Proof of Gelin-Cesaro Identity
By Dean Hanafy on Sunday, October 20, 2002 - 07:04 pm:

Fn 4 - Fn-2 Fn-1 Fn+1 Fn+2 = 1

Thanks
Deano

By Michael Doré on Sunday, October 20, 2002 - 08:15 pm:

Hint: square the identity Fn 2 = (-1)n+1 + Fn+1 Fn-1 .

By Michael Doré on Monday, October 21, 2002 - 12:18 am:

Well we get Fn 4 = 1 + 2(-1)n+1 Fn+1 Fn-1 + Fn+1 2 Fn-1 2 and we want this to equal Fn-2 Fn-1 Fn+1 Fn+2 + 1 so it suffices to show:

2(-1)n+1 Fn+1 Fn-1 + Fn+1 2 Fn-1 2 = Fn-2 Fn-1 Fn+1 Fn+2

i.e.

2(-1)n+1 + Fn+1 Fn-1 = Fn-2 Fn+2

But the right hand side is (Fn -Fn-1 )(Fn+1 +Fn ) so we just need:

2(-1)n+1 + Fn+1 Fn-1 = Fn Fn+1 - Fn-1 Fn+1 + Fn 2 - Fn-1 Fn

The first and last terms on the right hand side contribute Fn (Fn+1 -Fn-1 ) = Fn 2 so we're left needing to prove:

2(-1)n+1 + Fn+1 Fn-1 = 2Fn 2 - Fn-1 Fn+1

which is just

(-1)n+1 + Fn+1 Fn-1 = Fn 2

which we already know.

By Yatir Halevi on Sunday, October 20, 2002 - 09:11 pm:

Here is a very nice proof I found:
Fn 2 -Fn+1 Fn-1 =Fn (Fn-1 +Fn-2 )-Fn+1 Fn-1 =(Fn -Fn+1 )Fn-1 +Fn Fn-2
By using the rule of formation for the fibonacci sequence, the expressionin prentheses may be replace by the term -Fn-1 which gives us:
Fn 2 -Fn+1 Fn-1 =(-1)(Fn-1 2 -Fn Fn-2 )
The RHS is completely equal to the LHS except for the (-1) and the fact that the subscripts decreased by 1, so we can continue to use this formula on the RHS (with lowering the subscripts) n-1 more times to get the following result:
Fn 2 -Fn+1 Fn-1 =(-1)n-2 (F2 2 -F3 F1 )=(-1)n-2 (12 -2*1)=(-1)n-1


Yatir

By Marcos Charalambides on Monday, October 21, 2002 - 01:14 pm:

Okay here's my proof (there's probably a better way but here it goes):

Fn 4 - Fn-2 Fn-1 Fn+1 Fn+2 = 1
Using the identities:
Fn-2 = Fn - Fn-1
Fn+1 = Fn + Fn-1
Fn+2 = 2Fn + Fn-1
the original expression reduces to:
Fn 4 - 2Fn 3 Fn-1 - Fn 2 Fn-1 2 + 2Fn Fn-1 3 + Fn-1 4 = 1

By substituting values you can see that it works for the first few fibonacci numbers...

Now assume the case where n = k is true, we'll try see if based on this assumption (k + 1) is true.

Substituting n=(k+1) ⇒
Fk+1 4 - 2Fk+1 3 Fk - Fk+1 2 Fk 2 + 2Fk+1 Fk 3 + Fk 4 = 1

By using the identity Fk+1 = Fk + Fk-1 , expanding then collecting together we get:
Fk 4 - 2Fk 3 Fk-1 - Fk 2 Fk-1 2 + 2Fk Fk-1 3 + Fk-1 4 = 1
This is identical to our expression that we assumed to be true and so the case for (k + 1) follows from the case for k. Therefore (by induction) we have proved the identity...