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, $2024 \in A$.