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Problem 778

AMC 12 late, AIME early

Number theory Difficulty 4.3 Prove it Junior Turkish Mathematical Olympiad · Turkey

Find all triples $(x, y, p)$ satisfying
$x^2 - 3xy + p^2y^2 = 12p$
where $x, y$ are integers and $p$ is a prime number.

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

Since $x^2 + p^2y^2 \equiv 0 \pmod{3}$ , if $p \ne 3$ then $x^2 + y^2 \equiv 0 \pmod{3}$ , and consequently $x \equiv y \equiv 0 \pmod{3}$ . Then $9 \mid x^2 - 3xy + py^2$ but $9 \nmid 12$ , a contradiction. Thus, $p = 3$ and $x^2 - 3xy + 9y^2 = 36$ . Therefore, $3 \mid x$ . $x = 3k$ , and we get a second order equation $k^2 - ky + y^2 - 4 = 0$ . The discriminant $\Delta = 16 - 3y^2 = m^2$ . Therefore, $y^2 = 0$ or $y^2 = 4$ . Solutions are: $(-6, 0, 3)$ , $(0, -2, 3)$ , $(0, 2, 3)$ , $(6, -2, 3)$ , $(6, 0, 3)$ , $(6, 2, 3)$ .

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