5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$

Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive integers $p, q$, such that
$a_{l}+k a_{0}=a_{m}+k a_{q^{*}} \quad$ (Xiong Bin)
When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.

For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin{array}{l}
\text { then } a_{2 n} \equiv a_{0}=0(\bmod k), \\
a_{2 n+1} \equiv a_{1}=1(\bmod k)(n=0,1, \cdots) .
\end{array}
$$

If there exist $l, m \in \mathbf{N}, p, q \in \mathbf{N}_{+}, l \neq m$, satisfying $a_{l}+k a_{p}=a_{m}+k a_{q}$, without loss of generality, assume $l0$, then
$$
\begin{array}{l}
a_{l}+k a_{p} \leqslant k a_{p}+a_{p-1}=a_{p+1} \\
\leqslant a_{m}\frac{a_{l}+k a_{p}-a_{m}}{k}=a_{q} \text {. }
\end{array}
$$

From $k a_{q}+a_{m}=k a_{p}+a_{l} \geqslant k a_{p}$,
we know $a_{q} \geqslant a_{p}-\frac{a_{m}}{k} \geqslant a_{p}-\frac{a_{p}}{k}=\frac{k-1}{k} a_{p}$.
Noting that $a_{p} \geqslant k a_{p-1}$. Thus,
$a_{p}>a_{q} \geqslant \frac{k-1}{k} a_{p} \geqslant(k-1) a_{p-1} \geqslant a_{p-1}$.
From equation (4), we know $a_{q}=a_{p-1}$.
Therefore, the equalities in equations (1), (2), and (3) must all hold.
From equations (2) and (3), we get
$m=p, p=2$.
Thus, $a_{q}=a_{p-1}=a_{1}=1$.
From equation (1), we know $l=0$.
Therefore, from $a_{l}+k a_{p}=a_{m}+k a_{q}$, we get $k^{2}=k+k$, i.e., $k=2$, which is a contradiction.
Thus, $k \geqslant 3$ does not satisfy the problem's conditions.
Hence, $k$ can only be 2.