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:

Tue, 05 May 2026 00:36:03 +0000

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