Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Prove it Slovenia

Let $x$ , $y$ and $z$ be non-zero real numbers such that $3x + 2y = z$ and $\frac{3}{x} + \frac{1}{y} = \frac{2}{z}$ . Prove that $5x^2 - 4y^2 - z^2$ is always an integer.

Solution

The second equation implies $2xy = 3yz + xz$ . Multiplying the first equation respectively by $z$ , $x$ and $y$ , we get $z^2 = 3xz + 2zy$ , $3x^2 = zx - 2xy$ and $2y^2 = zy - 3xy$ . So,
$5x^2 - 4y^2 - z^2 = \frac{5}{3}(zx - 2xy) - 2(zy - 3xy) - (3xz + 2zy) = \frac{4}{3}(2xy - 3zy - xz) = 0.$

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