n+2}\cdot F_n-F_{n+1}^2$$
and expand $F_{n+2}=F_{n}+F_{n+1}$ and $F_{n+1}=F_{n-1}+F_{n}$ and rearrange to get the expression
$$-(F_{n+1}\cdot F_{n-1}-F_n^2)$$
which you can, using the inductive hypothesis, change to
$$-(-1)^n=(-1)^{n+1}$$
completing the proof. Thus, you prove the important equality through purely algebraic means and only use the inductive hypothesis at the end.
A: Note: Based on your answers and on what I have learnt, I will try to give my complete easy answer to this question/problem.
Ignoring the base case, which I think all we agree that I proved in my question, I will pass directly to the inductive step:
I will assume that $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true for $n$, and I will try to prove that also $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$ is true for $n + 1$.
The most important thing to note (immediately from the beginning) is that if $(-1)^n$ is even, then $(-1)^{n+1}$ is odd, and vice-versa, that is if $(-1)^n$ is odd, then $(-1)^{n+1}$ is even. This because, remember, a negative number raised to the power of an even number, becomes a positive number, but if raised to the power of an odd number, remains a negative number.
That said, we can try to verify if the following equation is true:
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n+2}\cdot F_n-F_{n+1}^2)$$
Now, we must have some imagination, or try some times to come out with something useful. In this case, I replaced $F_{n + 2}$ on the right side expression with $(F_{n + 1} + F_n)$, from the definition given in my question of $F_n$ (I must admit this step is one of the most difficult steps):
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -((F_{n + 1} + F_n)\cdot F_n-F_{n+1}^2)$$
Now, I will just distribute the right side expression:
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1} \cdot F_n + F^2_n - F_{n+1}^2)$$
In the following step, I will just exchange the position of $F_{n+1}^2$ with $F^2_n$, because it comes out that is the right thing to do (magical step, eh? this is why maths is considered difficult, because it's a lot of imagination and providences of god)
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1} \cdot F_n - F_{n+1}^2 + F^2_n)$$
Picking from the first 2 terms of the right side expression $F_{n + 1}$, we have:
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -(F_{n + 1}\cdot (F_n - F_{n+1}) + F^2_n)$$
Now, try to guess what $(F_n - F_{n+1})$ is? 
(yes, don't be mad with me, I did not invent this) 
From the definition of what $F_n$ is in my question, we arrive to say that:
$$F_n - F_{n+1} = -F_{n- 1}$$
$$-F_{n+1} = -F_{n- 1} - F_n $$
$$F_{n+1} = F_n + F_{n - 1} $$
Thus, $(F_n - F_{n+1})$ can be replaced with $-F_{n- 1}$ in my right side expression, and we obtain:
$$F_{n+1}\cdot F_{n-1}-F_n^2 = -( -F_{n + 1}\cdot F_{n- 1} + F^2_n)$$
Simplifying the minuses, we have:
$$F_{n+1}\cdot F_{n-1}-F_n^2 = F_{n + 1}\cdot F_{n- 1} - F^2_n$$
which is true, so we proved that if $P(n)$ is true, then also $P(n + 1)$, for all $n > 1$.
