, 2023

For each $k = 1, 2, \dots, n$, denote $b_k = \frac{a_1 a_2 \dots a_n}{a_k^2}$ and consider the following polynomial
$$P(x) = (x - b_1)(x - b_2) \dots (x - b_n) = x^n + c_{n-1} x^{n-1} + \dots + c_1 x + c_0.$$
Based on Vieta's theorem, one can see that $c_{n-1}$ is an integer, and for any $k$ integers $i_1, i_2, \dots, i_k \in \{1, 2, \dots, n\}$, $b_{i_1} b_{i_2} \dots b_{i_k}$ is also an integer, which implies that all coefficients $c_k$ are integers. On the other hand, $P(x)$ is a monic polynomial which implies that all of its roots $b_1, b_2, \dots, b_n$ are also integers. Hence,
$$\frac{a_1 a_2 \dots a_n}{a_1^2}, \frac{a_1 a_2 \dots a_n}{a_2^2}, \dots, \frac{a_1 a_2 \dots a_n}{a_n^2} \in \mathbb{Z}.$$