Three, (Full marks 50 points) Let $m$ be an integer, $|m| \geq 2$. The integer sequence $a_{1}, a_{2}, \cdots$ satisfies: $a_{1}, a_{2}$ are not both zero, and for any positive integer $n$, we have $a_{n+2}=a_{n+1}-m a_{n}$.
Proof: If there exist integers $r, s (r > s \geq 2)$ such that $a_{r} = a_{s} = a_{1}$, then $r - s \geq |m|$.
Proof: Without loss of generality, assume $a_{1}, a_{2}$ are coprime (otherwise, if $\left(a_{1}, a_{2}\right) = d > 1$, then $\frac{a_{1}}{d}$ and $\frac{a_{2}}{d}$ are coprime, and replacing $a_{1}, a_{2}, a_{3}, \cdots$ with $\frac{a_{1}}{d}, \frac{a_{2}}{d}, \frac{a_{3}}{d}, \cdots$ does not change the conditions or the conclusion).
From the recurrence relation of the sequence, we have
$$
a_{2} \equiv a_{3} \equiv a_{4} \equiv \cdots (\bmod |m|).
$$

We will prove that for any integer $n \geq 3$, we have
$$
a_{n} \equiv a_{2} - \left(a_{1} + (n-3) a_{2}\right) m (\bmod m^{2}).
$$

In fact, when $n = 3$, (2) is clearly true. Assume (2) holds for $n = k$ (where $k$ is some integer greater than 2). Noting (1), we have $m a_{k-1} \equiv m a_{2} (\bmod m^{2})$, and combining this with the induction hypothesis, we get
$$
\begin{aligned}
a_{k+1} = & a_{k} - m a_{k-1} \equiv a_{2} - \left(a_{1} + (k-3) a_{2}\right) m - m a_{2} \\
& \equiv a_{2} - \left(a_{1} + (k-2) a_{2}\right) (\bmod m^{2}),
\end{aligned}
$$

which means (2) also holds for $n = k+1$. Therefore, (2) holds for any integer $n \geq 3$.
20 points
Note that when $a_{1} = a_{2}$, (2) also holds for $n = 2$.
Let integers $r, s (r > s \geq 2)$ satisfy $a_{r} = a_{s} = a_{1}$.
If $a_{1} = a_{2}$, by (2) holding for $n \geq 2$, we have
$$
a_{2} - \left(a_{1} + (r-3) a_{2}\right) m \equiv a_{r} = a_{s} \equiv a_{2} - \left(a_{1} + (s-3) a_{2}\right) m (\bmod m^{2}),
$$

which implies $a_{1} + (r-3) a_{2} \equiv a_{1} + (s-3) a_{2} (\bmod |m|)$, or
$$
(r-s) a_{2} \equiv 0 (\bmod |m|).
$$

If $a_{1} \neq a_{2}$, then $a_{r} = a_{s} = a_{1} \neq a_{2}$, so $r > s \geq 3$. In this case, since (2) holds for $n \geq 3$, we similarly find that (3) still holds.
We will prove that $a_{2}$ and $m$ are coprime.
In fact, if $a_{2}$ and $m$ have a common prime factor $p$, then by (1), $p$ is a common factor of $a_{2}, a_{3}, a_{4}, \cdots$, but since $a_{1}$ and $a_{2}$ are coprime, $p \nmid a_{1}$, which contradicts $a_{r} = a_{s} = a_{1}$.
Therefore, by (3), we have $r - s \equiv 0 (\bmod |m|)$. Since $r > s$, it follows that $r - s \geq |m|$.
50 points