If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then $a + b = c + d$ I came across this problem today. I would be interested to see if anyone knows a proof for it:
If $a^3+b^3 = c^3+d^3$ and $a^2+b^2 = c^2 + d^2$, then show that $a + b = c + d$.

The statement has a natural geometric interpretation when $a, b, c, d \geq 0$. A circle $x^2 + y^2 = r$ and a curve $x^3 + y^3 = s$ for $x, y > 0$ are symmetric about the line $x = y$ and if they intersect, they either intersect at a point $(x,x)$ or two symmetric points $(x,y)$ and $(y,x)$ for $x \neq y$. Thus if $a^2 + b^2 = c^2 + d^2$ and $a^3 + b^3 = c^3 + d^3$ for $a, b, c, d \geq 0$, then $(a,b) = (c,d)$ or $(a,b) = (d,c)$. In either case, $a + b = c + d$.
Why the curves intersect the way they do: If $p > q$ then $x^p + y^p = 1$ is "fatter" than $x^q + y^q = 1$, and so scaled versions of these curves will intersect as above. Maybe to prove this might require algebra like in Sivaram's answer, but the picture seems clear (to me anyway).
