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Fibonacci Mystery Proof


Thanks to an anonymous reader, who we shall just call Brendan, we finally have this proof for our Fibonacci Mystery.
  1. 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
  2. Next, we assume when n is any odd number:

    F(n)*F(n+2) - F(n+1)^2 = 1         (a) for odd n
  3. 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)
  4. 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.
  5. 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.

  6. 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
  7. We assume when n is any even number:

    F(n)*F(n+2) - F(n+1)^2 = -1         (c) for even n
  8. 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
Done!

by Phil for Humanity
on 10/10/2011

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