Competition problem — Number theory

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Number theory Difficulty 5.2 AIME, harder Prove it Ukraine

Sequence of positive integers $a_1, a_2, a_3, \dots$ is defined by $a_{n+1} = a_n^2 + 2018$ , where $a_1$ is some positive integer. Prove that in this sequence no more than one number can be a cube of a positive integer number.

Solution

Suppose there is more than one number, which is a cube of a positive integer. Let $a_k$ be the smallest of all possible cubes. Then, it gives $0, \pm 1$ modulo $9$ , hence, $a_k^2 \equiv 0, 1 \pmod{9}$ . From now on, we write all remainders modulo $9$ . Let us list all the cases.

$a_k^2 \equiv 0$ and $a_{k+1} \equiv 2 \Rightarrow a_{k+1}^2 \equiv 4$ and $a_{k+2} \equiv 6 \Rightarrow a_{k+2}^2 \equiv 0$ and $a_{k+3} \equiv 2$ , and there cannot be any more cubes.

$a_k^2 \equiv 1$ and $a_{k+1} \equiv 3 \Rightarrow a_{k+1}^2 \equiv 0$ and $a_{k+2} \equiv 2$ , and there cannot be any more cubes.

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