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 $a + b = c + d$.

If negative values of the variables are allowed then there are counterexamples. The simplest is the following:
$$a=2,\quad b=2,\quad c=\sqrt{3}-1-\sqrt{2\sqrt{3}}\doteq -1.129,\quad d=\sqrt{3}-1+\sqrt{2\sqrt{3}}\doteq2.593$$
with $a+b=4$, $c+d\doteq1.464\ .$ 
In the following we assume $0\leq a\leq b$ and $0\leq c\leq d$. Then icobes' conjecture is true:
Since $a^2+b^2=c^2+d^2=:r^2>0$ we may write
$$a=r\sin\bigl({\pi\over4}-\alpha\bigr), \quad
b=r\sin\bigl({\pi\over4}+\alpha\bigr), \quad
c=r\sin\bigl({\pi\over4}-\beta\bigr), \quad
d=r\sin\bigl({\pi\over4}+\beta\bigr)$$
for some $\alpha, \beta\in\ \bigl[0,{\pi\over4}\bigr]$. It follows that $$a+b=\sqrt{2}r\cos\alpha,\quad c+d=\sqrt{2}r\cos\beta$$
and therefore
$$2(a^3+b^3)=3(a+b)(a^2+b^2)-(a+b)^3=\sqrt{2}r^3(3\cos\alpha-2\cos^3\alpha)=:\sqrt{2}r^3 f(\alpha)\ ;$$
and similarly $2(c^3+d^3)=\sqrt{2}r^3f(\beta)$.
Now $f'(t)=3\sin t\cos(2t)>0$ for $0<t<{\pi\over 4}$, whence $f$ is strictly increasing on $\bigl[0,{\pi\over4}\bigr]$. It follows that $a^3+b^3=c^3+d^3$ implies $\alpha=\beta$, and this in turn implies $a=c$, $b=d$.
