2024/25 AHSMC Part 1 Problem 3

Rewriting $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{4}$ as $(x-4)(y-4) = 16$ with $x \ne 0$, $y \ne 0$, the factor $x-4$ is one of $\{\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\}$. This gives $10$ pairs, but $x-4 = -4$ yields $x = 0$, which is not allowed. The other nine values are valid. The answer is (C).