a) Every number in set is $p\sqrt{2023}$ so $(a,b)=(p\sqrt{2023},q\sqrt{2023}) \to a^2+b \sqrt{2023}=2023p^2+2023q$ is rational b) Let $S$ has at least 3 numbers $a,b,c$ From $c^2+b\sqrt{2023},c^2+a\sqrt{2023}$ are rationals follows $(a-b)\sqrt{2023}$ is rational so $a=p\sqrt{2023}+b$ for some rational $p \neq 0$ $(a^2+b\sqrt{2023})-(b^2+a\sqrt{2023})=(a-b)(a+b-\sqrt{2023})=p\sqrt{2023} (2b+(p-1)\sqrt{2023})=2023p(p-1)+2p*b\sqrt{2023}$ is rational, so $b\sqrt{2023}$ is rational So enough $k=3$ And for $k=2$ we can choose $b$ as root $b^2-b\sqrt{2023}=1 $ and $a=\sqrt{2023}-b$ Then $a^2+b\sqrt{2023}= 2023+b^2-b\sqrt{2023}=2024$ and $b^2+a\sqrt{2023}=2024$