Problem 6. Find all real numbers $x$ such that both numbers $x+\sqrt{3}$ and $x^{2}+\sqrt{3}$ are rational.
Answer: $\frac{1}{2}-\sqrt{3}$.

Solution. Let the number $x+\sqrt{3}$ be denoted by $a$, and the number $x^{2}+\sqrt{3}$ by $b$, then $x=a-\sqrt{3}, b=$ $(a-\sqrt{3})^{2}+\sqrt{3}=a^{2}-2 a \sqrt{3}+3+\sqrt{3}$. From this we get $\left(a^{2}+3-b\right)+(1-2 a) \sqrt{3}=0$. On the right side, we have the rational number 0, the term $a^{2}+3-b$ is also rational, so $(1-2 a) \sqrt{3}$ must be rational, which implies $1-2 a=0$, that is, $a=\frac{1}{2}$. Then $x=a-\sqrt{3}=\frac{1}{2}-\sqrt{3}$.