Proof. We prove that $C = 8$ satisfies the requirement.

(1) First consider the case where $a_1$ and $a_2$ are coprime. Let $d = a_2 - a_1$ be the common difference. For any prime $p$, if $p \mid d$, then $p \nmid a_1$ and hence $p$ does not divide any $a_n$. If $p \nmid d$, then any consecutive $p^2$ terms of the sequence $a_n$ form a complete set of residues modulo $p^2$, among which exactly one term is divisible by $p^2$.

Let $N = 4a_2$. We prove the existence of $1 \le n \le N$ such that $a_n$ has no square factors. If $1 \le m \le N$ and $a_m$ has a square factor, then there exists a prime $p$ such that $p^2 \mid a_m$. Thus, $p \nmid d$ and $p \le \sqrt{a_N}$. Moreover, the number of terms in $a_1, a_2, \dots, a_N$ that are divisible by $p^2$ is at most $\lceil \frac{N}{p^2} \rceil$. Therefore, the number $M$ of terms in $a_1, a_2, \dots, a_N$ that have square factors satisfies:
$$M \le \sum_{p \le \sqrt{a_N}} \left\lceil \frac{N}{p^2} \right\rceil < \sum_{p \le \sqrt{a_N}} \left( \frac{N}{p^2} + 1 \right) \le N \sum_{p \le \sqrt{a_N}} \frac{1}{p^2} + \sqrt{a_N} < \frac{N}{2} + \sqrt{a_N}.$$
The last inequality is due to
$$\sum_{p} \frac{1}{p^2} < \frac{1}{4} + \sum_{k=2}^{\infty} \frac{1}{(2k-1)^2} < \frac{1}{4} + \sum_{k=2}^{\infty} \frac{1}{(2k-2)2k} = \frac{1}{2}.$$
When $N = 4a_2$, we have $\sqrt{a_N} < \sqrt{Na_2} = 2a_2 = \frac{N}{2}$. Thus, $M < N$. Therefore, there exist numbers in $a_1, a_2, \dots, a_N$ that have no square factors.

(2) Assume $\gcd(a_1, a_2) = \gcd(a_1, d) = q = q_1 \cdots q_l > 1$, where $q_1, \cdots, q_l$ are pairwise distinct prime factors. Note that every $a_i$ is divisible by $q$. For each prime factor $q_i$, there exists an index $t_i$ such that $v_{q_i}(a_{t_i}) = 1$. In fact, we can take $t_i \in 1, 2$ since $a_1$ and $a_2$ cannot both be divisible by $q_i^2$. By Chinese Remainder Theorem, there exists $k \in 1, 2, \cdots, q$ such that $k \equiv t_i \pmod{q_i}$ for $1 \le i \le l$. Thus, for $1 \le i \le l$, we have $a_k \equiv a_{t_i} \pmod{q_id}$, which implies $v_{q_i}(a_k) = 1$.

Consider the subsequence $a_k, a_{k+q}, a_{k+2q}, \dots$, which has common difference $qd$ and is divisible by $q^2$. By construction, $q_i^2 \nmid a_k$ for $1 \le i \le l$, i.e., each term in this subsequence has no square factors that are equal to $q_1, q_2, \dots, q_l$. Let $b_i = \frac{a_{k+(i-1)q}}{q}$, then $b_1 < b_2 < \dots$ is an arithmetic sequence with common difference $d$, and $b_1$ is coprime to $d$. By the conclusion of (1), there exists $i \le 4b_2$ such that $b_i$ has no square factor, and since $q$ is coprime to $b_i$, $a_{k+(i-1)q} = qb_i$ also has no square factor. Finally,
$$k + (i - 1)q \le iq \le 4b_2q = 4a_{k+q} < 4(k + q)a_2 \le 8da_2 < 8a_2^2.$$