HMMT February, 2018

Number theory Difficulty 4.8 AIME Prove it United States

Problem:
A triple of integers $(a, b, c)$ satisfies $a + b c = 2017$ and $b + c a = 8$ . Find all possible values of $c$ .

Solution

Solution:
Add and subtract the two equations to find
$\begin{aligned} & (b + a)(c + 1) = 8 + 2017 \\ & (b - a)(c - 1) = 2017 - 8 \end{aligned}$
We see that $c$ is even and then that every integer $c$ with $c + 1 \mid 2025$ , $c - 1 \mid 2009$ works. We factor and solve.
The full solutions are $(2017, 8, 0)$ , $(-667, 1342, 2)$ , $(-59, -346, -6)$ , $(-31, 256, 8)$ .

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