<s>[INST] Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number. 
Let $F(n)$ denotes the $n$th number in Fibonacci sequence.  Then for all $n\in\mathbb{N}$,
  $$F(n)^2+F(n+1)^2=F(2n+1).$$

I know how to prove it by using the formula 
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt{5}},$$
but is there a way to prove it by induction?
I am year 12 standard so please don't go too deep.
 [/INST] \begin{align} &F(x)=F(x-1)+F(x-2) = F(2)F(x-1)+F(1)F(x-2). \ \\ \ \\ &F(x-1)=F(x-2)+F(x-3) \\ &\Rightarrow F(x)=F(2)(F(x-2)+F(x-3))+F(1)F(x-2)=(F(1)+F(2))F(x-2) \\ &+F(2)F(x-3)=F(3)F(x-2)+F(2)F(x-3). \ \\ \ \\ &F(x-2)=F(x-3)+F(x-4). \\ &\Rightarrow F(x)=F(3)(F(x-3)+F(x-4))+F(2)F(x-3)=(F(2)+F(3))F(x-3) \\ &+F(3)F(x-4)=F(4)F(x-3)+F(3)F(x-4). \\ &\cdot \\ &\cdot \\ &\cdot \\ &\therefore F(x)=F(k+1)F(x-k)+F(k)F(x-k-1). \\ &x=2k+1; \ F(2k+1)=F(k)^2+F(k+1)^2. \\ \ \\ &\therefore F(x)^2+F(x+1)^2=F(2x+1).\end{align}
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