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Problem 498

AMC 12 late, AIME early

Number theory Difficulty 4.9 Prove it 25th ASU · Soviet Union

Problem:
Find all integers $a$ , $b$ , $c$ , $d$ such that $ab - 2cd = 3$ , $ac + bd = 1$ .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Solution:

$11 = (ab - 2cd)^2 + 2(ac + bd)^2 = (a^2 + 2d^2)(b^2 + 2c^2)$ , so we must have either (1) $a^2 + 2d^2 = 1$ , $b^2 + 2c^2 = 11$ , or (2) $a^2 + 2d^2 = 11$ , $b^2 + 2c^2 = 1$ .

(1) gives $a = \pm 1$ , $d = 0$ , $b = \pm 3$ , $c = \pm 1$ . If $a = 1$ and $d = 0$ , then $ac + bd = 1$ implies $c = 1$ , and $ab - 2cd = 3$ implies $b = 3$ . Similarly, if $a = -1$ , then $c = -1$ , and $b = -3$ .

Similarly, (2) gives $(a, b, c, d) = (3, 1, 0, 1), (-3, -1, 0, -1)$ .

$(a, b, c, d) = (1, 3, 1, 0), (-1, -3, -1, 0), (3, 1, 0, 1), (-3, -1, 0, -1)$

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.