Show $x+y, xy \in \mathbb{Q}$ Let $x,y \in \mathbb{R}$ and $x^{2}+y^{2}, x^{3}+y^{3}, x^{4}+y^{4} \in \mathbb{Q}$.

Show $x+y, xy \in \mathbb{Q}$ 

My attempt: 


*

*Since $x^{2}+y^{2}, x^{4}+y^{4} \in \mathbb{Q}$, Thus $ (x^{2}+y^{2})^{2} - (x^{4}+y^{4}) = 2x^{2}y^{2} \in \mathbb{Q}$. Thus $$x^{2}y^{2} \in \mathbb{Q} \quad (1)$$. 

*By using $(1)$, we have $(x^{2}+y^{2})^{3} - 3x^{2}y^{2}(x^{2} + y^{2}) = x^{6} + y^{6} \in \mathbb{Q} \quad (2)$.

*By using $(2)$, we have $(x^{3}+y^{3})^{2} - (x^{6}+y^{6}) = 2x^{3}y^{3} \in \mathbb{Q}$. Thus $$x^{3}y^{3} \in \mathbb{Q} \quad (3)$$

*By using $(1), (2)$ we have $\frac{x^{3}y^{3}}{x^{2}y^{2}} = xy \in \mathbb{Q}$.



For the other case ($x+y \in \mathbb{Q}$), I don't have any idea.

Hint:
$$x^3+y^3-(x+y)(x^2+y^2)=?$$
Or
$$x^3+y^3=(x+y)(x^2+y^2-xy)$$
