74th Romanian Mathematical Olympiad

Since $1+(-1)=0\in A$, according to (b) we obtain $1\in B$ and, since $1=\sqrt{0^2-0+1}\in B$, we infer from (c) that $2\in A$, from which $\sqrt{3}\in B$.

Since $\sqrt{2^2-2+1}=\sqrt{3}\in B$, it follows from (c) that $2+2=4\in A$ and, from (b), we infer $\sqrt{13}\in B$.

Since $\sqrt{4^2-4+1}=\sqrt{13}\in B$, we further infer that $2+4=6\in A$ and thus $\sqrt{1+5+5^2}=\sqrt{31}\in B$.

Using the equality $1+x+x^2=(x+1{)}^2-(x+1)+1$, we have that if $1+x\in A$, then $3+x\in A$. Since $0\in A$, $2\in A$ and it follows that the set $A$ contains all the even numbers. In particular, $2024\in A$.