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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.
\( x = 0 \): \( y^2 = \frac{144}{16} = 9 \) → \( y = \pm 3 \) → valid
\( x = \pm 1 \): \( 9(1) = 9 \), \( 144 - 9 = 135 \), \( y^2 = 135/16 \) → not integer
\( x = \pm 2 \): \( 9(4) = 36 \), \( 144 - 36 = 108 \), \( y^2 = 108/16 = 27/4 \) → not integer
