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 $A_k^{\prime }=A\cup \{m+1,m+2,\dots ,m+k\}$. Then there is a positive integer $k$ such that $$\begin{array}{rl}-D(A)& <(m+1{)}^k-(k^3+2m{k}^2+m^{2}k)\\ & =(m+1{)}^k-k(m+k{)}^2\le D(A_k^{\prime })-D(A)\end{array}$$ and hence $D(A_k^{\prime })>0$. Thus, it suffices to find a finite set $B$ containing $A_k^{\prime }$ 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 $$\begin{array}{rl}D(A_{k+1})& =\prod _{x\in A_{k+1}}x-\sum _{x\in A_{k+1}}{x}^2\\ & =\prod _{x\in A_k}x(\prod _{x\in A_k}x-1)-\sum _{x\in A_k}{x}^2-(\prod _{x\in A_k}x-1{)}^2\\ & =\prod _{x\in A_k}x-\sum _{x\in A_k}{x}^2-1\\ & =D(A_k)-1.\end{array}$$ 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.