$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$ 
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.

I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted. 
Now how do I proceed? 

Let $\alpha=a+b=c+d$, $\beta=a^3+b^3=c^3+d^3$.
Then $$ \alpha^3=(a+b)^3=a^3+3a^2b+3ab^2+b^3=\beta+\alpha ab.$$
It follows that $a,b$ (and likewise $c,d$) are the two solutions of 
$$ X^2-\alpha X+\frac{\alpha^3-\beta}{\alpha}=0,$$
i.e., $\{a,b\}=\{c,d\}$ and hence $$a^n+b^n=c^n+d^n $$
for arbitrary integers $n$. this argument fails only when we cannot divide by $\alpha$, i.e., when $b=-a$. However, in that case also $d=-c$ and so
$$ a^n+b^n=0=c^n+d^n$$
for arbitrary odd integer $n$.
