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Number theory Difficulty 4.7 AIME Prove it Slovenia

The integers $x$ and $y$ are such that $x + xy + y^2 = 1$ and $y(5 + x) \ge 0$ . What integer values can the expression $x - y$ take?

Solution

The equality gives us $x(1+y) = 1 - y^2 = (1+y)(1-y)$ . If $y = -1$ , the equality holds, and from the inequality we derive $-(5+x) \ge 0$ or $x \le -5$ . Hence $x - y = x + 1 \le -4$ .

If, on the other hand, $y \ne -1$ , the equality reduces to $x = 1 - y$ . Using the last relation in the inequality we derive $y(6-y) \ge 0$ , and hence $0 \le y \le 6$ . We conclude $x - y = 1 - 2y \in \{1, -1, -3, -5, -7, -9, -11\}$ .

The expression $x - y$ can take all values smaller or equal to $-4$ as well as values $-3$ , $-1$ and $1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.