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Problem 996

AMC 12 late, AIME early

Number theory Difficulty 4.6 Prove it HMMT November · United States · 2013

Find all triples of positive integers $(x, y, z)$ such that $x^{2} + y - z = 100$ and $x + y^{2} - z = 124$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Answer: $(12, 13, 57)$

Cancel $z$ to get $24 = (y - x)(y + x - 1)$ . Since $x, y$ are positive, we have $y + x - 1 \geq 1 + 1 - 1 > 0$ , so $0 < y - x < y + x - 1$ . But $y - x$ and $y + x - 1$ have opposite parity, so $(y - x, y + x - 1) \in \{(1, 24), (3, 8)\}$ yields $(y, x) \in \{(13, 12), (6, 3)\}$ .
Finally, $0 < z = x^{2} + y - 100$ forces $(x, y, z) = (12, 13, 57)$ .

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