<s>[INST] Prove Fibonacci Sequence Property: $x^2_n + x^2_{n+1}=x_{2n+1}$ For Fibonacci Sequence, We know that the recursive difference equation is:
$$
x_{n+2} = x_n + x_{n+1}\ \ \ \ n\ \geq 0
$$
And that the closed form solution is:
$$
x_n = \frac{1}{\sqrt{5}}\left[ \left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}  \right]\ \ \ \  n \geq 0
$$
But how to prove that this fibonacci property is true using this information?:
$$x^2_n + x^2_{n+1}=x_{2n+1}$$
Hint: substitute the closed form expression for the fibonacci sequence into difference equation and verify that its true.
well...I've tried this at least 3 different ways of doing exactly as hinted without much luck...the equation always blows up on me.  Any ideas how to prove  it?
 [/INST] The 'standard proof' uses strong induction, and I'll leave you to figure that out.
Let $\varphi$ and $\psi$ represent the two roots of $x^2-x-1=0$ where $\varphi > \psi$ (i.e. $\varphi,\psi=\frac{1\pm \sqrt{5}}{2}$). The given closed form solution is $$F_n=\frac{\varphi ^n-\psi ^n}{\sqrt{5}}$$ so we know $$F_n^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n} }{5}$$ and $$F_{n+1}^2=\frac{\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2}}{5}$$ so $$F_n^2+F_{n+1}^2=\frac{\varphi ^{2n}-2(\varphi \psi)^n+\psi ^{2n}+\varphi ^{2n+2}-2(\varphi \psi )^{n+1}+\psi ^{2n+2} }{5}$$But we know that $\varphi \psi =-1$, so either $(\varphi\psi)^n=1$ and $(\varphi\psi)^{n+1}=-1$ or $(\varphi\psi)^n=-1$ and $\varphi\psi)^{n+1}=1$. Either way, the two terms cancel each other, so$$F_n^2+F_{n+1}^2=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}$$It can be easily seen that $\frac{1+\varphi^2}{\sqrt{5}}=\varphi$ and $\frac{1+\psi^2}{\sqrt{5}}=-\psi$ so the above becomes \begin{align*}F_n^2+F_{n+1}^2&=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}\\ &=\frac{\varphi^{2n}\cdot \varphi\sqrt{5} + \psi^{2n}\cdot (-\psi\sqrt{5})}{5} \\ &=\frac{\sqrt{5}(\varphi^{2n+1}-\psi^{2n+1})}{5} \\ &=\frac{\varphi^{2n+1}-\psi^{2n+1}}{\sqrt{5}} \\ &= F_{2n+1}\end{align*}completing the proof.
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