USA IMO

For any finite set $S$ of positive integers, let
$$D(S) = \prod_{x \in S} x - \sum_{x \in S} x^2.$$
If $D(A) = 0$, then we take $B = A$.

If $D(A) < 0$, then let $m = \max A$. Write $A'_k = A \cup \{m+1, m+2, \dots, m+k\}$. Then there is a positive integer $k$ such that
$$\begin{aligned} -D(A) &< (m+1)^k - (k^3 + 2mk^2 + m^2k) \\ &= (m+1)^k - k(m+k)^2 \le D(A'_k) - D(A) \end{aligned}$$
and hence $D(A'_k) > 0$. Thus, it suffices to find a finite set $B$ containing $A'_k$ such that $D(B) = 0$, because then $B$ contains $A$ as well. This reduces the problem to the next and final case.

Assume that $D(A) > 0$, and write $A_0 = A$. Define $A_{k+1} = A_k \cup \{\prod_{x \in A_k} x - 1\}$ recursively for $k = 0, 1, \dots, D(A) - 1$. If $D(A_k) > 0$, we have $A_k \ne \{1\}$ and hence
$$\max A_k < \sum_{x \in A_k} x^2 = \prod_{x \in A_k} x - D(A_k) < \prod_{x \in A_k} x.$$
Therefore, $\prod_{x \in A_k} x - 1 > \max A_k$ and $A_{k+1}$ has one more element than $A_k$. It follows that
$$\begin{align*} D(A_{k+1}) &= \prod_{x \in A_{k+1}} x - \sum_{x \in A_{k+1}} x^2 \\ &= \prod_{x \in A_k} x (\prod_{x \in A_k} x - 1) - \sum_{x \in A_k} x^2 - (\prod_{x \in A_k} x - 1)^2 \\ &= \prod_{x \in A_k} x - \sum_{x \in A_k} x^2 - 1 \\ &= D(A_k) - 1. \end{align*}$$
Because $D(A_0) > 0$, it follows that $D(A_k) = D(A) - k > 0$ for $k < D(A)$ and that $D(A_{D(A)}) = 0$. Taking $B = A_{D(A)}$ completes the proof.