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Thread: Algebra Brainteaser

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

    Algebra Brainteaser

    $\displaystyle a^2+b^2=c^2+d^2$
    $\displaystyle a^3+b^3=c^3+d^3$
    Show that:
    $\displaystyle a+b=c+d$

    Obvi, factor the second equation and we get

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

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

    Can anyone crack it?
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  2. #2
    Jun 2011
    Right now it's the Earth, a strange place

    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):
    $\displaystyle c^2+d^2=2$
    $\displaystyle c^3+d^3=2$
    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:
    $\displaystyle (c+d)(c^2+d^2-cd)=2$
    $\displaystyle u(2-cd)=2$
    $\displaystyle cd=2-\frac{2}{u}$
    Adding twice to the first equation
    $\displaystyle c^2+2cd+d^2=2+4-\frac{4}{u}$
    $\displaystyle u^2=6-\frac{4}{u}$
    $\displaystyle u^3-6u+4=0$
    As, expected, one solution is u=c+d=1+1=2. But the other roots
    $\displaystyle u(u-2)(u+2)-2u+4=0$
    $\displaystyle (u-2)(u^2+2u-2)=0$
    are the roots of the quadratic:
    $\displaystyle u^2+2u-2=0$
    And those are $\displaystyle -1\pm\sqrt{3}$.
    So, c+d can be equal to those numbers, and not just to 2=1+1=a+b.
    For example:
    $\displaystyle c=\frac{a+\sqrt{a^2+4a}}{2}$
    $\displaystyle d=\frac{a-\sqrt{a^2+4a}}{2}$
    $\displaystyle a=-1+\sqrt{3}$
    one of the roots of the above quadratic.
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