$x=a-\sqrt{3}$ $x^2+\sqrt{3}=a^2+3-2a\sqrt{3}+\sqrt{3}$ $-2a+1=0\implies a=\frac12\implies x=\frac12-\sqrt{3}$

$y$ does not exist.

Please post your proof, not only your answer. One more question. b) Does there exist a real number $y$ such that $y+\sqrt{3}$ and $y^n+\sqrt{3}$ are both rationals for $n\geq 3$ being a positive integer.

$y=b-\sqrt{3}$ $y^3+\sqrt{3}-c=-3\sqrt{3}+\sqrt{3}-b^2\sqrt{3}$ $b=\pm\sqrt{2}\not{\in}\mathbb{Q}$

silouan wrote: One more question. b) Does there exist a real number $y$ such that $y+\sqrt{3}$ and $y^n+\sqrt{3}$ are both rationals for $n\geq 3$ being a positive integer.