Example 5. (22nd IMO selected problems) Let $\left\{f_{\mathrm{n}}\right\}$ be the Fibonacci sequence.
(a) Find all pairs of real numbers $(a, b)$ such that for each $n, a f_{0}+b f_{\mathrm{n}-1}$ is a term in the sequence $\left\{f_{\mathrm{n}}\right\}$.
(b) Find all pairs of positive real numbers $(u, v)$ such that for each $n, \| f_{u}^{2}+f_{\mathrm{a}+1}^{2}$ is a term in the sequence $\left\{f_{n}\right\}$.
(a) Let real numbers $a, b$ be such that for all integers $n, a f_{n}+b f_{n+1}$ is a term in the Fibonacci sequence. Then for $n=1, 2, 3$, we can set
$$
\begin{array}{l}
a+b=f_{m}, a+2 b=f_{m^{\prime}}, \\
2 a+3 b=f_{m^{\prime \prime}} .
\end{array}
$$

where $m, m^{\prime}, m^{\prime \prime}$ are some three natural numbers. Eliminating $a, b$ from the above three equations, we get $f_{m}+f_{m^{\prime}}=f_{m^{\prime \prime}}$. In the Fibonacci sequence, the sum of two adjacent terms equals the next term, i.e., for any natural number $k$, we have $f_{k}+f_{k+1}=f_{k+2}$. Therefore, there may be the following four cases:
(1) $\left\{\begin{array}{l}m=1, \\ m^{\prime}=3 \\ m^{\prime \prime}=4\end{array}\right.$
(2) $\left\{\begin{array}{l}m^{\prime}=1, \\ m=3, \\ m^{\prime \prime}=4\end{array}\right.$
(3) $\left\{\begin{array}{l}m^{\prime}=m+1, \\ m^{\prime \prime}=m+2\end{array}\right.$
(4) $\left\{\begin{array}{l}m^{\prime}=m-1, \\ m^{\prime \prime}=m+1\end{array}\right.$
For case (1), we get $a+b=2, a+2 b=1, 2 a+3 b=3$, i.e., $a=3, b=-1$. But $3 f_{n}-f_{n+1}=f_{n}+f_{n-2}$, unless $n=3$, otherwise, $3 f_{n}-f_{n+1}$ is not a term in the Fibonacci sequence, so $a=3, b=-1$ is not a solution.

From case (3), we get $a+b=f_{m}, a+2 b=f_{m+1}$, solving gives $a=f_{m-2}, b=f_{m-1}(m \geqslant 3)$. When $m=1$, we have $a=1, b=0$. When $m=2$, we have $a=0, b=1$. It is easy to prove that these results are all solutions.
In particular, using mathematical induction, we can prove
$$
\begin{aligned}
f_{m+n-1} & =f_{1} f_{m+n-3}+f_{2} f_{m+n-2} \\
& =f_{2} f_{m+\cdots-4}+f_{3} f_{m+n-3} \\
& =\cdots \cdots \\
& =f_{m-2} f_{n}+f_{m-1} f_{n+1} \\
& =a f_{n}+b f_{n+1} .
\end{aligned}
$$

From case (4), we get
$$
\begin{array}{l}
a+b=f_{m}, \\
a+2 b=f_{m-1} . \quad(m \geqslant 2)
\end{array}
$$

Solving gives
$$
\begin{array}{c}
a=f_{m-2}+f_{m}, \\
b=-f_{m-2}, \quad(m \geqslant 3) \\
\text { For } m=2, \text { we have }\left\{\begin{array}{l}
a+b=1 ; \\
a+2 b=1
\end{array}\right.
\end{array}
$$

yielding $a=1, b=0$.

However, solving.
$$
\begin{array}{l}
a f_{4}+b f_{5}=3 a+5 b \\
=3\left(f_{m-2}+f_{m}\right)-5 f_{m-2} \\
=3 f_{m}-2 f_{m-2} \\
=f_{m-1}+f_{m+1} \text {. } \\
\end{array}
$$

But for $m \geqslant 3$, $f_{m-1}+f_{m+1}$ is not a term in the Fibonacci sequence, so we know that these $a, b$ are not solutions.
In summary, all solutions are:
$$
\left\{\begin{array}{l}
a=0, \\
b=1,
\end{array} \left\{\begin{array}{l}
a=1, \quad a=f_{m-2}, \\
b=0, \quad b=f_{m-1} .
\end{array} \quad(m \geqslant 3)\right.\right.
$$
(b) It can be solved similarly to (a).