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Thread: Proof Method

  1. #1
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    Proof Method

    Hi all,

    I need the proof of the following:
    If $\displaystyle A, B , $ and $\displaystyle C$ are positive integers such that $\displaystyle C^2= A^2 + B^2$ then either $\displaystyle A$ or $\displaystyle B$ is an even integer.

    I could be grateful if you help me.

    Raed
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  2. #2
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    If A and B are both odd, then C^2 is a multiple of 4. (It can't be even but not divisible by 4, right?) Then if you represent A and B as 2 times an even number plus 1, then the right-hand side is not divisible by 4.
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  3. #3
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    Quote Originally Posted by raed View Post
    Hi all,

    I need the proof of the following:
    If $\displaystyle A, B , $ and $\displaystyle C$ are positive integers such that $\displaystyle C^2= A^2 + B^2$ then either $\displaystyle A$ or $\displaystyle B$ is an even integer.

    I could be grateful if you help me.

    Raed
    You could also try to prove that the squares of a pair of odd integers do not sum to an integer square...

    $\displaystyle A=2f+1,\;\;\;B=2j+1$

    $\displaystyle C$ will have to be even, since $\displaystyle Odd^2+Odd^2=Even$

    Odd squares are Odd and Even squares are Even, hence C is Even.

    $\displaystyle C=2k$

    $\displaystyle A^2+B^2=(2f+1)^2+(2g+1)^2=4f^2+4f+1+4g^2+4g+1$

    This needs to equal $\displaystyle 4k^2$

    $\displaystyle \Rightarrow\ 2\left(2f^2+2g^2+2f+2g+1\right)=2\left(2k^2\right)$

    $\displaystyle 2\left(f^2+g^2+f+g\right)+1=2k^2$

    However, this is contradictory as the LHS is Odd.
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