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Thread: Proving if something is even

  1. #1
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    Proving if something is even

    Let A,B,C be integers such that A^2 + B^2 = C^2. Prove at least one of A and B is even.

    Please help guide me through this problem so I can understand it better thanks.


    I have another question also
    let a and b be positive such that a^2 = b^3
    given 4 divides b, prove that 8 divides a

    Im hoping if someone can guide me through this problem also thanks.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Proving if something is even

    For the first question, have you taken a look at Euclid's method for generating Pythagorean triples?

    For the second question, I would let b = 4n...what do you find?
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  3. #3
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    Re: Proving if something is even

    for the first one i want to try to use a proof by contradiction but i am kind of stuck on how to do it

    and for the 2nd one when b = 4n
    b^3 = (4n)^3
    so a^2 = (4n)^3
    so a = sqrt((4n)^3)
    Last edited by gfbrd; Apr 17th 2013 at 08:16 PM.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Proving if something is even

    What is the square root of 4 cubed?
    Thanks from gfbrd
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  5. #5
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    Re: Proving if something is even

    Edit: Nevermind I figred everything out thanks a lot for your help!
    Last edited by gfbrd; Apr 17th 2013 at 09:19 PM.
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    Re: Proving if something is even

    Hello, gfbrd!

    $\displaystyle \text{Let }A,B,C\text{ be integers such that }A^2 + B^2 \,=\, C^2.$
    $\displaystyle \text{Prove that at least one of }A\text{ and }B\text{ is even.}$

    First, we will establish a theorem.

    Consider the square of an integer, $\displaystyle n^2.$

    If $\displaystyle n$ is even, we have: .$\displaystyle n = 2k$
    . . Hence: .$\displaystyle n^2 \:=\:(2k)^2 \:=\:4k^2$

    If $\displaystyle n$ is odd, we have: .$\displaystyle n = 2k+1$
    . . Hence: .$\displaystyle n^2 \:=\:(2k+1)^2 \:=\:4k^2 + 4k + 1 \:=\:4(k^2+k)+1$

    Therefore, the square of an integer is either:
    . . (1) a multiple of 4, or
    . . (2) one more than a multiple of 4.


    We are given: .$\displaystyle A^2 + B^2\,=\,C^2$

    Suppose $\displaystyle A$ and $\displaystyle B$ are both odd.

    We have: .$\displaystyle A \,=\,2h+1,\;B \,=\,2k+1$

    The we have:

    . . $\displaystyle A^2 + B^2\;=\:(2h+1)^2 + (2k+1)^2$

    . . . . . . . . .$\displaystyle =\:4h^2+4h+1 + 4k^2+4k+1$

    . . . . . . . . .$\displaystyle =\;4(h^2+h + k^2+k) + 2$


    Hence: $\displaystyle A^2+B^2$ is two more than a multiple of 4.
    . . . . . .It cannot equal a square, $\displaystyle C^2.$


    Therefore, at least one of $\displaystyle A$ and $\displaystyle B$ must be even.
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