Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:

Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{n + 2} $ are the length measures of the sides of a triangle whose area is $ \frac{1}{2} $ unit.
  (I also put this in the Wikipedia article about the Fibonacci Sequence).


The hypothesis is equivalent to $g(n)=0$ with$$g(n):=F_nF_{n+1}^2(3F_n+4F_{n+1})+1-F_n^2F_{n+2}^2-F_{n+1}^2F_{n+2}^2\\=-F_n^4-2F_n^3F_{n+1}+F_n^2F_{n+1}^2+2F_nF_{n+1}^3-F_{n+1}^4+1.$$Since $F_0=0,\,F_1=1$, we have the base case $g(0)=0$. For the inductive step use the abbreviations $a:=F_n,\,b:=F_{n+1}$, so$$g(n+1)-g(n)=a^4-b^4+2a^3b-2b^3(a+b)-a^2b^2+b^2(a+b)^2-2ab^3+2b(a+b)^3+b^4-(a+b)^4\\=\color{red}{a^4}\color{blue}{-b^4}\color{orange}{+2a^3b}\color{limegreen}{-2ab^3}\color{blue}{-2b^4}\color{hotpink}{-a^2b^2+a^2b^2}\color{limegreen}{+2ab^3}\color{blue}{+b^4}\color{limegreen}{-2ab^3}\color{orange}{+2a^3b}\color{hotpink}{+6a^2b^2}\color{limegreen}{+6ab^3}\color{blue}{+2b^4}\color{blue}{+b^4}\color{red}{-a^4}\color{orange}{-4a^3b}\color{hotpink}{-6a^2b^2}\color{limegreen}{-4ab^3}\color{blue}{-b^4}.$$I've colour-coded the terms progressing through the rainbow, except for yellow being changed to pink as yellow is hard to read, so red is $a^4$, orange is $a^3b$, pink is $a^2b^2$, green is $ab^3$ and blue is $b^4$. And you can verify each colour has zero coefficient overall.
