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Thread: Discrete!! Please Help. thanks in advance

  1. March 25th 2008, 08:49 AM #1
    jas05s
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    Discrete!! Please Help. thanks in advance

    Prove that there are no positive integer solutions to the equation x^2 + 8y = 3.
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  2. March 25th 2008, 09:23 AM #2
    iknowone
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    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.
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