59th Ukrainian National Mathematical Olympiad

Question

Sequence of positive integers a_1, a_2, a_3, 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.

Accepted Answer

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 (mod 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.