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Math Help - ab(a^2-b^2) conundrum

  1. #1
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    ab(a^2-b^2) conundrum

    Hi,

    If a, b and k are integers, a\not=b

    I can't find any examples where ab(a^2-b^2)=6k doesn't hold true, but can't figure why this is so.

    Anyone any ideas?

    Thanks
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  2. #2
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    Re: ab(a^2-b^2) conundrum

    Quote Originally Posted by gortwell View Post
    I can't find any examples where ab(a^2-b^2)=6k doesn't hold true
    Take a = 2, b = 1 and k = 2 By the way, the assumption a\ne b is unnecessary because 6 divides 0.

    To prove that 6 divides ab(a^2-b^2), prove that ab(a^2-b^2) is even by considering cases where a and b are even or odd. Also, prove that ab(a^2-b^2) is divisible by 3 by considering various remainders when a and b are divided by 3.
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  3. #3
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    Re: ab(a^2-b^2) conundrum

    Quote Originally Posted by emakarov View Post
    Take a = 2, b = 1 and k = 2


    Thanks for the kickstart

    Doing it altogether

    a Mod 6 b Mod 6 ab Mod 6 a+b Mod 6 a-b Mod 6 ab(a^2-b^2) Mod 6
    0 0 0 0 0 0
    0 1 0 1 5 0
    0 2 0 2 4 0
    0 3 0 3 3 0
    0 4 0 4 2 0
    0 5 0 5 1 0
    1 0 0 1 1 0
    1 1 1 2 0 0
    1 2 2 3 5 0
    1 3 3 4 4 0
    1 4 4 5 3 0
    1 5 5 0 2 0
    2 0 0 2 2 0
    2 1 2 3 1 0
    2 2 4 4 0 0
    2 3 0 5 5 0
    2 4 2 0 4 0
    2 5 4 1 3 0
    3 0 0 3 3 0
    3 1 3 4 2 0
    3 2 0 5 1 0
    3 3 3 0 0 0
    3 4 0 1 5 0
    3 5 3 2 4 0
    4 0 0 4 4 0
    4 1 4 5 3 0
    4 2 2 0 2 0
    4 3 0 1 1 0
    4 4 4 2 0 0
    4 5 2 3 5 0
    5 0 0 5 5 0
    5 1 5 0 4 0
    5 2 4 1 3 0
    5 3 3 2 2 0
    5 4 2 3 1 0
    5 5 1 4 0 0
    Last edited by gortwell; April 16th 2012 at 07:59 AM.
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  4. #4
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    Re: ab(a^2-b^2) conundrum

    A shorter proof. First, ab(a^2-b^2) is even. Indeed, this is obvious if a is even or b is even. If both a and b are odd, then a + b is even.

    Second, ab(a^2-b^2) is divisible by 3. This is obvious is a or b is divisible by 3 and when a and b have the same remainder when divided by 3 (in the latter case, 3 | a - b). If a ≡ 1 (mod 3) and b ≡ 2 (mod 3) (or vice versa), then a + b ≡ 0 (mod 3).
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  5. #5
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    Re: ab(a^2-b^2) conundrum

    Thanks emakarov
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