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?
Hint: If $a+b=c+d$ and $ab=cd$ then $\{a,b\}=\{c,d\}$. Use the identity $$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)^3-3xy(x+y).$$
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 $n$.
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.
$$(a+b)^3=(c+d)^3$$
$$\iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
If $a+b\ne0, ab=cd\ \ \ \ (1)$
$\iff\dfrac ad=\dfrac cb=k$(say) $\ \ \ \ (2)$
$a+b=c+d\implies dk+b=bk+d\iff d(k-1)=b(k-1)$
Either $d=b\iff a=c$
or $k=1$ use this in $(2)$
Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$
$c,d$ will be the roots of same equation $\implies\{a,b\}\equiv\{c,d\}$
In both bases we can prove $$a^n+b^n=c^n+d^n$$
If $a=-b$ so $c=-d$ and we are done.
Let $a+b\neq0.$
Thus, $$(a+b)^3-3ab(a+b)=(c+d)^3-3cd(c+d)$$ or $$ab(a+b)=cd(c+d)$$ or $$ab=cd$$ and the rest is smooth.
Hint: the case $a+b=c+d=0$ is obvious, so you can assume $a+b=c+d\ne0$; prove that $ab=cd$. Further hint: $x^2-xy+y^2=(x+y)^2-{\color{red}{?}}$.
Use that we get from $$a+b=c+d$$ so $$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$ and $$x^3+y^3=(x+y)(x^2-xy+y^2)$$
