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📅 March 5, 2026 👤 scraface

Mar 05, 2026

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For $(5,5)$: $k - y = 5$, $k + y = 5$ → $k = 5$, $y = 0$ → $x = 10$.
Negative pairs yield symmetric solutions. Total lattice points: $(\pm2...
Solution: We are given a recurrence:
T(1) = 4, \quad T(n) = 2T(n-1) + 3 \text{ for } n \geq 2.
We compute $ T(n) $ for $ n = 1 $ to $ 5 $:
$ T(1) = 4 $
$ T(2) = 2 \cdot 4 + 3 = 8 + 3 = 11 $
$ T(3) = 2 \cdot 11 + 3 = 22 + 3 = 25 $
$ T(4) = 2 \cdot 25 + 3 = 50 + 3 = 53 $
$ T(5) = 2 \cdot 53 + 3 = 106 + 3 = 109 $

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