If $a^4+b^4\in\mathbb Q$ and $a^3+b^3\in\mathbb Q$ and $a^2+b^2\in\mathbb Q$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$.

If $\begin{cases}a^4+b^4\in\mathbb Q\\ a^3+b^3\in\mathbb Q\\ a^2+b^2\in\mathbb Q\end{cases}$, prove that $a+b\in\mathbb Q$ and $ab\in\mathbb Q$. It is given that $a,b\in\mathbb R$.

The proof of the latter would simply follow from the former, and vice versa. So I think a better question would be:

Prove one of these statements: $a+b\in\mathbb Q$ or $ab\in\mathbb Q$.

The problem is from the selection to IMO.

I've tried a whole lot of things, including the identities: $$a^4+b^4=(a+b)(a^3+b^3)-ab(a^2+b^2)\\ a^3+b^3=(a+b)(a^2-ab+b^2)\\ a^2+b^2=(a+b)^2-2ab\\ (a+b)^3=a^3+b^3+3ab(a+b)\\ \text{etc...}$$

Even if one could solve the problem using these identities, doing it would most likely be quite tedious imho... Any observations would be greatly appreciated. Thanks.

• at this point all i could see is square of $ab$ is in $\mathbb{Q}$... which i thought is worth sharing as we are at least sure of what power of $ab$ is in $\mathbb{Q}$ – user87543 May 22 '14 at 13:02
• @PraphullaKoushik How have you figured that out? – user26486 May 22 '14 at 13:07
• $a^4+b^4=(a^2+b^2)^2-2a^2b^2\Rightarrow (ab)^2\in \mathbb{Q}$ – user87543 May 22 '14 at 13:08
• @PraphullaKoushik oh, I see. Thanks. – user26486 May 22 '14 at 13:09
• you are welcome!! – user87543 May 22 '14 at 13:10

(Note, added later: This answer was posted before the OP added the stipulation that $a,b\in\mathbb{R}$.)

This is more plodding than ronno's slick hint, but perhaps it shows more of the thought process:

$$(a^2+b^2)^2=a^4+2(ab)^2+b^4\implies (ab)^2\in\mathbb{Q}$$ $$(a^2+b^2)^3=a^6+3(ab)^2(a^2+b^2)+b^6\implies a^6+b^6\in\mathbb{Q}$$ $$(a^3+b^3)^2=a^6+2(ab)^3+b^6\implies (ab)^3\in\mathbb{Q}$$ $$(ab)^2\in\mathbb{Q}\land (ab)^3\in\mathbb{Q}\implies ab\in\mathbb{Q}$$

It would be nice to prove $a+b\in\mathbb{Q}$ as well, but you can't: $a=1+\sqrt{-3}, b=-1+\sqrt{-3}$ is a counterexample. What you can show is

$$a^3+b^3=(a+b)(a^2-ab+b^2)\implies a+b\in\mathbb{Q}\lor a^2-ab+b^2=0$$

which is to say, $a+b$ is rational unless $a^3=-b^3$ with $a\not=-b$.

• So do you mean that $ab\in\mathbb Q$ doesn't imply $a+b\in\mathbb Q$ whenever $a,b\in\mathbb C$? I've never seen any olympiad problem taking into account complex numbers when it is not specified. In this problem, complex numbers aren't even mentioned, so I suppose $a,b\in\mathbb R$. I remember losing a few points in an olympiad some time ago due to assuming the variables used in an equation were complex, not just real. – user26486 May 22 '14 at 20:17
• @mathh, I can't advise you as to what to assume you can assume in a contest problem, but my counterexample clearly shows that $ab\in\mathbb{Q}$ does not imply $a+b\in\mathbb{Q}$ if $a$ and $b$ are only assumed to be in $\mathbb{C}$. – Barry Cipra May 22 '14 at 20:28
• Wait, I'm sorry, the problem description does actually explicitly specify that $a,b\in\mathbb R$, I should've noted that. I'm editing my question. – user26486 May 22 '14 at 20:37
• I am glad you were attentive enough to note that the fact doesn't hold given the variables $a,b\in\mathbb C$. I guess you are used to assuming complex numbers are what is implied when nothing is specified about the type of variables being used... definitely deserves a +1 from me. The solution you've shown is a ton easier to think of given the level of my math skills so far. I'd never find out the solution ronno has used in my life given my currect knowledge, the identity looks too hardcore to think of. – user26486 May 22 '14 at 20:40

Hint: $(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = a^2b^2(a^2+b^2-2ab)$. Now use @Praphulla's comment.

• wow. That sure is a useful observation, not so sure how you've derived this identity, though. It seems like quite a hard one to figure out. – user26486 May 22 '14 at 13:30
• @mathh Have you seen a proof of the "The fundamental theorem of symmetric polynomials"? This sort of cancellation is very standard when dealing with symmetric polynomials. – ronno May 22 '14 at 13:33
• Thanks, I haven't. It looks too advanced for my level. I'll definitely check it out later on when I get more educated. – user26486 May 22 '14 at 13:37
• well done!! happy that my comment was of some help... – user87543 May 22 '14 at 14:37
• I am sorry for not accepting your answer; this is because there's absolutely no way I could've figured out that identity myself. However, I am still very grateful for bringing up the fundamental theorem of symmetric polynomials. I am sure it will greatly help me out in the future. – user26486 May 24 '14 at 14:11

Maybe this useful let $x_{n}=a^n+b^n$,then $$x_{n}=(a+b)x_{n-1}-abx_{n-2}$$

• I've used this observation on factoring both $a^4+b^4$, $a^3+b^3$ and $a^2+b^2$ as you can see on my question details. – user26486 May 22 '14 at 12:59
• The generalization is ($a,b,c\in\mathbb R$, $n\in\mathbb N$ and $n\ge 4$ (the conditions I've outlined are not as sharp as possible imo, but the other ones are less likely to be required)) $$a^n+b^n+c^n=(a+b+c)(a^{n-1}+b^{n-1}+c^{n-1})-(ab+bc+ca)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3})$$ – user26486 May 23 '14 at 0:05