Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational.

Question

Find all $x \in \mathbb R$ such that $x + \sqrt{3}$ and $x^2 + \sqrt{3}$ are rational. I have started by assuming that $x + \sqrt{3} = \frac{a}{b}$ and substituting $x = \frac{a}{b} - \sqrt{3}$ into $x^2 + \sqrt{3}$. It led me to $\frac{a^2}{b^2} + 3 + \sqrt{3} \cdot \frac{b - 2a}{b}$. I don't know how to continue.

Answer 1

Well, since $x^2 + \sqrt{3}$ is assumed to be rational, then your work implies that

$$b - 2a = 0.$$

Make sure that you understand why this is the case. Once you have this, $2a = b$ and

$$x = \frac 1 2 - \sqrt{3}$$

is the only solution.

Answer 2

Let $x+\sqrt3=r$ and $x^2+\sqrt3=q$.

Thus, $$(r-\sqrt3)^2=q-\sqrt3$$ or $$r^2-2\sqrt3r+3=q-\sqrt3,$$ which gives $2r=1$ and $r^2+3=q$ and $x=\frac{1}{2}-\sqrt3.$