SELECTION TESTS FOR THE 2019 BMO AND IMO

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 $a_1<a_2<\cdots <a_k$ 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.