# Induction proof on Fibonacci sequence: $F(n-1) \cdot F(n+1) - F(n)^2 = (-1)^n$

I can't seem to solve this problem. It is:

The Fibonacci numbers $F(0), F(1), F(2),\dots$ are defined as follows:

\begin{align} F(0) &::= 0 \\ F(1) &::= 1 \\ F(n) &::= F(n-1) + F(n-2)\qquad(\forall n \ge 2)\end{align}

Thus, the first Fibonacci numbers are $0, 1, 1, 2, 3, 5, 8, 13,$ and $21$. Prove by induction that $\forall n \ge1$,

$$F(n-1) \cdot F(n+1) - F(n)^2 = (-1)^n$$

I'm stuck, as I my induction hypothesis was the final equation, and I replaced n in it with n+1, which gave me:

$$F(n) \cdot F(n+2) - F(n+1)^2 = (-1)^{n+1}$$

I then tried simplifying this using the first equation, which gave me: $$[(F(n-1) + F(n-2)]\cdot F(n+2) - F(n+1)^2 = (-1)^{n+1}$$

I then tried replacing $n$ in the first equation with $n+1$, but that just gave me

$$2F(n-1) + F(n-2)$$

I'm really not sure how to proceed, and I was hoping for some help. I'm new to induction and I'm hoping this is just an algebra problem and not a problem with the method, but any help would be greatly appreciated.

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You've written the wrong thing as a sum. $F_n\cdot F_{n+2} - F_{n+1}^2 = F_n(F_{n+1}+F_n) - F_{n+1}(F_n + F_{n-1})$. – Daniel Fischer Oct 12 '13 at 22:07

Just to be contrary, here's a (more instructive?) proof that isn't directly by induction:

Lemma. Let $A$ be the $2\times 2$ matrix $\begin{pmatrix}1&1\\1&0\end{pmatrix}$. Then $A^n= \begin{pmatrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix}$ for every $n\ge 1$.

This can be proved by induction on $n$ since $$A\begin{pmatrix}F_n & F_{n-1} \\ F_{n-1} & F_{n-2}\end{pmatrix} = \begin{pmatrix}F_n+F_{n-1} & F_{n-1}+F_{n-2} \\ F_n & F_{n-1}\end{pmatrix} = \begin{pmatrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{pmatrix}$$

Now, $F_{n+1}F_{n-1}-F_n^2$ is simply the determinant of $A^n$, which is $(-1)^n$ because the determinant of $A$ is $-1$.

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Basis: $n = 1$

$$F_{n-1} \cdot F_{n+1} - F_{n}^2 = (-1)^n$$ $$F_{0} \cdot F_{2} - F_{1}^2 = (-1)^n$$ $$0 \cdot 1 - 1 = -1$$ $$-1 = -1 \text{, which is true}$$

Inductive hypothesis: $n=k$

We assume that the statement holds for some number $k$

$$F_{k-1} \cdot F_{k+1} - F_{k}^2 = (-1)^k$$

Inductive step: $n = k+1$

We need to prove that the following statement holds:

$$F_{k} \cdot F_{k+2} - F_{k+1}^2 = (-1)^{k+1}$$

Starting from the inductive hypothesis we have:

$$F_{k-1} \cdot F_{k+1} - F_{k}^2 = (-1)^k$$

Multiply both sides by $-1$:

$$F_{k}^2 - F_{k-1} \cdot F_{k+1}= (-1)^{k+1}$$

Using the property on Fibonacci numbers we have:

$$F_{k}^2 - (F_{k+1} - F_{k}) \cdot F_{k+1}= (-1)^{k+1}$$

$$F_{k}^2 + F_{k} \cdot F_{k+1} - F_{k+1}^2 = (-1)^{k+1}$$

$$F_{k}(F_{k} + F_{k+1}) - F_{k+1}^2 = (-1)^{k+1}$$

$$F_{k} \cdot F_{k+2} - F_{k+1}^2 = (-1)^{k+1}$$

Q.E.D.

