# Math Help - Discrete!! Please Help. thanks in advance

1. ## Discrete!! Please Help. thanks in advance

Prove that there are no positive integer solutions to the equation x^2 + 8y = 3.

2. Let $x$ be some positive integer. Then $x^2$ is a positive perfect square (integer). If $x > 1$ then $y = \frac{3-x^2}{8} < 0$. If $x = 0 \text{ or } 1, y$ is not an integer.