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Math Help - Algebra Brainteaser

  1. #1
    Senior Member sfspitfire23's Avatar
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    Oct 2009
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    Algebra Brainteaser

    Given:
    a^2+b^2=c^2+d^2
    a^3+b^3=c^3+d^3
    Show that:
    a+b=c+d


    Obvi, factor the second equation and we get


    (a+b)(a^2+b^2-ab)=(c+d)(c^2+d^2-cd)


    So, we only have to show that ab=cd.


    Can anyone crack it?
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  2. #2
    Newbie
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    Re: Algebra Brainteaser

    Say, in what domain are we solving this? If it's the reals, than this assumption doesn't hold. Take for example the way less general (a=1, b=1):
    c^2+d^2=2
    c^3+d^3=2
    So your assumption would imply that c+d=2, but one can find other c and d that solve the above system. Let u=c+d, so from the second:
    (c+d)(c^2+d^2-cd)=2
    u(2-cd)=2
    cd=2-\frac{2}{u}
    Adding twice to the first equation
    c^2+2cd+d^2=2+4-\frac{4}{u}
    u^2=6-\frac{4}{u}
    u^3-6u+4=0
    As, expected, one solution is u=c+d=1+1=2. But the other roots
    u(u-2)(u+2)-2u+4=0
    (u-2)(u^2+2u-2)=0
    are the roots of the quadratic:
    u^2+2u-2=0
    And those are -1\pm\sqrt{3}.
    So, c+d can be equal to those numbers, and not just to 2=1+1=a+b.
    For example:
    c=\frac{a+\sqrt{a^2+4a}}{2}
    d=\frac{a-\sqrt{a^2+4a}}{2}
    Where
    a=-1+\sqrt{3}
    one of the roots of the above quadratic.
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