SAUDI ARABIAN IMO Booklet 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 $$P(x)=(x-b_1)(x-b_2)\dots (x-b_n)=x^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0.$$ 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, $$\frac{a_1{a}_2\dots a_n}{a_1^2},\frac{a_1{a}_2\dots a_n}{a_2^2},\dots ,\frac{a_1{a}_2\dots a_n}{a_n^2}\in \mathbb{Z}.$$