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