Solution: Rewrite the equation as $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Possible integer values for $y$ are $-2, -1, 0, 1, 2$. For each $y$, solve for $x$:

["Solution: Rewriting the Equation to Find Integer Solutions", "When analyzing conic sections like ellipses, rewriting the standard form equation is a crucial step toward identifying key properties—such as intercepts—and finding integer solutions. Consider the given ellipse equation:", "[\n\frac{x^2}{16} + \frac{y^2}{4} = 1\n]", "This is an ellipse centered at the origin with semi-major axis along the $x$-axis and semi-minor axis along the $y$-axis.", "---", "### Step 1: Understand the Constraints", "Since both denominators are positive, the right-hand side equals 1. For real values of $x$ and $y$, both terms $\frac{x^2}{16}$ and $\frac{y^2}{4}$ must be at most 1. Therefore:", "- $ \frac{x^2}{16} \leq 1 \Rightarrow x^2 \leq 16 \Rightarrow -4 \leq x \leq 4 $\n- $ \frac{y^2}{4} \leq 1 \Rightarrow y^2 \leq 4 \Rightarrow -2 \leq y \leq 2 $", "This matches the given integer values for $y$: $y = -2, -1, 0, 1, 2$.", "---", "### Step 2: Solve for $x$ for Each Integer $y$", "We now substitute each integer $y$ into the equation and solve for possible integer $x$ values.", "#### Case $y = -2$:\n[\n\frac{y^2}{4} = \frac{(-2)^2}{4} = \frac{4}{4} = 1\n]\nSubstitute into the equation:\n[\n\frac{x^2}{16} + 1 = 1 \Rightarrow \frac{x^2}{16} = 0 \Rightarrow x^2 = 0 \Rightarrow x = 0\n]\nSolution: $(0, -2)$", "---", "#### Case $y = -1$:\n[\n\frac{y^2}{4} = \frac{1}{4}\n]\n[\n\frac{x^2}{16} = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow x^2 = 16 \cdot \frac{3}{4} = 12 \Rightarrow x = \pm \sqrt{12} = \pm 2\sqrt{3}\n]\nBut $\sqrt{3}$ is irrational, so no integer $x$.\nNo integer solution", "---", "#### Case $y = 0$:\n[\n\frac{y^2}{4} = 0\n]\n[\n\frac{x^2}{16} = 1 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4\n]\nSolutions: $(4, 0), (-4, 0)$", "---", "#### Case $y = 1$:\n[\n\frac{y^2}{4} = \frac{1}{4}\n]\nSame as $y = -1$:\n[\n\frac{x^2}{16} = \frac{3}{4} \Rightarrow x^2 = 12 \Rightarrow x = \pm 2\sqrt{3}\n]\nAgain, no integer $x$.\nNo integer solution", "---", "#### Case $y = 2$:\n[\n\frac{y^2}{4} = \frac{4}{4} = 1\n]\n[\n\frac{x^2}{16} = 0 \Rightarrow x = 0\n]\nSolution: $(0, 2)$", "---", "### Summary of Integer Solutions", "| $y$ | $x^2 = 16(1 - \frac{y^2}{4})$ | Integer $x$ found? | Solution(s) |\n|-----|-------------------------------|--------------------|-------------|\n| -2 | $x^2 = 0$ | Yes (0) | $(0, -2)$ |\n| -1 | $x^2 = 12$ | No | — |\n| 0 | $x^2 = 16$ | Yes ((\pm4)) | $(4,0), (-4,0)$ |\n| 1 | $x^2 = 12$ | No | — |\n| 2 | $x^2 = 0$ | Yes (0) | $(0,2)$ |", "---", "### Why Rewriting Is Key", "Rewriting the equation in standard form transforms complex geometric information into a clear algebraic framework. This allows precise determination of valid domain bounds and straightforward computation of intercepts and symmetric points—essential for identifying all integer-coordinate points on the ellipse.", "---", "### Final Answer", "The equation\n[\n\frac{x^2}{16} + \frac{y^2}{4} = 1\n]\nhas integer solutions at points:\n$$\n(0, -2),\ (4, 0),\ (-4, 0),\ (0, 2)\n$$\nfor integer $y = -2, -1, 0, 1, 2$. For each valid $y$, substitution led directly to solving for $x$, leveraging the standard ellipse form to efficiently isolate and compute possible integer values.", "---", "Keywords: ellipse equation, integer solutions, rewrite equation, conic section, parametric solution for $y$ values, algebraic approach, solve for $x$, standard form."]