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?
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 $\varphi &gt; \psi$ (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&amp;=\frac{\varphi^{2n}(1+\varphi ^2)+\psi ^{2n}(1+\psi^2)}{5}\\ &amp;=\frac{\varphi^{2n}\cdot \varphi\sqrt{5} + \psi^{2n}\cdot (-\psi\sqrt{5})}{5} \\ &amp;=\frac{\sqrt{5}(\varphi^{2n+1}-\psi^{2n+1})}{5} \\ &amp;=\frac{\varphi^{2n+1}-\psi^{2n+1}}{\sqrt{5}} \\ &amp;= F_{2n+1}\end{align*}completing the proof.
Based on the title of the topic and the hint, you need to plug the Closed form expression into your equality

$$x_{n}^2+x_{n+1}^2-x_{2n+1}$$

and make sure it equals zero identically. 

Best way is to taky any CAS and plug the figures. For Wolfram Mathematica


  x[n_] := 1/Sqrt[5]*(((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n);
  
  x[a]^2 + x[a + 1]^2 - x[2*a + 1]
  
  Simplify[Out[]]


Then you get something like this

$$\frac{1}{5} \left(\left(\frac{2}{\sqrt{5}+1}\right)^{-a-1}-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{a+1}\right)^2+\frac{1}{5} \left(\left(\frac{1}{2} \left(\sqrt{5}+1\right)\right)^a-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^a\right)^2-\frac{\left(\frac{2}{\sqrt{5}+1}\right)^{-2 a-1}-\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{2 a+1}}{\sqrt{5}}$$

You need to carefully open up all brackets, cancel out the terms to get zero.