Selection Tests for the BMO and IMO, 2019

Letting $f_n(a_1, a_2, \dots, a_n) = a_1^2 a_2^2 \cdots a_n^2 - 4(a_1^2 + a_2^2 + \cdots + a_n^2)$, the conclusion is a straightforward consequence of the following two facts:
(1) For infinitely many positive integers $n$ there exist positive integers $a_1, a_2, \dots, a_n$ such that $f_n(a_1, a_2, \dots, a_n)$ is the square of an integer, and $a_1 < a_2$; and
(2) Given an integer $n \ge 3$, if $a_1, a_2, \dots, a_n$ are positive integers such that $f_n(a_1, a_2, \dots, a_n)$ is the square of an integer, and $a_1 < a_2 < \dots < a_k$ for some index $k$, $1 < k < n$, then there exists an integer $a > a_k$ such that $f_n(a_1, a_2, \dots, a_k, a, a_{k+2}, \dots, a_n)$ is the square of an integer.

To prove (1), notice that, if $a_1 < a_2$ are arbitrarily large positive integers, then $f_2(a_1, a_2)$ is an arbitrarily large positive integer congruent to $0$ or $1$ modulo $4$. Subtraction of a suitable number of $4$'s then yields $0$ or $1$, each of which is a square. Letting that suitable number of $4$'s be $n-2$, which is clearly arbitrarily large, $f_n(a_1, a_2, 1, \dots, 1)$ is the square of an integer.

To prove (2), write $f_n(a_1, a_2, \dots, a_n) = b^2$ and notice that $(a_{k+1}, b)$ solves the Pell equation
$$(a_1^2 \cdots a_k^2 a_{k+2}^2 \cdots a_n^2 - 4)x^2 - y^2 = 4(a_1^2 + \cdots + a_k^2 + a_{k+2}^2 + \cdots + a_n^2).$$
The latter has therefore infinitely many solutions in positive integers. In particular, it has a solution $(a, c)$ such that $a > a_{k+1}$. This establishes (2) and concludes the proof.