Claim 1: The sequence is bounded. Proof: $a_i \mid a_ia_{i+1} \mid k-a_i^2$ so $a_i \mid k$, giving $a_i \le k$. Claim 2: $\text{rad}(a_i)=\text{rad}(a_{i+1})$ for $i \ge N$ Proof: $a_{i+1} \mid a_ia_{i+1} \mid k-a_{i}^2$ so $a_{i+1} \mid a_i^2$ which is enough as it proves that $\text{rad}(a_i)$ is a decreasing sequence in positive integers so at some point it must become estable and just become constant. Now let $k=\prod_{i=1}^{\theta} p_i^{\alpha_i}$ be it's canonic decomposition, notice that from Claim 1 and Claim 2 as well that a subset of all the prime factors of $k$ divides all $a_i$'s for $i \ge N$. WLOG let this subset of primes be $p_1<p_2< \cdots <p_{\ell}$. Claim 3: $\nu_{p_i}(k) \ge 2$ for all $1 \le i \le \ell$ Proof: $p_i^2 \mid a_ia_{i+1} \mid k-a_i^2$ so $p_i^2 \mid k$ being equivalent to the Claim. Now consider the sequence $b_i=\frac{a_{i+N}}{p_1p_2 \cdots p_{\ell}}$ for all $i \in \mathbb Z_{>0}$, then notice that we now have $b_ib_{i+1} \mid k'-b_i^2$, and that we can repeat this process, over and over. All the three Claims above will also be true for this new sequence just now with $k'=\frac{k}{(p_1p_2 \cdots p_{\ell})^2}$, so a bunch of things may happen if you keep this: -Either $k$ eventually runs out of square factors and becomes square-free while none of the $a_i$'s has hit the value of $1$ yet, this would mean that the next time you do this process, you get a contradiction. -Or a value of the $a_i$'s eventually becomes $1$ which means that all of the next following values will become $1$ as well as then $a_{i+1} \mid k-1$ but $\gcd(k,k-1)=1$ so $a_{i+1}=1$ and so on. Scaling back this means that at some point there will be some $a_M=\beta$ for which $a_j=\beta$ for all $j \ge M$, which is what we wanted thus we are done .

here is my idea

We uploaded our solution https://calimath.org/pdf/MEMO2024-T8.pdf on youtube https://youtu.be/IPdIoYsGRL4.