HMMT February, 2019

Number theory Difficulty 5.3 AIME, harder Prove it United States

Problem:
How many pairs of integers $(x, y)$ are there such that $|x^{2}-2 y^{2}| \leq 1$ and $|3 x-4 y| \leq 1$ ?

Solution

Solution:
Note that if $(x, y)$ is a solution, so is $(-x,-y)$ . Thus, we consider $x \geq 0$ .

When $x \equiv 0 \pmod{4}$ , $y = 3x/4$ by inequality 2. Inequality 1 gives $|x^{2}/9| \leq 1$ , so $x \leq 3$ , so $x = 0$ .

When $x \equiv 1 \pmod{4}$ , $y = (3x+1)/4$ by inequality 2. Beyond $x = 1$ , $2y^{2} - x^{2} > 1$ , so there are no more solutions.

When $x \equiv 2 \pmod{4}$ , there are no solutions for $y$ .

When $x \equiv 3 \pmod{4}$ , $y = (3x-1)/4$ by inequality 2. Beyond $x = 7$ , $2y^{2} - x^{2} > 1$ , so there are no more solutions.

Thus, the solutions are $(0,0), (1,1), (3,2), (7,5)$ , and the negations of the latter three, giving 7 solutions.

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