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Math Help - Beginning Discrete Proof

  1. #1
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    Beginning Discrete Proof

    Hello everyone,
    Got stuck on this one and wanted to see if anyone could help me here.
    Thank you.

    For all integers n, if n is a perfect square, then n+2 is NOT a perfect square.

    I keep running in circles, can anyone help me prove this one?
    Thanks!
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  2. #2
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    Quote Originally Posted by jthrock View Post
    For all integers n, if n is a perfect square, then n+2 is NOT a perfect square.
    If N = K^2 \;\& \;N + 2 = J^2 \; \Rightarrow \;J > K. These are positive integers.
    That means 2 = J^2  - K^2  = \left( {J - K} \right)\left( {J + K} \right).
    There is contradiction hiding there. Where is it?
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  3. #3
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    I know, this is where we got stuck in class. The prof was unsure about it either so we decided this would be a "quiz" in the future.
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  4. #4
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    Both (J-K)\;\&\;(J+K) are positive integers.
    Both factors of 2. What are the factors of 2?
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  5. #5
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    I'm sorry but I'm a bit confused, the factors of 2 are 2 *1. But I do not understand where you are going with this. By the way, thank you for your help with this.
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