<s>[INST] $a+b=c+d$ , $a^3+b^3=c^3+d^3$ , 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? 
 [/INST] Prove: for all $n$: $a^{2n+1}+b^{2n+1}=c^{2n+1}+d^{2n+1}$
If $a+b\ne 0$ then we get $$a^2-ab+b^2 =c^2-cd+d^2\implies (a+b)^2-3ab = (c+d)^2-3cd$$
so $$ab = cd$$
Induction step: $n-1,n\to n+1$
By I. H. we have $$a^{2n-1}+b^{2n-1} = c^{2n-1}+d^{2n-1}\;\;\; $$
now we multiply this with $a^2+b^2 = c^2+d^2$ we get $$ a^{2n+1}+\color{red}{b^{2n-1}a^2+ a^{2n-1}b^2}+b^{2n+1}  =c^{2n+1}+\color{red}{c^2d^{2n-1} +c^{2n-1}d^2}+d^{2n+1}\;\;\; $$
Since (again by I.H.)$$ a^2b^2(\color{red}{b^{2n-3}+ a^{2n-3}}) =c^2d^2(\color{red}{d^{2n-3} +c^{2n-3}})$$
and we are done.
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