, 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 $xy \neq 0$. For $n = 4$ we have $x^2 + y^2$, $x^3 + y^3$, and $x^4 + y^4$ are rational. From the equality $(x^2 + y^2)^2 = (x^4 + y^4) + 2x^2y^2$ 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:
$$\begin{cases} x + y = 2 - \sqrt{2}, \\ xy = \frac{2}{3} - 2\sqrt{2}, \end{cases}$$
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 $x^3 + y^3 = a(a^2 - 3b) = 4 = q$.