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.

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.

(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 $a\not=-b$.

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