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Home » Thus, the minimum value is $ \boxed5 $. - Portal da AcÃºstica

Thus, the minimum value is $ \boxed5 $. - Portal da AcÃºstica

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 Question: How many integer solutions \( (x, y) \) lie on the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) with \( |x| \leq 4 \) and \( |y| \leq 3 \)?

📌 The ellipse is \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \). Since \( |x| \leq 4 \), \( x^2 \leq 16 \), so \( \frac{x^2}{16} \leq 1 \), and similarly for \( y \).

📌 We seek integer \( x, y \) such that:

📌 \frac{x^2}{16} + \frac{y^2}{9} = 1, \quad |x| \leq 4, \quad |y| \leq 3

📌 Multiply both sides by \( 144 = 16 \cdot 9 \):

📌 \cdot \left( \frac{x^2}{16} + \frac{y^2}{9} \right) = 9x^2 + 16y^2 = 144

📌 So we solve \( 9x^2 + 16y^2 = 144 \) in integers with \( |x| \leq 4 \), \( |y| \leq 3 \).

📌 We try all integer \( x \) from \( -4 \) to \( 4 \), compute \( 9x^2 \), then \( 16y^2 = 144 - 9x^2 \), so \( y^2 = \frac{144 - 9x^2}{16} \), must be a perfect square and integer.

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