50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

We show that for $n=4$ it cannot hold.

Suppose that $xy=0$, or $y=0$. For $n=2$ it is possible, as an example we can take $x=\sqrt{2}\in \mathbb{R}\setminus \mathbb{Q}$, $x^2=2\in \mathbb{Q}$. But, if $x^2$ and $x^3$ are rational, then $\frac{x^3}{x^2}=x$ is also rational.

Let us consider the case $xy\ne 0$. For $n=4$ we have $x^2+y^2$, $x^3+y^3$, and $x^4+y^4$ are rational. From the equality $(x^2+y^2{)}^2=(x^4+y^4)+2x^2{y}^2$ it follows that $(xy{)}^2\in \mathbb{Q}$. But then, $(x^6+y^6)=(x^2+y^2)((x^4+y^4)-(xy{)}^2)\in \mathbb{Q}$, and the equality $(x^3+y^3{)}^2=(x^6+y^6)+2(xy{)}^3$ implies that $(xy{)}^3\in \mathbb{Q}$, hence, $\left(\frac{(xy{)}^3}{(xy{)}^2}\right)=xy\in \mathbb{Q}$, and finally we have $x+y=\frac{x^3+y^3}{(x^2+y^2)-xy}=\alpha \in \mathbb{Q}$ which provides a contradiction.

We now show that for $n=3$ such numbers do exist. Denote $a=x+y$, $b=xy$. $x^2+y^2=(x+y{)}^2-2xy=(a^2-2b)\in \mathbb{Q}$, $x^3+y^3=(x+y)((x+y{)}^2-3xy)=a(a^2-3b)\in \mathbb{Q}$. Now we find irrational $a,b$, that satisfy the last equalities. Let $(a^2-2b)=p$, $a(a^2-3b)=q$, then $b=\frac{1}{2}(a^2-p)\Rightarrow a(3p-a^2)=2q$, in this way, we have the following cubic equation

for $a$: $a^3-3ap+2q=0$. Take $a=2-\sqrt{2}$ and find $p$ and $q$: $p=\frac{14}{3}$, $q=4$. Then, we get $b=\frac{2}{3}-2\sqrt{2}$. To find $x,y$, we have to solve the following system of equations: $$\{\begin{array}{l}x+y=2-\sqrt{2},\\ xy=\frac{2}{3}-2\sqrt{2},\end{array}$$ hence, they are roots of quadratic equation: $t^2-at+b=0$. We have: $D^2=a^2-4b=\frac{10}{3}+4\sqrt{2}>0$ which gives us that our roots are, indeed, real. Finally, we have $x^2+y^2=(a^2-2b)=\frac{14}{3}=p$ and $x^3+y^3=a(a^2-3b)=4=q$.