Hey sorry for the weird title. First I had to find $\phi\in\mathbb{R}$ so that $x_{n+2}=x_{n+1}+x_n$ with $x_n=\phi^2$ and $\phi_1<\phi_2$ is satisfied. That part I figured out. For people coming here through Google: You can write this as a polynomial $(\phi^2-\phi-1)\phi^n$ which will give you give $\phi_1=\frac{1-\sqrt 5}{2}$ and $\phi_2=\frac{1+\sqrt 5}{2}$ as the desired solutions.
Now I need to show that $x_{n+2}=x_{n+1}+x_n$ also holds true for the linear combination (with $a,b\in\mathbb{R}$) $a\phi_1^n+b\phi_2^n$, so I think I need to show that $$x_n=a(\frac{1-\sqrt 5}{2})^n+b(\frac{1+\sqrt 5}{2})^n.$$
But to be quite frank I don't even know where to start. This has nothing (at least yet — I know you can use this to construct the numbers) to do with Fibonacci, right? Please help.
Can you show that if $(x_n)_n$ and $(y_n)_n$ satifies the given recurring relation, then $(ax_n+by_n)_n$ does for any choice of reals $a$ and $b$? To do so, try to express the following in terms of $x_n,x_{n+1},y_n,y_{n+1}$: $$ax_{n+2}+by_{n+2}=?$$
If so, you are done, since you already know that $({\phi_1}^n)_n$ and $({\phi_2}^n)_n$ are solutions. Indeed, you have proved: $${\phi_1}^2=\phi_1+1$$ therefore for all integer $n$, one has: $${\phi_1}^{n+2}={\phi_1}^n(\phi_1+1)={\phi_1}^{n+1}+{\phi_1}^n,$$ so that $({\phi_1}^n)_n$ is a solution and the same holds for $({\phi_2}^n)_n$.