, 2014

We first show that if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 \le x_3 x_4$, then $x_1 + x_2 \le x_3 + x_4$. Suppose, if possible, that $x_1 + x_2 > x_3 + x_4$. Let $n = \lfloor \sqrt{x} \rfloor$, $x \in S$, and write $x_k = n^2 + w_k$, where the $w_k$ are non-negative integers less than $2n+1$, to deduce that $w_1 + w_2 - w_3 - w_4 \ge 1$. The condition $x_1 x_2 \le x_3 x_4$ yields $(w_1 + w_2 - w_3 - w_4)n^2 \le w_3 w_4 - w_1 w_2$, so $w_3 > 0$ and
$$\begin{aligned} n^2 &\le (w_1 + w_2 - w_3 - w_4)n^2 \le w_3 w_4 - w_1 w_2 < w_3(w_1 + w_2 - w_3) - w_1 w_2 \\ &= (w_1 - w_3)(w_3 - w_2) \le ((w_1 - w_3) + (w_3 - w_2))^2 / 4 = (w_1 - w_2)^2/4 \le n^2, \end{aligned}$$
which is a contradiction.

Thus, if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 = x_3 x_4$, then $x_1 + x_2 = x_3 + x_4$, so $x_1^2 + x_3 x_4 = x_1(x_1 + x_2) = x_1(x_3 + x_4)$, i.e., $(x_1 - x_3)(x_1 - x_4) = 0$ whence $x_1 = x_3$ or $x_1 = x_4$. The conclusion now follows at once.