, 2025

Let $c=100!$. Suppose that $n \geq m+2$. Then $a_{m+n}=a_{(m+1)+(n-1)}$ divides both $c\left(a_{m}+a_{m+1}+\cdots+a_{n-1}+a_{n}\right)$ and $c\left(a_{m+1}+\cdots+a_{n-1}\right)$, so it also divides the difference $c\left(a_{m}+a_{n}\right)$. Notice that if $n=m+1$ then $a_{m+n}$ divides $c\left(a_{m}+a_{m+1}\right)=c\left(a_{m}+a_{n}\right)$, and if $n=m$ then $a_{m+n}$ divides both $c a_{m}$ and $2 c a_{m}=c\left(a_{m}+a_{n}\right)$. In either cases; $a_{m+n}$ divides $c\left(a_{m}+a_{n}\right)$.
Analogously, one can prove that if $m>n, a_{m-n}=a_{(m-1)-(n-1)}$ divides $c\left(a_{m}-a_{n}\right)$, as it divides both $c\left(a_{n+1}+\cdots+a_{m}\right)$ and $c\left(a_{n}+\cdots+a_{m-1}\right)$.
From now on, drop the original divisibility statement and keep the statements " $a_{m+n}$ divides $c\left(a_{m}+a_{n}\right)$ " and " $a_{m-n}$ divides $c\left(a_{m}-a_{n}\right)$." Now, all conditions are linear, and we can suppose without loss of generality that there is no integer $D>1$ that divides every term of the sequence; if there is such an integer $D$, divide all terms by $D$.
Having this in mind, notice that $a_{m}=a_{m+n-n}$ divides $c\left(a_{m+n}-a_{n}\right)$ and also $c\left(a_{m+n}-a_{m}-a_{n}\right)$; analogously, $a_{n}$ also divides $c\left(a_{m+n}-a_{m}-a_{n}\right)$, and since $a_{m+n}$ divides $c\left(a_{m}+a_{n}\right)$, it also divides $c\left(a_{m+n}-a_{m}-a_{n}\right)$. Therefore, $c\left(a_{m+n}-a_{m}-a_{n}\right)$ is divisible by $a_{m}, a_{n}$, and $a_{m+n}$, and therefore also by $\operatorname{lcm}\left(a_{m}, a_{n}, a_{m+n}\right)$. In particular, $c a_{m+n} \equiv c\left(a_{m}+a_{n}\right)\left(\bmod \operatorname{lcm}\left(a_{m}, a_{n}\right)\right)$.
Since $a_{m+n}$ divides $c\left(a_{m}+a_{n}\right), a_{m+n} \leq c\left(a_{m}+a_{n}\right)$.
From now on, we divide the problem in two cases.

