Problem

Source: 2017 Argentina OMA Finals L3 p4

Tags: Perfect Square, perfect cube, number theory



For a positive integer $n$ we denote $D_2(n)$ to the number of divisors of $n$ which are perfect squares and $D_3(n)$ to the number of divisors of $n$ which are perfect cubes. Prove that there exists such that $D_2(n)=999D_3(n).$ Note. The perfect squares are $1^2,2^2,3^2,4^2,…$ , the perfect cubes are $1^3,2^3,3^3,4^3,…$ .