prove that: $\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}$ Pythagoras's constant in Fibonacci number!
How do I show that?
$$\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}\tag1$$
Where $F_n$ is Fibonacci sequence. 
$$2(F_n^4+F_{n+1}^4+F_{n+2}^4)^2=F_n^2+F_{n+1}^2+F_{n+2}^2$$
$$2(F_n^8+F_{n+1}^8+F_{n+2}^8+2F_n^2F_{n+1}^2+2F_n^2F_{n+2}^2+F_{n+1}^2F_{n+2}^2)=F_n^2+F_{n+1}^2+F_{n+2}^2$$
This is getting too messy. I hope there is an easy way of proving this formula.
 A: Let $x=F_{n+2}$,$y=F_{n+1}$ and $z=F_{n}$. Squaring both sides, we have:
$$2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2=0$$
Factoring out, we have:
$$(x-y-z)(x+y-z)(x-y+z)(x+y+z)=0$$
Now, by the product cancellation law, we can say that all the product is equal to zero when: $$x-y-z=0\leftrightarrow x=y+z$$
Remebring the substituitons done, we are in the context of Fibonacci's sequence, namely:
$$x=y+z\rightarrow F_{n+2}=F_{n+1}+F_n=F_k=F_{k-1}+F_{k-2}$$for $k=n-2, n\geq 2$.
Note that this proof holds for all different sequences such that $F_n=F_{n-1}+F_{n-2}$. For ecample, consider this case:$$\sqrt2=\frac{1^2+2^2+3^2}{\sqrt{1^4+2^4+3^4}}$$
A: Using the same notation as Matteo $x=F_{n}$ and $y=F_{n+1}$ we have \begin{eqnarray}F_n^4+F_{n+1}^4+F_{n+2}^4&=&x^4+y^4+(x+y)^4 \\
&=& 2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\\
 &=&2(x^4+2x^3y+3x^2y^2+2xy^3+y^4)\\
&=&2(x^2+y^2+xy)^2
\end{eqnarray}
On the other hand we have \begin{eqnarray}F_n^2+F_{n+1}^2+F_{n+2}^2&=& x^2+y^2+(x+y)^2\\ &=& 2(x^2+xy+y^2)\end{eqnarray} and we are done.
