v_Enhance wrote:

The condition implies that the difference $\frac{a_{n+1}}{a_1} - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}}$ is an integer for all $n > N$ .

We proceed by $p$ -adic valuation only henceforth.

Claim: If $p \nmid a_1$ , then $\nu_p(a_{n+1}) \le \nu_p(a_n)$ for $n \ge N$ .

Proof. The third term in the difference must have nonnegative $\nu_p$ . $\blacksquare$

Claim: If $p \mid a_1$ , then $\nu_p(a_n)$ is eventually constant.

Proof. By hypothesis $\nu_p(a_1) > 0$ . We consider two cases,

First assume $\nu_p(a_k) \ge \nu_p(a_1)$ for some $k > N$ . We claim that for any $n \ge k$ (by induction) we have: $\nu_p(a_1) \le \nu_p(a_{n+1}) \le \nu_p(a_n).$ Note in this case that $\nu(\frac{a_n}{a_1}) \ge 0$ , so we must have $\nu_p\left( \frac{a_{n+1}}{a_1} + \frac{a_n}{a_{n+1}} \right) \ge 0$ which implies the displayed inequality (since otherwise exactly one of the summands has nonnegative $\nu_p$ ). Thus once we reach this case, $\nu_p(a_n)$ is monotic but bounded below by $\nu_p(a_1)$ , and so it is eventually constant.
Now assume $\nu_p(a_k) < \nu_p(a_1)$ for every $k > N$ . Take any $n > N$ then. We have $\nu_p\left(\frac{a_{n+1}}{a_1}\right) < 0$ , and also $\nu_p\left(\frac{a_n}{a_1}\right) < 0$ , so among the three summands two must have equal $p$ -adic valuation. We consider all three possibilities: $\begin{align*} \nu_p\left(\frac{a_{n+1}}{a_1}\right) = \nu_p\left(\frac{a_n}{a_1}\right) &\implies \boxed{\nu_p(a_{n+1}) = \nu_p(a_{n})} \\ \nu_p\left(\frac{a_{n+1}}{a_1}\right) = \nu_p\left(\frac{a_n}{a_{n+1}}\right) &\implies \boxed{\nu_p(a_{n+1}) = \frac{\nu_p(a_n) + \nu_p(a_1)}{2}} \\ \nu_p\left(\frac{a_{n}}{a_1}\right) = \nu_p\left(\frac{a_n}{a_{n+1}}\right) &\implies \nu_p(a_{n+1}) = \nu_p(a_1),\text{ but this is impossible}. \end{align*}$ Thus, eventually $\nu_p(a_{n+1}) \ge \nu_p(a_n)$ in either possible situation, but $\nu_p(a_n)$ is bounded above by $\nu_p(a_1) - 1$ , so in this case we must also stabilize. \qedhere

$\blacksquare$

Since the latter claim is applied only to finitely many primes, after some time $\nu_p(a_n)$ is fixed for all $p \nmid a_n$ . Afterwards, the sequence satisfies $a_{n+1} \mid a_n$ for each $n$ , and thus must be eventually constant.

Remark: This solution is almost completely $p$ -adic, in the sense that I think a similar result if one replaces $a_n \in {\mathbb Z}$ by $a_n \in {\mathbb Z}_p$ for any particular prime $p$ . In other words, the primes almost do not talk to each other.

There is one caveat: if $x_n$ is an integer sequence such that $\nu_p(x_n)$ is eventually constant for each prime then $x_n$ may not be constant. For example, take $x_n$ to be the $n$ th prime! That's why in the first claim (applied to co-finitely many of the primes), we need the stronger non-decreasing condition, rather than just eventually constant.

This is basically my solution. This "caveat" at the end is easy to get around if you notice that since $\frac{a_{n+1}^2-a_na_{n+1}+a_na_1}{a_1a_{n+1}}$ is an integer, $a_{n+1} \mid a_na_1$ , so eventually $\frac{a_n}{(a_n,a_1)}$ converges to a constant. Then you basically do this analysis only on primes that divide $a_1$ , of which there are finitely many. I think this is the more intuitive way to deal with it.