1. ## 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?

2. ## 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.