Hellenic Mathematical Olympiad

a. Let $x+\sqrt{3}=q$, $x^2+\sqrt{3}=p$ with $p,q\in \mathbb{Q}$. Then $$x=q-\sqrt{3}\Rightarrow x^2=q^2-2q\sqrt{3}+3$$ so substituting in the second one gives: $$(q^2-2q\sqrt{3}+3)+\sqrt{3}=p\iff -\sqrt{3}(2q-1)=p-q^2-3$$ It follows that $2q-1=0\iff q=\frac{1}{2}$. In this case, $p=q^2+3\Rightarrow p=\frac{1}{4}+3=\frac{13}{4}$ and $x=\frac{1}{2}-\sqrt{3}$.

b. Let $y+\sqrt{3}=q$, $y^3+\sqrt{3}=p$ with $p,q\in \mathbb{Q}$. Then $$y=q-\sqrt{3}\Rightarrow y^3=q^3-3q^2\sqrt{3}+9q-3\sqrt{3}$$ so substituting in the second one gives: $$\begin{gathered} (q^3-3q^2\sqrt{3}+9q-3\sqrt{3})+\sqrt{3}=p\iff -\sqrt{3}(3q^2+2)=p-q^3-9q\iff \\ \sqrt{3}=\frac{q^3+9q-p}{3q^2+2}\in \mathbb{Q} \end{gathered}$$ which is absurd.