So $x^2 = 144(k^2 + 1) = (12)^2(k^2 + 1)$, hence $x = 12\sqrtk^2 + 1$. For $x$ to be integer, $k^2 + 1$ must be a perfect square. Let $k^2 + 1 = m^2$, so: - Portal da Acústica