How can i show that$(F_{n-1}F_{n+2})^2+(2F_nF_{n+1})^2=(F_{2n+2})^2$ by combinatorial proof?

Question

How can i show that $$(F_{n-1}F_{n+2})^2+(2F_nF_{n+1})^2=(F_{2n+2})^2$$ by combinatorial proof? Here $F_n$ is the $n$-th Fibonacci number. I got stuck on this problem. I want to solve it by combinatorial proof. plz help ㅜㅜ

Answer 1

Using Axel Kemper's hint we get that $$ (F_{n-1}F_{n+2})^2 + (2F_nF_{n+1})^2 = ((F_{n+1}-F_n)(F_{n+1}+F_n))^2 + 4F_n^2F_{n+1}^2 $$ $$ = (F_{n+1}^2 - F_n^2)^2 + 4F_n^2F_{n+1}^2 = (F_{n+1}^2 + F_n^2)^2. $$ Thus it suffices to show that $$ F_{n+1}^2+F_n^2 = F_{2n+2}. $$ Now take a tiling of a $1 \times (2n+2)$ board with $1 \times 1$ and $1 \times 2$ tiles. The number of such tilings is $F_{2n+2}$. Distinguish two cases, the first case is if there is a $2 \times 1$ tile in the middle and the second case is if there is no such tile.

I think you will be able to finish the proof yourself.