4. (CAN) Let $\left\{f_{n}\right\}$ be the Fibonacci sequence $\{1,1,2,3,5, \ldots\}$. (a) Find all pairs $(a, b)$ of real numbers such that for each $n, a f_{n}+b f_{n+1}$ is a member of the sequence. (b) Find all pairs $(u, v)$ of positive real numbers such that for each $n$, $u f_{n}^{2}+v f_{n+1}^{2}$ is a member of the sequence.
4. We shall use the known formula for the Fibonacci sequence
$$ f_{n}=\frac{1}{\sqrt{5}}\left(\alpha^{n}-(-1)^{n} \alpha^{-n}\right), \quad \text { where } \alpha=\frac{1+\sqrt{5}}{2} . $$
(a) Suppose that $a f_{n}+b f_{n+1}=f_{k_{n}}$ for all $n$, where $k_{n}>0$ is an integer depending on $n$. By (1), this is equivalent to $a\left(\alpha^{n}-(-1)^{n} \alpha^{-n}\right)+$
$b\left(\alpha^{n+1}+(-1)^{n} \alpha^{-n-1}\right)=\alpha^{k_{n}}-(-1)^{k_{n}} \alpha^{-k_{n}}$, i.e.,
$$ \alpha^{k_{n}-n}=a+b \alpha-\alpha^{-2 n}(-1)^{n}\left(a-b \alpha^{-1}-(-\alpha)^{n-k_{n}}\right) \rightarrow a+b \alpha $$
as $n \rightarrow \infty$. Hence, since $k_{n}$ is an integer, $k_{n}-n$ must be constant from some point on: $k_{n}=n+k$ and $\alpha^{k}=a+b \alpha$. Then it follows from (2) that $\alpha^{-k}=a-b \alpha^{-1}$, and from (1) we conclude that $a f_{n}+b f_{n+1}=$
$f_{k+n}$ holds for every $n$. Putting $n=1$ and $n=2$ in the previous relation and solving the obtained system of equations we get $a=f_{k-1}$, $b=f_{k}$. It is easy to verify that such $a$ and $b$ satisfy the conditions.
(b) As in (a), suppose that $u f_{n}^{2}+v f_{n+1}^{2}=f_{l_{n}}$ for all $n$. This leads to
$$ \begin{aligned} u+v \alpha^{2}-\sqrt{5} \alpha^{l_{n}-2 n}= & 2(u-v)(-1)^{n} \alpha^{-2 n} \\
& -\left(u \alpha^{-4 n}+v \alpha^{-4 n-2}+(-1)^{l_{n}} \sqrt{5} \alpha^{-l_{n}-2 n}\right) \\
\rightarrow & 0 \end{aligned} $$
as $n \rightarrow \infty$. Thus $u+v \alpha^{2}=\sqrt{5} \alpha^{l_{n}-2 n}$, and $l_{n}-2 n=k$ is equal to a constant. Putting this into the above equation and multiplying by $\alpha^{2 n}$ we get $u-v \rightarrow 0$ as $n \rightarrow \infty$, i.e., $u=v$. Finally, substituting $n=1$ and $n=2$ in $u f_{n}^{2}+u f_{n+1}^{2}=f_{l_{n}}$ we easily get that the only possibility is $u=v=1$ and $k=1$. It is easy to verify that such $u$ and $v$ satisfy the conditions.