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📅 May 12, 2026 👤 scraface

May 12, 2026

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Then $y = 5(0) = 0$, and $x^2 = 144(0 + 1) = 144 \Rightarrow x = \pm 1...
Thus, the only lattice points are $(\pm 12, 0)$.
Check: $\frac{144}{144} - \frac{0}{25} = 1$, valid.
No other $k$ gives $k^2 + 1$ square. Try small $y$ directly:
$y = 0$: $25x^2 = 3600 \Rightarrow x^2 = 144$, $x = \pm 12$.
$y = \pm1$: $25x^2 = 144 + 3600 = 3744$, $x^2 = 3744/25 = 149.76$, not...
$y = \pm5$: $25x^2 = 144(25) + 3600 = 3600 + 3600 = 7200$, $x^2 = 288$...
Eventually, only solutions are $(\pm12, 0)$.
Thus, exactly 2 lattice points.
Question: A high-performance computing simulation requires a grid size...

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