Fibonacci's Fourth Power Is Fire

For integers n 2 n\geq 2 , evaluate F n 4 F n 2 F n 1 F n + 1 F n + 2 . \large F^{4}_{n}-F_{n-2}F_{n-1}F_{n+1}F_{n+2} . Notation

F n F_{n} denotes the n th n^\text{th} Fibonacci number .


The answer is 1.

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3 solutions

Shaun Leong
Dec 28, 2015

Firstly, we make two claims, which we use induction to prove.

Claim 1: F n 2 F n 1 F n + 1 = ( 1 ) n 1 F_n^2-F_{n-1}F_{n+1} = (-1)^{n-1} for n 2 n \geq 2 .

The base case n = 2 n=2 gives 1 2 ( 1 ) ( 2 ) = 1 1^2-(1)(2) = -1 , which is certainly true.

Next, assume the claim holds for a certain n = k n=k .

When n = k + 1 n=k+1 ,

F k + 1 2 F k F k + 2 F_{k+1}^2-F_{k}F_{k+2}

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

= F k + 1 2 F k 2 F k F k + 1 =F_{k+1}^2-F_{k}^2-F_{k}F_{k+1}

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

(Using F k 2 = F k 1 F k + 1 + ( 1 ) k 1 F_k^2=F_{k-1}F_{k+1} + (-1)^{k-1} from the induction hypothesis)

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

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

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

= ( 1 ) k =(-1)^k

Hence if the claim holds for n = k n=k , then we have shown that it holds for n = k + 1 n=k+1 , completing the induction.

Claim 2: F n 2 F n 2 F n + 2 = ( 1 ) n F_n^2-F_{n-2}F_{n+2} = (-1)^n for n 3 n \geq 3 .

The base case n = 3 n=3 gives 2 2 ( 1 ) ( 5 ) = 1 2^2-(1)(5) = -1 , which is certainly true.

Next, assume the claim holds for a certain n = k n=k . To make induction easier, we rewrite the equation in terms of F n 2 F_{n-2} and F n F_{n} by noting that F n + 2 = 3 F n F n 2 F_{n+2}=3F_{n}-F_{n-2} .

The equivalent induction hypothesis is then F k 2 + F k 2 2 3 F k F k 2 = ( 1 ) k F_{k}^2+F_{k-2}^2-3F_{k}F_{k-2} = (-1)^k .

Also, note that F k + 1 = 2 F k F k 2 F_{k+1} = 2F_{k}-F_{k-2} , F k 1 = F k F k 2 F_{k-1}=F_{k}-F_{k-2} and F k + 3 = 5 F k 2 F k 2 F_{k+3}=5F_{k}-2F_{k-2} .

When n = k + 1 n=k+1 ,

F k + 1 2 F k 1 F k + 3 F_{k+1}^2 - F_{k-1}F_{k+3}

= ( 2 F k F k 2 ) 2 ( F k F k 2 ) ( 5 F k 2 F k 2 ) =(2F_{k}-F_{k-2})^2-(F_{k}-F_{k-2})(5F_{k}-2F_{k-2})

= 4 F k 2 4 F k F k 2 + F k 2 2 5 F k 2 + 2 F k F k 2 + 5 F k F k 2 2 F k 2 =4F_{k}^2-4F_{k}F_{k-2}+F_{k-2}^2-5F_{k}^2+2F_{k}F_{k-2}+5F_{k}F_{k-2}-2F_{k}^2

= F k 2 + 3 F k F k 2 F k 2 2 =-F_{k}^2+3F_{k}F_{k-2}-F_{k-2}^2

= ( F k 2 + F k 2 2 3 F k F k 2 ) =-(F_{k}^2+F_{k-2}^2-3F_{k}F_{k-2})

= ( 1 ) k + 1 =(-1)^{k+1}

Hence if the claim holds for n = k n=k , then we have shown that it holds for n = k + 1 n=k+1 , completing the induction.

Now, we can compute the expression in the question:

F n 4 F n 2 F n 1 F n + 1 F n + 2 F_{n}^4-F_{n-2}F_{n-1}F_{n+1}F_{n+2}

= F n 4 ( F n 2 F n + 2 ) ( F n 1 F n + 1 ) =F_{n}^4-(F_{n-2}F_{n+2})(F_{n-1}F_{n+1})

= F n 4 ( F n 2 ( 1 ) n ) ( F n 2 + ( 1 ) n ) =F_{n}^4-(F_{n}^2-(-1)^n)(F_{n}^2+(-1)^n)

= F n 4 F n 4 + ( 1 ) 2 n =F_{n}^4-F_{n}^4+(-1)^{2n}

= 1 =\boxed {1} for n 3 n \geq 3 .

Nice, you proved Cassini's identity!

Pi Han Goh - 5 years, 5 months ago

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Oh lol I had never heard of that before. I don't really know much about Fibonacci identities so I wanted to prove it from scratch. It was fun!

Shaun Leong - 5 years, 5 months ago
Chew-Seong Cheong
Dec 28, 2015

The problem is the Gelin-Cesàro identity as mentioned by Rohit Udaiwal .

F n 4 F n 2 F n 1 F n + 1 F n + 2 = 1 F_n^4 - F_{n-2}F_{n-1}F_{n+1}F_{n+2} = 1

It can be proven using Catalan's identity which is as follows:

F n 2 F n r F n + r = ( 1 ) n r F r 2 F n r F n + r = F n 2 ( 1 ) n r F r 2 \begin{aligned} F_n^2 - F_{n-r}F_{n+r} & = (-1)^{n-r}F_r^2 \\ \Rightarrow F_{n-r}F_{n+r} & = F_n^2 - (-1)^{n-r}F_r^2 \end{aligned}

Therefore,

F n 4 F n 2 F n 1 F n + 1 F n + 2 = F n 4 F n 2 F n + 2 F n 1 F n + 1 = F n 4 ( F n 2 ( 1 ) n 2 F 2 2 ) ( F n 2 ( 1 ) n 1 F 1 2 ) = F n 4 ( F n 2 ± 1 ) ( F n 2 1 ) = F n 4 ( F n 4 1 ) = 1 \begin{aligned} F_n^4 - F_{n-2}F_{n-1}F_{n+1}F_{n+2} & = F_n^4 - F_{n-2}F_{n+2}F_{n-1}F_{n+1} \\ & = F_n^4 - \left(F_n^2 - (-1)^{n-2}F_2^2\right) \left(F_n^2 - (-1)^{n-1}F_1^2\right) \\ & = F_n^4 - \left(F_n^2 \pm 1 \right) \left(F_n^2 \mp 1 \right) \\ & = F_n^4 - \left(F_n^4 - 1 \right) \\ & = \boxed{1} \end{aligned}

This identity is called Gelin-Cesàro Identity.

Rohit Udaiwal - 5 years, 5 months ago

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Thanks, I have amended the solution.

Chew-Seong Cheong - 5 years, 5 months ago

Like a cheater, we can use2,3,5,8,13,21,34 to solve the problem. Now putting into the problem we get=8^4-(3×5×13×21)=1. As problem states its a general constant so answer is=1.

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