Note that his identity is called Cassini identity for Fibonacci Numbers, which is a generalization of the Catalan identity for Fibonacci Numbers, which states:

$$F_n^2 -F_{n-r}F_{n+r} = (-1)^{n-r}F_r^2$$

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Seems wrong to me. You are assuming what you want to prove, and then deriving that $(-1)^{k+1}=(-1)^{k+1}$. You need to start with the induction assumption for $k$ and prove it for $k+1$. Perhaps your math can be rearranged to provide this proof. – marty cohen Oct 13 '13 at 1:18
I don't think it's wrong at all. We started and using the hypothesis and algebraic transformation we reached something which is true, meaning that we proved the inductive step. Anyway I edit the answer and I hope it's better and clearer now. – Stefan4024 Oct 13 '13 at 1:49

You have written the wrong Fibonacci number as a sum. You know something about $F_{n-1},\, F_n$ and $F_{n+1}$ by the induction hypothesis, while $F_{n+2}$ is new. So you should write $F_{n+2} = F_{n+1} + F_n$. And in the other summand, write one factor too as a sum,

$$F_n\cdot F_{n+2} - F_{n+1}^2 = F_n(F_{n+1} + F_n) - F_{n+1}(F_n + F_{n-1})$$

can be easily and fruitfully related to the induction hypothesis.

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@Ralph I'm not sure why that edit was approved, but I do not feel it should have been. You added far more to the Answer than was originally written; Daniel's voice was essentially drowned out by your own. If you feel it is worthwhile to include an entire proof (but there is already Stefan4024's answer, so...) , I think you should write your own answer. – pjs36 Jun 10 at 21:31
Sure I can do that. I was surprised it let me edit it in the first place. The Stefan answer seems to be done backwards. – Ralph Yozzo Jun 10 at 23:12

Fibonacci Recurrence Definition: $$F_{n} = F_{n-1} + F_{n-2} \, (where \, n \ge 2), \, and \, F_{0}=0, F_{1}=1$$

Theorem: $$\forall n\ge 1: (F_{n+1}\cdot F_{n-1}) - F_{n}^2 = (-1)^n$$

Proof by Induction:

Base step: $n = 1$

$$F_{2} \cdot F_{0} - F_{1}^2 = (-1)^n$$ $$1 \cdot 0 - 1 = -1$$ $$-1 = -1 \text{, which is true}$$

Inductive hypothesis: $n=k$

We assume that the statement holds for some number $k$

$$(F_{k+1}\cdot F_{k-1}) - F_{k}^2 = (-1)^k$$

Inductive step: $n = k+1$

We need to prove that the following statement holds:

$$(F_{k+2}\cdot F_{k}) - F_{k+1}^2 = (-1)^{(k+1)}$$

But this can be rewritten as:

$$(F_{k}\cdot (F_{k+1} + F_{k})) - (F_{k+1} \cdot F_{k+1}) = (-1)^{(k+1)}$$

$$(F_{k}\cdot (F_{k+1} + F_{k})) - (F_{k+1} \cdot (F_{k} + F_{k-1})) = (-1)^{(k+1)}$$

$$(F_{k}^2) - (F_{k+1} \cdot F_{k-1}) = (-1)^{(k+1)}$$

since $\forall n:-1^{n+1} \cdot -1 = -1^{n}$

$$(F_{k+1} \cdot F_{k-1}) -F_{k}^2 = (-1)^{k}$$

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The inductive step is easiest to do by considering: $$(F_n F_{n +2} - F_{n + 1}^2) + (F_{n - 1} F_{n + 1} - F_n^2)$$ I.e., adding up cases $n$ and $n + 1$. Massaging this with the Fibonacci recurrence $F_{n + 1} = F_{n + 2} - F_n$ reduces to zero, so you know they have the same absolute value and alternating signs.

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