Find all pairs $(a,b)\in \mathbb{R^2}$ such that $af_n^2+bf_{n+1}^2 $ is a member of the Fibonacci sequence for all $n\in\mathbb{N}$. Question : Let$\{f_n\}$ be the Fibonacci sequence $\{1,1,2,3,...\}$. Find all pairs $(a,b)\in \mathbb{R^2}$ such that $af_n^2+bf_{n+1}^2 $ is a meber of the sequence for all $n\in\mathbb{N}$.

I used  $f_n=\frac {1}{\sqrt 5}[\phi^n-(-1)^n\phi^{-n}]$ with $\phi=\frac {1+\sqrt 5}{2}$. To reduce the given expression but then was not getting any helpful ideas to  proceed. So any help/hint will be appreciated. Can we find such $a,b$ so that $af_n^k+bf_{n+1}^k$ for some $k&gt;2$ in positive integers?

Thank You.
Start with

$$f_{n}=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n}\phi^{-n}\right]$$

Then

$$f_{n}^{2} =\frac{1}{5}\left[\phi^{2n}-2(-1)^{n}+\phi^{-2n}\right]$$
$$f_{n+1}^{2} =\frac{1}{5}\left[\phi^{2(n+1)}-2(-1)^{n+1}+\phi^{-2(n+1)}\right]$$
$$af_{n}^{2}+bf_{n+1}^{2} =\frac{1}{5}\left[(a+b\phi^{2})\phi^{2n}-2(a-b)(-1)^{n}+(a+b\phi^{-2})\phi^{-2n}\right]$$

This will be a Fibonacci number if $a=b$ (to cancel the second term) and if

$$a+b\phi^{+2}=+\sqrt{5}\phi^{+k}$$
$$a+b\phi^{-2}=-(-1)^{2n+k}\sqrt{5}\phi^{-k}$$

where k is an integer. Setting $b=a$ and multiplying gives
$$a^2(1+\phi^{2})(1+\phi^{-2})=-5(-1)^{k}$$

This has real solutions for $a$ when $k$ is odd. Then
$$a^2(1+\phi^{2})(1+\phi^{-2})=5$$

Which leads to $a=1, b=1$.
Start with

$$f_{n}=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n}\phi^{-n}\right]$$

Then

$$f_{n}^{2} =\frac{1}{5}\left[\phi^{2n}-2(-1)^{n}+\phi^{-2n}\right]$$
$$f_{n+1}^{2} =\frac{1}{5}\left[\phi^{2(n+1)}-2(-1)^{n+1}+\phi^{-2(n+1)}\right]$$
$$af_{n}^{2}+bf_{n+1}^{2} =\frac{1}{5}\left[(a+b\phi^{2})\phi^{2n}-2(a-b)(-1)^{n}+(a+b\phi^{-2})\phi^{-2n}\right]$$

This will be a Fibonacci number if $a=b$ (to cancel the second term) and if

$$a+b\phi^{+2}=+\sqrt{5}\phi^{+k}$$
$$a+b\phi^{-2}=-(-1)^{2n+k}\sqrt{5}\phi^{-k}$$

where k is an integer. Setting $b=a$ and multiplying gives
$$a^2(1+\phi^{2})(1+\phi^{-2})=-5(-1)^{k}$$

This has real solutions for $a$ when $k$ is odd. Then
$$a^2(1+\phi^{2})(1+\phi^{-2})=5$$

Which leads to $a=1, b=1$.