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📅 March 5, 2026 👤 scraface

Mar 05, 2026

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19y^2 \equiv 0 \pmod{25} \Rightarrow y^2 \equiv 0 \pmod{25} \Rightarro...
So let $y = 5k$. Substitute:
25x^2 = 144(25k^2) + 3600 = 3600k^2 + 3600 = 3600(k^2 + 1).
Divide both sides by 25:
x^2 = 144(k^2 + 1).
So $x^2 = 144(k^2 + 1) = (12)^2(k^2 + 1)$, hence $x = 12\sqrt{k^2 + 1}...
m^2 - k^2 = 1 \Rightarrow (m - k)(m + k) = 1.
Integer solutions: $m = 1, k = 0$ or $m = -1, k = 0$. So only $k = 0$.
Then $y = 5(0) = 0$, and $x^2 = 144(0 + 1) = 144 \Rightarrow x = \pm 1...
Thus, the only lattice points are $(\pm 12, 0)$.

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