Competition problem — Combinatorics

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Combinatorics Difficulty 4.9 AIME Prove it Romania

Consider two sets $A$ and $B$ of real numbers that have the following properties:
a. $0 \in A$ ;
b. if $1+x \in A$ , then $\sqrt{1+x+x^2} \in B$ ;
c. if $\sqrt{x^2 - x + 1} \in B$ , then $2+x \in A$ .
Prove that $\sqrt{3}$ , $\sqrt{13}$ , $\sqrt{31}$ are elements of the set $B$ and $2024 \in A$ .

Solution

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, $2024 \in A$ .

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