Fibonacci's Fourth Power Is Fire

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

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

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

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

When $n=k+1$ ,

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

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

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

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

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

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

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

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

$=(-1)^k$

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

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

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

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

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

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

When $n=k+1$ ,

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

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

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

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

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

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

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

Now, we can compute the expression in the question:

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

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

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

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

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

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

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

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

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

Therefore,

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

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.