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Knowing that $a+b=c+d$ and $a^3+b^3=c^3+d^3$, prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$. Can you guys please help me complete this proof, as I was attempting it yesterday, without getting far. This question is relevant to the community, as it will broaden the mathematical horizons, of the fellow users of this website, which see and understand the answer

Thanking you in advance

Kevin

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closed as off-topic by quid Apr 14 at 19:41

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    $\begingroup$ If you could prove that $a$ and $b$ were the same as $c$ and $d$ (or $d$ and $c$) then you'd be laughing. $\endgroup$ – Lord Shark the Unknown Apr 14 at 18:56
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    $\begingroup$ Assuming $a+b\neq 0$. Can you see how to recover $ab$ from $a+b$ and $a^3+b^3$? $\endgroup$ – user10354138 Apr 14 at 18:56
  • $\begingroup$ sorry but I can't $\endgroup$ – kenith Apr 14 at 18:57
  • $\begingroup$ Can someone please put this question off hold, as I believe this question is useful for other community members $\endgroup$ – kenith Apr 15 at 10:41
  • $\begingroup$ This is old hat. Essentially a duplicate of this. See, for example, my answer there for the reason. $\endgroup$ – Jyrki Lahtonen May 2 at 11:11
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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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Hint: If $a+b=c+d$ and $ab=cd$ then $\{a,b\}=\{c,d\}$. Use the identity $$x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)^3-3xy(x+y).$$

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    $\begingroup$ @uniquesolution Woops, I meant to type $\{a,b\}=\{c,d\}$, corrected. $\endgroup$ – Servaes Apr 14 at 19:00
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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.

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  • $\begingroup$ Why was this downvoted? (Wish I didn't reach my daily voting limit) $\endgroup$ – Mr Pie Apr 14 at 19:04
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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 $n$.

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