Prove that there are no positive integer solutions to the equation x^2 + 8y = 3.
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Let $\displaystyle x$ be some positive integer. Then $\displaystyle x^2$ is a positive perfect square (integer). If $\displaystyle x > 1$ then $\displaystyle y = \frac{3-x^2}{8} < 0$. If $\displaystyle x = 0 \text{ or } 1, y$ is not an integer.
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