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D_2(n)=999D_3(n), number of divisors of n that are perfect squares / cubes

Source: 2017 Argentina OMA Finals L3 p4

January 16, 2023
Perfect Squareperfect cubenumber theory

Problem Statement

For a positive integer nn we denote D2(n)D_2(n) to the number of divisors of nn which are perfect squares and D3(n)D_3(n) to the number of divisors of nn which are perfect cubes. Prove that there exists such that D2(n)=999D3(n).D_2(n)=999D_3(n).
Note. The perfect squares are 12,22,32,42,…1^2,2^2,3^2,4^2,… , the perfect cubes are 13,23,33,43,…1^3,2^3,3^3,4^3,… .
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