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: - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$So $x^2 = 144(k^2 + 1) = (12)^2(k^2 + 1)$, hence $x = 12\sqrt{k^2 + 1}$. For $x$ to be integer, $k^2 + 1$ must be a perfect square. Let $k^2 + 1 = m^2$, so:$

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