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).

 A: 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.
A: To prove 
$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 would write it as
$F^2_{n + 2}(F_n ^ 2 + F_ {n + 1}^ 2)
 = F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1 
$
or
$(F_{n+1}+F_n)^2(F_n ^ 2 + F_ {n + 1}^ 2)
 = F_n F_ {n + 1} ^ 2(3 F_n +4 F_ {n + 1}) + 1 
$.
For $n=0$, 
$F_n = 0, F_1 = 1$,
this becomes
$(1)^2(1)
 = 1 
$,
which is true.
For $n=1$, 
$F_n = 1, F_{n+1} = 1$,
this becomes
$(2)^2(2)
 = (3+4) + 1 
$,
which is true.
For $n=2$, 
$F_n = 1, F_{n+1} = 2$,
this becomes
$(2+1)^2(1 ^ 2 + 2^ 2)
 = 2 ^ 2(3  +4\cdot 2) + 1 
$,
or
$45
=45
$
which is true.
For $n=3$, 
$F_n = 2, F_{n+1} = 3$,
this becomes
$(3+2)^2(2 ^ 2 + 3^ 2)
 = 2\ 3 ^ 2(3\ 2+4\ 3) + 1 
$.
or
$25\ 13
=18(18)+1
$
or
$325 = 325
which is true.
So, it looks good.
The next step would be
to expand the sides,
eliminate any common terms,
and try to prove it by induction.
I'll leave it at this.
A: Using the notations and function $g$ given by J.G., define $$f(a,b)=a^4+2a^3b-a^2b^2-2ab^3+b^4.$$ It suffices to prove that $f(a,b)=1$. But $$f(a,b)=(a^2+ab-b^2)^2$$ and $$a^2+ab-b^2=F_n^2+F_nF_{n+1}-F_{n+1}^2$$ $$=F_n(F_n+F_{n+1})-F_{n+1}^2=F_nF_{n+2}-F_{n+1}^2=(-1)^{n+1},$$ by Cassini's identity which is a special case of Catalan's identity. So the result is clear.
A: This is probably the least effective method possible, but  there is a direct way to show this without induction.
Let $f:\Bbb{R}\to\Bbb{C}$ where...
$$f(x)=\frac{\phi^x-(-\Phi)^x}{\sqrt{5}}\qquad:\qquad\phi=\frac{1+\sqrt{5}}{2},\quad\Phi=\frac1\phi$$
Then, $F_n=f(n)$ for all $n\in\Bbb{N}$, and...
$$F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}}\iff f(n+2)=\sqrt{\frac{f(n+1)^2f(n)(4f(n+1)+3f(n))+1}{f(n+1)^2+f(n)^2}}$$
...for all $n\in\Bbb{N}$.
Plug in the formula for $f$ in the second equation and solve for $n$. You should get $n=\Bbb{Z}$ or $n=K\subseteq\Bbb{Z}[i\pi]$ (one of these is right, but I'm not sure which). Since the Fibonacci sequence is defined for natural numbers, restrict the solution set to the naturals, and you have your answer: $n=\Bbb{Z}[i\pi]\cap\Bbb{N}=\Bbb{N}$. In other words, the set of natural numbers for which your equation holds is the set of all natural numbers - your equation is an identity.
Now, I wouldn't recommend doing this, but you could if you wanted to - it's kind of like using Cramer's rule to solve a large system of equations.
