Chinese Olympiad 2019 Round 2 Question 3 I came across this question and the AOPS website has two response: one is written in Chinese  which I have no clue about and there's another approach which I'm not familiar with. Can anyone suggest any other way to approach this question? I have tried attempting this question but my method is not making any sense. Thanks in advance!
Let $m$ be an integer where $|m|\ge 2$. Let $a_1,a_2,\cdots$ be a sequence of integers such that $a_1,a_2$ are not both zero, and for any positive integer $n$, $a_{n+2}=a_{n+1}-ma_n$.
Prove that if positive integers $r>s\ge 2$ satisfy $a_r=a_s=a_1$, then $r-s\ge |m|$
 A: (Note: This is the official solution in Chinese which I had translated. It is not my original solution)
Assume that $a_1, a_2$ are coprime, (otherwise $(a_1,a_2)=d>1,\frac{a_1}{d}$ and $\frac{a_2}{d}$ are coprime, we can substitute $\frac{a_1}{d}, \frac{a_2}{d}, \frac{a_3}{d}, \cdots$ with $a_1, a_2, a_3, \cdots$ and the conclusion remains unchanged.)
We know that $a_2\equiv a_3 \equiv a_4\equiv \cdots \pmod{|m|}$. ----(1)
Using induction, we will show that $a_n\equiv a_2-(a_1+(n-3)a_2)m \pmod{m^2}$ is true for any integer $n\ge3$----(2)
Base case $n=3$: it is obviously true.
Assuming it holds for $n=k$, where k is some integer $>2$,
From (1), $ma_{k-1}\equiv ma_2 \pmod{m^2}$
$a_{k+1}=a_k-ma_{k-1}\equiv a_2-(a_1+(k-3)a_2)m-ma_2\equiv a_2-(a_1+(k-2)a_2)m \pmod{m^2}$
$\therefore$ (2) is true for all integers $n \ge 3$.
If $a_1=a_2$, (2) is true for $n=2$ as well. 
$\qquad$$a_2-(a_1+(r-3)a_2)m \equiv a_r \equiv a_s \equiv a_2-(a_1+(s-3)a_2)m \pmod{m^2}$
$\qquad$Since $a_1+(r-3)a_2 \equiv a_1+(s-3)a_2 \pmod{|m|}$, $$(r-s)a_2 \equiv 0 \pmod{|m|}$$ ----(3)
Else if $a_1\neq a_2,a_r=a_s=a_1\neq a_2, \therefore r>s\ge3$,
$\qquad$ We will prove $a_2$ and $m$ are coprime.
$\qquad$ if they have common prime factor $p$, $p$ is also a common prime factor of $a_2, a_3, a_4, \cdots$. Since $a_1, a_2$ are coprime, so $p\nmid a_1$, which would contradict $a_r=a_s=a_1$, therefore not possible
Hence from (3) $r-s\equiv 0 \pmod{|m|}$, and since $r>s$, $\therefore r-s\ge|m|$