Case 1: there exist $m, n$ such that $\operatorname{lcm}\left(a_{m}, a_{n}\right)>c^{2}\left(a_{m}+a_{n}\right)$.
If $\operatorname{lcm}\left(a_{m}, a_{n}\right)>c^{2}\left(a_{m}+a_{n}\right)>c\left(a_{m}+a_{n}\right)$ then both $c\left(a_{m}+a_{n}\right)$ and $c a_{m+n}$ are less than $c a_{m+n} \leq c^{2}\left(a_{m}+a_{n}\right)<\operatorname{lcm}\left(a_{m}, a_{n}\right)$. This implies $c a_{m+n}=c\left(a_{m}+a_{n}\right) \Longleftrightarrow a_{m+n}=a_{m}+a_{n}$. Now we can extend this further: since $\operatorname{gcd}\left(a_{m}, a_{m+n}\right)=\operatorname{gcd}\left(a_{m}, a_{m}+a_{n}\right)=\operatorname{gcd}\left(a_{m}, a_{n}\right)$, it follows that
$$\begin{aligned} \operatorname{lcm}\left(a_{m}, a_{m+n}\right) & =\frac{a_{m} a_{m+n}}{\operatorname{gcd}\left(a_{m}, a_{n}\right)}=\frac{a_{m+n}}{a_{n}} \operatorname{lcm}\left(a_{m}, a_{n}\right)>\frac{c^{2}\left(a_{m}+a_{n}\right)^{2}}{a_{n}} \\ & >\frac{c^{2}\left(2 a_{m} a_{n}+a_{n}^{2}\right)}{a_{n}}=c^{2}\left(2 a_{m}+a_{n}\right)=c^{2}\left(a_{m}+a_{m+n}\right) . \end{aligned}$$
We can iterate this reasoning to obtain that $a_{k m+n}=k a_{m}+a_{n}$, for all $k \in \mathbb{Z}_{>0}$. In fact, if the condition $\operatorname{lcm}\left(a_{m}, a_{n}\right)>c^{2}\left(a_{m}+a_{n}\right)$ holds for the pair $(n, m)$, then it also holds for the pairs $(m+n, m),(2 m+n, m), \ldots,((k-1) m+n, m)$, which implies $a_{k m+n}=a_{(k-1) m+n}+a_{m}= a_{(k-2) m+n}+2 a_{m}=\cdots=a_{n}+k a_{m}$.
Similarly, $a_{m+k n}=a_{m}+k a_{n}$.
Now, $a_{m+n+m n}=a_{n+(n+1) m}=a_{m+(m+1) n} \Longrightarrow a_{n}+(n+1) a_{m}=a_{m}+(m+1) a_{n} \Longleftrightarrow m a_{n}= n a_{m}$. If $d=\operatorname{gcd}(m, n)$ then $\frac{n}{d} a_{m}=\frac{m}{d} a_{n}$.
Therefore, since $\operatorname{gcd}\left(\frac{m}{d}, \frac{n}{d}\right)=1, \frac{m}{d}$ divides $a_{m}$ and $\frac{n}{d}$ divides $a_{n}$, which means that
$$a_{n}=\frac{m}{d} \cdot t=\frac{t}{d} m \quad \text{ and } \quad a_{m}=\frac{n}{d} \cdot t=\frac{t}{d} n, \quad \text{ for some } t \in \mathbb{Z}_{>0}$$
which also implies
$$a_{k m+n}=\frac{t}{d}(k m+n) \quad \text{ and } \quad a_{m+k n}=\frac{t}{d}(m+k n), \quad \text{ for all } k \in \mathbb{Z}_{>0} .$$
Now let's prove that $a_{k d}=t k=\frac{t}{d}(k d)$ for all $k \in \mathbb{Z}_{>0}$. In fact, there exist arbitrarily large positive integers $R, S$ such that $k d=R m-S n=(n+(R+1) m)-(m+(S+1) n)$ (for instance, let $u, v \in \mathbb{Z}$ such that $k d=m u-n v$ and take $R=u+Q n$ and $S=v+Q m$ for $Q$ sufficiently large.)
Let $x=n+(R+1) m$ and $y=m+(S+1) n$. Then $k d=x-y \Longleftrightarrow x=y+k d, a_{x}=\frac{t}{d} x$, and $a_{y}=\frac{t}{d} y=\frac{t}{d}(x-k d)=a_{x}-t k$. Thus $a_{x}$ divides $c\left(a_{y}+a_{k d}\right)=c\left(a_{k d}+a_{x}-t k\right)$, and therefore also $c\left(a_{k d}-t k\right)$. Since $a_{x}=\frac{t}{d} x$ can be arbitrarily large, $a_{k d}=t k=\frac{t}{d}(k d)$. In particular, $a_{d}=t$, so $a_{k d}=k a_{d}$.
Since $k a_{d}=a_{k d}=a_{1+(k d-1)}$ divides $c\left(a_{1}+a_{k d-1}\right), b_{k}=a_{k d-1}$ is unbounded. Pick $p>a_{k d-1}$ a large prime and consider $a_{p d}=p a_{d}$. Then
$$\operatorname{lcm}\left(a_{p d}, a_{k d-1}\right) \geq \operatorname{lcm}\left(p, a_{k d-1}\right)=p a_{k d-1}$$
We can pick $a_{k d-1}$ and $p$ large enough so that their product is larger than a particular linear combination of them, that is,
$$\operatorname{lcm}\left(p, a_{k d-1}\right)=p a_{k d-1}>c^{2}\left(p a_{d}+a_{k d-1}\right)=c^{2}\left(a_{p d}+a_{k d-1}\right)$$
Then all the previous facts can be applied, and since $\operatorname{gcd}(p d, k d-1)=1, a_{k}=a_{k \operatorname{gcd}(p d, k d-1)} =k a_{\operatorname{gcd}(p d, k d-1)}=k a_{1}$, that is, the sequence is linear. Also, since $a_{1}$ divides all terms, $a_{1}=1$.

Case 2: $\operatorname{lcm}\left(a_{m}, a_{n}\right) \leq c^{2}\left(a_{m}+a_{n}\right)$ for all $m, n$.
Suppose that $a_{m} \leq a_{n}$; then $\operatorname{lcm}\left(a_{m}, a_{n}\right)=M a_{n} \leq c^{2}\left(a_{m}+a_{n}\right) \leq 2 c^{2} a_{n} \Longrightarrow M \leq 2 c^{2}$, that is, the factor in the smaller term that is not in the larger term is at most $2 c^{2}$.
We prove that in this case the sequence must be bounded. Suppose on the contrary; then there is a term $a_{m}$ that is divisible by a large prime power $p^{d}$. Then every larger term $a_{n}$ is divisible by a factor larger than $\frac{p^{d}}{2 c^{2}}$. So we pick $p^{d}>\left(2 c^{2}\right)^{2}$, so that every large term $a_{n}$ is divisible by the prime power $p^{e}>2 c^{2}$. Finally, fix $a_{k}$ for any $k$. It follows from $a_{k+n} \mid c\left(a_{k}+a_{n}\right)$ that $\left(a_{n}\right)$ is unbounded, so we can pick $a_{n}$ and $a_{k+n}$ large enough such that both are divisible by $p^{e}$. Hence any $a_{k}$ is divisible by $p$, which is a contradiction to the fact that there is no $D>1$ that divides every term in the sequence.