# Fibonacci Mystery Proof

Thanks to an anonymous reader, who we shall just call Brendan, we finally have this proof for our Fibonacci Mystery.

- First, we define F(n) as the nth number in the Fibonacci Series. We know when n=1, then F(n)=F(1)=1 and F(n+1)=F(2)=1 and finally F(n+2)=F(3)+2. Therefore, this is also true:

F(n)*F(n+2) - F(n+1)^2 = 1

F(1)*F(3) - F(2)^2 = 1

1*2 - 1^2 = 1

1 = 1 - Next, we assume when n is any odd number:

F(n)*F(n+2) - F(n+1)^2 = 1 (a) for odd n - Therefore, we will show that this is also true when n is an odd number.

F(n+2)*F(n+4) - F(n+3)^2 = 1 (b) - We start off with (a):

F(n)*F(n+2) - F(n+1)^2 = 1

We expand F(n+2), since we know F(n+2)=F(n+1)+F(n).

F(n)*( F(n+1)+F(n) ) - F(n+1)^2 = 1

Multiply F(n) on the left side of the equation.

F(n)^2 + F(n)*F(n+1) - F(n+1)^2 = 1 (a)

For now, we will put this equation (a) aside. We will use this equation (a) in the future.

- If we return to equation (b):

F(n+2)*F(n+4) - F(n+3)^2 = 1

First, we expand F(n+3), because we know F(n+3)=F(n+2)+F(n+1).

F(n+2)*F(n+4) - (F(n+2)+F(n+1))^2 = 1

Take the square.

F(n+2)*F(n+4) - (F(n+2)^2 + 2*F(n+2)*F(n+1) + F(n+1)^2) = 1

Next, we expand F(n+4), because we know F(n+4)=F(n+3)+F(n+2).

F(n+2)*(F(n+3)+F(n+2)) - (F(n+2)^2 + 2*F(n+2)*F(n+1) + F(n+1)^2) = 1

We multiple F(n+2) to this new part of the equation.

F(n+2)*F(n+3) + F(n+2)^2 - (F(n+2)^2 + 2*F(n+2)*F(n+1) + F(n+1)^2) = 1

We again expand F(n+3), because we know F(n+3)=F(n+2)+F(n+1).

F(n+2)*(F(n+2)+F(n+1)) + F(n+2)^2 - (F(n+2)^2 + 2*F(n+2)*F(n+1) + F(n+1)^2) = 1

We multiple F(n+2) to this new part of the equation.

F(n+2)^2 + F(n+2)*F(n+1) + F(n+2)^2 - (F(n+2)^2 + 2*F(n+2)*F(n+1) + F(n+1)^2) = 1

Simplify.

2*F(n+2)^2 + F(n+2)*F(n+1) - F(n+2)^2 - 2*F(n+2)*F(n+1) - F(n+1)^2 = 1

Simplify some more.

F(n+2)^2 - F(n+2)*F(n+1) - F(n+1)^2 = 1

We again expand F(n+2), because we know F(n+2)=F(n+1)+F(n).

(F(n+1)+F(n))^2 - F(n+2)*F(n+1) - F(n+1)^2 = 1

We expand the square.

F(n+1)^2 + 2F(n+1)*F(n) + F(n)^2 - F(n+2)*F(n+1) - F(n+1)^2 = 1

We simplify again.

2F(n+1)*F(n) + F(n)^2 - F(n+2)*F(n+1) = 1

We again expand F(n+2), because we know F(n+2)=F(n+1)+F(n).

2F(n+1)*F(n) + F(n)^2 - (F(n+1)+F(n))*F(n+1) = 1

Multiply.

2F(n+1)*F(n) + F(n)^2 - (F(n+1)^2+F(n)*F(n+1)) = 1

Simplify.

2F(n+1)*F(n) + F(n)^2 - F(n+1)^2 - F(n)*F(n+1) = 1

Simplify.

F(n+1)*F(n) + F(n)^2 - F(n+1)^2 = 1

And this equals to:

(a) = 1

QED for all odd integers.

- We can use induction on even n numbers.

We know that when n=2, we get:

F(n)*F(n+2) - F(n+1)^2 = -1

F(2)*F(4) - F(3)^2 = -1

1*3 - 2^2 = -1

- We assume when n is any even number:

F(n)*F(n+2) - F(n+1)^2 = -1 (c) for even n - Therefore, we can show that:

F(n+2)*F(n+4) - F(n+3)^2 = -1 (d)

Using the same sequence of steps as above that we used to derive equation (a) with the final line being

(c) = -1

Thus, this is also true:

(d) = -1

by Phil for Humanity

on 10/10/2011